[4] Large deviations for the empirical distribution of Brownian motion
process.

Let $X_{n},n\geq 0$ be a Markov process in discrete time. Consider

$M_{N}(f)=\mathbb{E}exp(\sum_{n=0}^{N}f(X_{n}))$

Then, given $X_{0}=x$

$M_{N}(f)=(\pi_{f})^{N}(1)$

where $\pi$ is the transition probability operator kernel:

$\pi(g)(x)=\int\pi(x,dy)g(y)$

and

$\pi_{f}(g)(x)=\int exp(f(x))\pi(x,dy)g(y)$

ie

$\pi_{f}(x,dy)=exp(f(x))\pi(x,dy)$

If we assume that the state space $E$ is discrete, then $\pi(x,y)$ is a
stochastic matrix and $\pi_{f}(x,y)=exp(f(x))\pi(x,y)$. Let $D_{f}$ denote the
diagonal matrix $diag[f(x):x\in E]$. Then

$\pi_{f}=exp(D_{f})\pi$

and

$M_{N}(f)=(exp(D_{f})\pi)^{N}(1)$

The Gartner-Ellis limiting logarithmic moment generating function is then

$\Lambda(f)=lim_{N\rightarrow\infty}N^{-1}.log((exp(D_{f})\pi)^{N}(1))=log(\lambda_{m}(f))$

where $\lambda_{m}(f)$ is the maximum eigenvalue of the matrix
$exp(D_{f})\pi$. Now let

$I(\mu)=sup_{u>0}\sum_{x}\mu(x)log((\pi u)(x))/u(x))$

where $\mu$ is a probability distribution on $E$. We wish to show that

$I(\mu)=sup_{f}(\sum_{x}f(x)\mu(x)-\lambda_{m}(f))$
$=sup_{f}(<f,\mu>-\lambda_{m}(f))$

###  10.1 End-Semester Exam, duration:four hours, Pattern Recognition

[1] [a] Consider $N$ iid random vectors ${\bf X}_{n},n=1,2,...,N$ where ${\bf
X}_{n}$ has the probability density $\sum_{k=1}^{K}\pi(k)N({\bf
x}|\mu_{k},\Sigma_{k})$, ie, a Gaussian mixture density. Derive the optimal ML
equations for estimating the parameters
$\theta=\\{\pi(k),\mu_{k},\sigma_{k}):k=1,2,...,K\\}$ from the measurements
${\bf X}=\\{{\bf X}_{n}:n=1,2,...,N\\}$ by maximizing

$ln(p({\bf X}|\theta))=\sum_{n=1}^{N}ln(\sum_{k=1}^{K}\pi(k)N({\bf
X}_{n}|\mu_{k},\Sigma_{k}))$

[b] By introducing latent random vectors ${\bf
z}(n)=\\{z(n,k):k=1,2,...,K\\},n=1,2,...,N$ which are iid with
$P(z(n,k)=1)=\pi(k)$ such that for each $n$, exactly one of the
$z(n,k)^{\prime}s$ equals one and the others are zero, cast the ML parameter
estimation equation in the recursive EM form, namely as

$\theta_{m+1}=argmax_{\theta}Q(\theta,\theta_{m})$

where

$Q(\theta,\theta_{m})=\sum_{\bf z}ln(p({\bf X,Z}|\theta)).p({\bf Z}|{\bf
X},\theta_{m})$

with ${\bf Z}=\\{{\bf z}(n):n=1,2,...,N\\}$. Specifically, show that

$Q(\theta,\theta_{m})=\sum_{n,k}\gamma_{nk}({\bf X},\theta_{m})ln(\pi(k)N({\bf
X}_{n}|\mu(k),\Sigma_{k})))p({\bf Z}|{\bf X},\theta_{m})$

where

$\gamma_{n,k}({\bf X},\theta_{m})=\mathbb{E}(z(n,k)|{\bf X},\theta_{m})$
$=\sum_{\bf Z}z(n,k)p({\bf Z}|{\bf
X},\theta_{m})=\sum_{z(n,k)}z(n,k)p(z(n,k)|{\bf X},\theta_{m})$

Derive a formula for $p(z(n,k)|{\bf X},\theta_{m})$. Derive the optimal
equations for $\theta_{m+1}$ by maximizing $Q(\theta,\theta_{m})$.

[2] Let ${\bf X}_{n},n=1,2,...,N$ be vectors in $\mathbb{R}^{N}$. Choose
$K<<N$ and vectors $\\{\mu(k):k=1,2,...,K\\}$ and a partition

$\mathcal{P}=E_{1}\cup E_{2}\cup...\cup E_{K}$

of $\\{1,2,...,N\\}$ into $K$ disjoint non-empty subsets such that

$\mathcal{E}=\sum_{k=1}^{K}\sum_{n\in E_{k}}\parallel{\bf
X}_{n}-\mu(k)\parallel^{2}$

is a minimum. Interpret this ”Clustering principle” from the viewpoint of data
compression and image segmentation. Explain how you would obtain an iterative
solution to this clustering problem by defining

$r(n,k)=1,n\in E_{k},r(n,k)=0,n\notin E_{k}$

and expressing

$\mathcal{E}=\sum_{n,k}r(n,k)\parallel{\bf X}_{n}-\mu(k)\parallel^{2}$

Relate this clustering problem to the problem of calculating the
$\mu(k)^{\prime}s$ and the $\pi(k)^{\prime}s$ in the Gaussian mixture model.
Specifically, compare the equations for calculating $r(n,k),\mu(k)$ in this
clustering problem to the equations for calculating the same in the GMM.
Explain how the GMM model is a ”smoothened version” of the clustering model by
remarking that the event ${\bf X}_{n}\in E_{k}$ is to be interpreted in the
GMM context by saying that ${\bf X}_{n}$ has the $N({\bf
x}|\mu(k),\Sigma_{k})$ distribution and that
$\sum_{n=1}^{N}r(n,k)/N=card(E_{k})/N$ corresponds to the probability $\pi(k)$
of the $k^{th}$ component in the GMM occurring. Derive the explicit value that
$r(n,k)$ corresponds to in the GMM by showing that the optimal $\pi(k)$ in the
GMM satisfies

$\pi(k)=\sum_{n=1}^{N}\frac{\pi(k)N({\bf
X}_{n}|\mu_{k},\Sigma_{k})}{\sum_{m=1}^{K}\pi(m)N({\bf
X}_{n}|\mu(m),\Sigma_{m})}$

Compare this with the quantity
$\pi(k)=\sum_{n=1}^{N}(r(n,k)/N),\sum_{n,k}r(n,k)=N$ in the clustering
problem.

[3] Let $\\{Z(n):n=1,2,...,N\\}$ be a Markov chain in discrete state space and
let $\pi(Z(n),Z(n+1)|\theta)$ be its one step transition probability
distribution. Let $\pi_{0}(Z(0))$ denote its initial distribution. Let ${\bf
X}=({\bf X}(n):n=0,1,...,N)$ be the HMM emission random variables with the
conditional probability density $p(X(n)|Z(n),\theta)$ defined.

[a] Show that the EM algorithm for estimating $\theta$ recursively amounts to
maximizing $Q(\theta,\theta_{0})$ where

$Q(\theta,\theta_{0})=$ $\sum_{n=0}^{N}\mathbb{E}(ln(p(X(n)|Z(n),\theta))|{\bf
X},\theta_{0})$
$+\sum_{n=0}^{N-1}\mathbb{E}(ln(\pi(Z(n),Z(n+1)|\theta_{0}))|{\bf
X},\theta_{0})$

[b] Show that for evaluating the conditional expectation in [a], we require
the conditional probabilities $p(Z(n)|{\bf X},\theta_{0})$ and
$p(Z(n),Z(n+1)|{\bf X},\theta_{0})$. Explain how you would evaluate these
conditional probabilities recursively by first proving the relations

$p({\bf
X}|Z(n),\theta_{0})=p(X(N),X(N-1),...,X(n)|Z(n),\theta_{0}).p(X(n-1),X(n-2),...,X(0)|Z(n),\theta_{0})$

and

$p({\bf X}|Z(n+1),Z(n),\theta_{0})=$
$p(X(N),...,X(n+1)|Z(n+1,\theta_{0}).p(X(n-1),...,X(0)|Z(n),\theta_{0})$

4 [a] How would you approximate (based on matching only first and second order
statistics) an $N$-dimensional random vector ${\bf X}$ by a $p<<N$ dimensional
random vector ${\bf S}$ using the linear model

${\bf X}\approx{\bf WS}+\mu+{\bf V}=\hat{\bf X}+{\bf V},\hat{\bf X}={\bf
WS}+\mu$

where ${\bf S,V}$ are zero mean mutually uncorrelated random vectors with
covariances $\Sigma_{S},\Sigma_{V}$ respectively and ${\bf W}$ is an $N\times
p$ matrix to be determined. To solve this problem, you must assume that
$\mathbb{E}({\bf X})$ is well approximated by $\mu$ and $Cov({\bf X})$ is well
approximated by ${\bf W}\Sigma_{S}{\bf W}^{T}+\Sigma_{V}$.

[b] Now consider the maximum likelihood version of the above PCA in which the
data is modeled by an iid sequence:

${\bf X}_{n}={\bf W}{\bf S}_{n}+\mu+{\bf V}_{n},n=1,2,...,N$

where the ${\bf S}_{n}^{\prime}s$ are iid $N({\bf 0},\Sigma_{S})$ random
vectors, the ${\bf V}_{n}^{\prime}s$ are iid $N({\bf 0},\Sigma_{v})$ random
vectors with the two sequences being mutually independent. Calculate the
equations that determine the optimal choice of ${\bf W},\mu$ that would
maximize $p({\bf X}_{n},n=1,2,...,N|{\bf W},\mu)$. Also by assuming the ${\bf
S}_{n}^{\prime}s$ to be latent random vectors, derive the EM algorithm
recursive equations for estimating ${\bf W},\mu$ from the ${\bf
X}_{n},n=1,2,...,N$.

Solution:

$p(\\{{\bf X}_{n}\\}_{n=1}^{N},\\{{\bf S}_{n}\\}_{n=1}^{N}|{\bf W},\mu)$
$=p(\\{{\bf X}_{n}\\}_{n=1}^{N}|\\{{\bf S}_{n}\\},{\bf W},\mu)p(\\{{\bf
S}_{n}\\}_{n=1}^{N})$ $=\Pi_{n=1}^{N}[f_{\bf V}({\bf X}_{n}-{\bf
WS}_{n}-\mu)f_{\bf S}({\bf S}_{n})]$

So in the EM algorithm, the function to be maximized is

$Q(\theta,\theta_{0})$

where

$\theta=({\bf W},\mu),\theta_{0}=({\bf W}_{0},\mu_{0})$

and

$Q(\theta,\theta_{0})=\sum_{n=1}^{N}\mathbb{E}[log(f_{\bf V}({\bf X}_{n}-{\bf
WS}_{n}-\mu))|\\{{\bf X}_{m}\\}_{m=1}^{N},{\bf W}_{0},\mu_{0})]$
$+\sum_{n=1}^{N}log(f_{\bf S}({\bf S}_{n}))|\\{{\bf X}_{m}\\}_{m=1}^{N},{\bf
W}_{0},\mu_{0})$ $=\sum_{n=1}^{N}\mathbb{E}[log(f_{\bf V}({\bf X}_{n}-{\bf
WS}_{n}-\mu))|{\bf X}_{n},{\bf W}_{0},\mu_{0})]$ $+\sum_{n=1}^{N}log(f_{\bf
S}({\bf S}_{n}))|{\bf X}_{n},{\bf W}_{0},\mu_{0})$

Note that only the first sum in this expression must be maximized since the
second one does not depend on ${\bf W},\mu$. However, if
$\Sigma_{S},\Sigma_{v}$ also have to be estimated then the parameter vector
would be

$\theta=({\bf W},\mu,\Sigma_{S},\Sigma_{v})$

and then the second term would also count. Note further, that

$p(\\{{\bf S}_{n}\\}_{n=1}^{N}|\\{{\bf X}_{n}\\}_{n=1}^{N},{\bf W},\mu)=$
$[\Pi_{n=1}^{N}f_{\bf V}({\bf X}_{n}-{\bf WS}_{n}-\mu)f_{\bf S}({\bf
S}_{n})]/\Pi_{n=1}^{N}f_{\bf X}({\bf X}_{n}|{\bf W},\mu)$

with $f_{\bf X}$ being $N(\mu,{\bf W}\Sigma_{s}{\bf W}^{T}+\Sigma_{v})$

[c] If $\Sigma_{V}=\sigma^{2}{\bf I}$, then derive the EM recursions for
estimating ${\bf W},\mu,\Sigma_{S},\sigma^{2}$ and compare with the exact
maximum likelihood equations. Explain why this process is called principle
component analysis and explain how much data compression we have achieved by
this process.

[d] Consider now the following data matrix version of PCA. ${\bf X}$ is a
given $N\times M$ real matrix. Show that it has a singular value decomposition
(SVD)

${\bf X}={\bf U}\Sigma{\bf V}$

where ${\bf U}$ and ${\bf V}$ are respectively $N\times N$ and $M\times M$
real orthogonal matrices and $\Sigma$ is a diagonal $N\times M$ matrix with
positive diagonal entries either $N$ or $M$ in number, depending on whether
$N$ or $M$ is smaller. Let $r<min(N,M)$ and then using the SVD, construct an
$N\times M$ matrix ${\bf Y}$ of rank $r$ so that $\parallel{\bf X}-{\bf
Y}\parallel_{F}^{2}$ is a minimum where $\parallel
Z\parallel_{F}=\sqrt{Tr({\bf Z}^{T}{\bf Z})}$ is the Frobenius norm of the
real matrix ${\bf Z}$.

[5] Consider the binary hypothesis testing problem in which under the
hypothesis $H_{1}$, each sample of the iid sequence $X(n),n\geq 1$ has the
probability density $p_{1}(X)$ while under $H_{0}$, each sample of this iid
sequence has the probability density $p_{0}(X)$. Show that the Neyman-Pearson
criterion of decision based on the observations $X(n),n=1,2,...,N$ leads to
the test of deciding $H_{1}$ in the case when
$N^{-1}\sum_{n=1}^{N}L(X(n))\geq\eta$ and deciding $H_{0}$ in the case when
$N^{-1}\sum_{n=1}^{N}L(X(n))<\eta$ where

$L(X)=ln(p_{1}(X)/p_{0}(X))$

and the threshold $\eta(N)$ is such that

$0<\alpha=\int_{Z_{1}}\Pi_{n=1}^{N}(p_{0}(X(n))dX(n))=P(N^{-1}\sum_{n=1}^{N}L(X(n))>\eta|H_{0})$

Here, $Z_{1}$ is the region where $H_{1}$ is decided. Estimate this
probability by the LDP and calculate using the LDP the value of $\eta$ for
given $\alpha$. Hence use this threshold $\eta$ to calculate the other error
probability

$\beta(N)=\int_{Z_{1}^{c}}\Pi_{n=1}^{N}(p_{1}(X(n))dX(n))=P(N^{-1}\sum_{n=1}^{N}L(X(n))<\eta|H_{1})$

and show that for any $\alpha>0$ however small, $\beta(N)$ if $\eta=\eta(N)$
is obtained as above for the given $\alpha$, then $\beta(N)$ converges to zero
at the rate of $D(p_{0}|p_{1})$, ie,

$N^{-1}log(\beta(N))\rightarrow D(p_{0}|p_{1}),N\rightarrow\infty$

where

$D(p_{0}|p_{1})=\int p_{0}(X).log(p_{0}(X)/p_{1}(X)).dX$

[6] This problem deals with non-parametric density estimation. Let
$p_{1}(X),p_{0}(X)$ be two different probability densities of a random vector
$X\in\mathbb{R}^{n}$. To estimate $p_{1}$ at a given point $X$, we take
samples from this distribution and draw a sphere of minimum radius $R_{1}$
with centre at $X$, for which the first sample point falls within this sphere.
Let $V(R_{1})$ denote the volume of such a sphere. Then, estimate $p_{1}(X)$
as $1/V(R_{1})$. Likewise do so for estimating $p_{0}(X)$. Let the minimum
radius for this be $R_{0}$. Justify using the law of large numbers that such a
non-parametric density estimate is reliable. Show that given a sample point
$X$, in order to test whether $X$ came from $p_{1}$ or from $p_{0}$, we use
the ML criterion, namely decide $p_{1}$ if $\hat{p}_{1}(X)\geq\hat{p}_{0}(X)$
and decide $p_{0}$ otherwise, where $\hat{p}$ denotes the estimate of $p$
obtained using the above scheme. Show that this MLE method is the same as the
minimum distance method, in the sense that if the first sample point that we
decide $p_{1}$ if $R_{1}(X)<R_{0}(X)$ and decide $p_{0}$ otherwise where
$R_{1}(X)$ is the distance of the first point taken from the $p_{1}$ sample
from $X$ and likewise $R_{0}(X)$ is the distance of the first point taken from
the $p_{0}$ sample from $X$.

[7] Let a group $G$ act on a set $\mathcal{M}$ and let
$f_{0},f_{1}:\mathcal{M}\rightarrow\mathbb{R}$ be two image fields defined on
the set $\mathcal{M}$ such that

$f_{1}(x)=f_{0}(g_{0}^{-1}x)+w(x),x\in\mathcal{M}$

where $g_{0}\in G$ and $w$ is a zero mean Gaussian noise field. Choose a point
$x_{0}\in\mathcal{M}$ and replace the above model by

$\tilde{f}_{1}(g)=\tilde{f}_{0}(g_{0}^{-1}g)+w(g),g\in G$

where

$\tilde{f}_{1}(g)=f_{1}(gx_{0}),\tilde{f}_{0}(g)=f_{0}(gx_{0})$

Let $\pi$ be a unitary representation of $G$ in a the vector space
$\mathbb{C}^{p}$. Define the Fourier transform of a function $F$ on $G$
evaluated at $\pi$ by

$\mathcal{F}(F)(\pi)=\int_{G}F(g)\pi(g)dg$

where $dg$ is the left invariant Haar measure on $G$. Show that the above
image model on $G$ gives on taking Fourier transforms

$\mathcal{F}(\tilde{f}_{1})(\pi)=\pi(g_{0})\mathcal{F}(\tilde{f}_{0})(\pi)+\mathcal{F}(w)(\pi)$

Let $H$ be the subgroup of $G$ consisting of all $h\in G$ for which
$hx_{0}=x_{0}$. Show that $\tilde{f}_{k}(gh)=\tilde{f}_{k}(g),k=1,2,g\in
g,h\in H$. Hence, deduce that if

$P_{H}=\int_{H}\pi(h)dh$

where $dh$ denotes the Haar measure on $H$, then $P_{H}$ is the orthogonal
projection onto the set of all vectors $v\in\mathbb{C}^{p}$ for which
$\pi(h)v=v\forall h\in H$ provided that $H$ is unimodular, ie, its left
invariant Haar measure coincides with its right invariant Haar measure. Deduce
further that

$\mathcal{F}(\tilde{f}_{k})(\pi)=\tilde{F}(\tilde{f}_{k})(\pi)P_{H},k=0,1$

Hence show that the above model for the image Fourier transforms is equivalent
to

$\mathcal{F}(\tilde{f}_{1})(\pi)P_{H}=\pi(g_{0})\mathcal{F}(\tilde{f}_{0})(\pi)P_{H}+\mathcal{F}(w)(\pi)P_{H}$

From this model, explain how by varying the representation $\pi$, we can
estimate the transformation $g_{0}$ by minimizing

$\mathcal{E}(g_{0})=\sum_{k}W(k)\parallel\mathcal{F}(\tilde{f}_{1})(\pi_{k})P_{H}-\pi_{k}(g_{0})\mathcal{F}(\tilde{f}_{0})(\pi_{k})P_{H}\parallel^{2}$

where $\parallel.\parallel$ is the Frobenius norm and $W(k),k\geq 1$ are
positive weights. By starting with an initial guess estimate $g_{00}$ for
$g_{0}$ and writing $g_{0}=(1+\delta X).g_{00}$ where $\delta X$ is small and
belongs to the Lie algebra of $G$, derive a set of linear equations for
$\delta X$ and solve them. For this, you can use the approximate identity

$\pi((1+\delta X)g_{00})=\pi(exp(\delta X)g_{00})=exp(d\pi(\delta
X))\pi(g_{00})=(1+d\pi(\delta X))\pi(g_{00})$

where $d\pi$ is the differential of $\pi$, ie, it is the representation of the
Lie algebra of $G$ corresponding to the representation $\pi$ of $G$. Note that
$d\pi$ is a linear map and hence if $\\{L_{1},...,L_{N}\\}$ denotes a basis
for the Lie algebra of $G$ then

$\delta X=\sum_{m=1}^{N}x(k)L_{k},d\pi(\delta
X)=\sum_{m=1}^{N}x(k)d\pi(L_{k})$

The problem of estimating $\delta X$ is then equivalent to estimating
$x(k),k=1,2,...,N$. Also explain how you would approximately evaluate the mean
and covariance of the random variables $x(k),k=1,2,...,N$ in terms of the
noise correlations.

Estimating the EEG parameters from the speech parameters using the EKF. The
EEG model is

$X_{E}(t+1)=A(\theta)X_{E}(t)+W_{E}(t+1)\in\mathbb{R}^{N}$

where

$A(\theta)=A_{0}+\sum_{k=1}^{p}\theta(k)A_{k}$
$\theta=(\theta(k):k=1,2,...,p)^{T}\in\mathbb{R}^{p}$

is the EEG parameter vector to be estimated. The speech model on the other
hand is

$X_{S}(t)=\sum_{k=1}^{m}rs(k)X_{S}(t-k)+r_{s}(m+1)+W_{S}(t)\in\mathbb{R}$

The speech parameter vector is

${\bf r}_{s}=(r_{s}(k);k=1,2,...,m+1)^{T}\in\mathbb{R}^{m+1}$

We assume that the speech and EEG parameters are correlated, ie, there exists
an affine relationship connecting the two of the form

${\bf r}_{s}={\bf C}\theta+{\bf d}$

The regression matrices ${\bf C,d}$ are assumed to be fixed, ie, they do not
vary from patient to patient. These matrices can be estimated using the
following procedure: Choose a bunch of $K$ patients numbered $1,2,...,K$
having various kinds of brain diseases. For each patient, collect his EEG and
speech data and apply the EKF separately to these data to using the above
models and noisy measurement models of the form

${\bf Z}_{E}(t)={\bf X}_{E}(t)+{\bf V}_{E}(t),Z_{S}(t)=X_{S}(t)+V_{S}(t)$

Denote the parameter estimates for the $k^{th}$ patient by
$\hat{\theta}_{k},\hat{r}s_{k}$ for $k=1,2,...,,p$. Then the regression
matrices ${\bf C,d}$ are estimated by minimizing

$E({\bf C,d})=\sum_{k=1}^{K}\parallel\hat{r}s_{k}-{\bf C}\hat{\theta}_{k}-{\bf
d}\parallel^{2}$

For convenience of notation, we write

$\theta_{k}=\hat{\theta}_{k},rs_{k}=\hat{r}s_{k}$

and then the optimal equations for the regression matrices obtained by setting
the gradient of $E({\bf C,d})$ to zero are given by

$\sum_{k=1}^{K}(rs_{k}-{\bf C}\theta_{k}-{\bf d})\theta_{k}^{T}=0,$
$\sum_{k=1}^{K}(rs_{k}-{\bf C}\theta_{k}-{\bf d})=0$

which can be expressed as

${\bf C}{\bf R}_{\theta}+{\bf d}\mu_{\theta}^{T}={\bf R}_{s\theta},$ ${\bf
C}\mu_{\theta}+{\bf d}=\mu_{s}$

where

${\bf
R}_{\theta}=K^{-1}.\sum_{k=1}^{K}\theta_{k}.\theta_{k}^{T}\in\mathbb{R}^{p\times
p}$ $\mu_{\theta}=K^{-1}\sum_{k=1}^{K}\theta_{k}\in\mathbb{R}^{p\times 1}$
${\bf
R}_{s\theta}=K^{-1}\sum_{k=1}^{K}rs_{k}\theta_{k}^{T}\in\in\mathbb{R}^{(m+1)\times(m+1)},$
$\mu_{s}=K^{-1}\sum_{k=1}^{K}rs_{k}\in\mathbb{R}^{(m+1)\times 1}$

Eliminating ${\bf d}$ from these equations gives

${\bf CR}_{\theta}+(\mu_{s}-{\bf C}\mu_{\theta})\mu_{\theta}^{T}={\bf
R}_{s\theta}$

giving

${\bf C}=({\bf R}_{s\theta}-\mu_{s}\mu_{\theta}^{T})({\bf
R}_{\theta}-\mu_{\theta}\mu_{\theta}^{T})^{-1}$

and

$\mu_{s}-{\bf C}\mu_{\theta}$

This solves the problem of determining the regression coefficient matrices.
Now given a fresh patient, we wish to estimate his EEG parameters from only
his speech data. To this end, we substitute into the speech model the
expression for the speech parameters in terms of the EEG parameters using our
regression scheme: First transform the speech model into a state variable
model by introducing state variables:

$\xi_{1}(t)=X_{S}(t-1),...,\xi_{m}(t)=X_{S}(t-m)$

and

$\xi(t)=[\xi_{1}(t),...,\xi_{m}(t)]^{T}$

The speech model then becomes in state variable form

$\xi(t+1)={\bf B}(r_{s})\xi(t)+u.rs_{m+1}+bW_{s}(t)$

where

${\bf B}(rs)=\left(\begin{array}[]{ccccc}rs1&r2&rs3&...&rsm\\\ 1&0&0&...&0\\\
..&..&..&...&..\\\ 0&0&0..&1&0\end{array}\right)$

We can write

${\bf B}(rs)=\sum_{k=1}^{m}rs_{k}{\bf B}_{k}$

and hence our speech model becomes

$\xi(t+1)=(\sum_{k=1}^{m}rs_{k}{\bf B}_{k})\xi(t)+u.rs_{m+1}+bW_{s}(t)$

Now

$rs_{k}=({\bf C}\theta)_{k}+d_{k},k=1,2,...,m+1$

Writing

$F_{1}(\xi,\theta)=\sum_{k=1}^{m}rs_{k}{\bf B}_{k}\xi+u.rs_{m+1}$
$=\sum_{k=1}^{m}(({\bf C}\theta)_{k}+d_{k}){\bf B}_{k}\xi+u({\bf
C}\theta)_{m+1}+d_{m+1})$

we find that

$\partial F_{1}/\partial\xi=[\sum_{k=1}^{m}((C\theta)_{k}+d_{k}){\bf B}_{k}]$
$\partial F_{1}/\partial\theta=[\sum_{k=1}^{m}C_{k1}{\bf
B}_{k}\xi+C_{m+1,1}{\bf u},\sum_{k=1}^{m}C_{k2}{\bf B}_{k}\xi+C_{m+1,2}{\bf
u},...,\sum_{k=1}^{m}C_{kp}{\bf B}_{k}\xi+C_{m+1,p}{\bf u}]$

###  10.2 Quantum Stein’s theorem

We wish to discriminate between two states $\rho,\sigma$ using a sequence of
iid measurements, ie, measurements taken when the states become $\rho^{\otimes
n}$ and $\sigma^{\otimes n},n=1,2,...$. Let $0\leq T_{n}\leq 1$ be the
detection operators. If the state is $\rho^{\otimes n}$, then the probability
of a wrong decision is $P_{1}(n)=Tr(\rho^{\otimes n}(1-T_{n}))$ while if the
state is $\sigma^{\otimes n}$, the probability of a wrong decision is
$P_{2}(n)=Tr(\sigma^{\otimes n}T_{n})$. The theorem that we prove is the
following: If the measurement operators $\\{T_{n}\\}$ are such that
$P_{2}(n)\rightarrow 0$, then $P_{1}(n)$ cannot go to zero at a rate faster
than $exp(-nD(\sigma|\rho))$, ie,

$liminf_{n\rightarrow}n^{-1}.log(P_{1}(n))\geq-D(\sigma|\rho)$

and conversely, for any $\delta>0$, there there exists a sequence
$\\{T_{n}\\}$ of detection operators such that $P_{2}(n)\rightarrow 0$ and
simultaneously

$limsup_{n}n^{-1}.log(P_{1}(n))\leq-D(\sigma|\rho)+\delta$

To prove this, we first use the monotonicity of the Renyi entropy to deduce
that for any detection operator $T_{n}$,

$(Tr(\rho^{\otimes n}(1-T_{n})))^{s}Tr(\sigma^{\otimes n}(1-T_{n}))^{1-s}$
$\leq Tr(\rho^{\otimes ns)}\sigma^{\otimes n(1-s)})$

for any $s\leq 0$. Therefore, if

$Tr(\sigma^{\otimes n}T_{n})\rightarrow 0$

then

$Tr(\sigma^{\otimes n}(1-T_{n}))\rightarrow 1$

and we get from the above on taking logarithms,

$limsupn^{-1}.log(Tr(\rho^{\otimes n}(1-T_{n}))\geq
log(Tr(\rho^{s}\sigma^{1-s}))/s$

or equivalently writing

$\alpha=limsupn^{-1}.log(P_{1}(n))$

we get

$\alpha\geq s^{-1}Tr(\rho^{s}\sigma^{1-s})\forall s\leq 0$

Taking $lims\uparrow 0$ in this expression gives us

$\alpha\geq-D(\sigma|\rho)=-Tr(\sigma(log(\sigma)-log(\rho)))$

This proves the first part. For the second part, we choose $T_{n}$ in
accordance with the optimal Neyman-Pearson test. Specifically, choose

$T_{n}=\\{exp(nR)\rho^{\otimes n}\geq\sigma^{\otimes n}\\}$

where the real number $R$ will be selected appropriately.

###  10.3 Question paper, short test, Statistical signal processing, M.Tech

[1] Consider the AR(p) process

$x[n]=-\sum_{k=1}^{p}a[k]x[n-k]+w[n]$

where $w[n]$ is white noise with autocorrelation

$\mathbb{E}(w[n]w[n-k]]=\sigma_{w}^{2}\delta[n-kj$

Assume that

$A(z)=1+\sum_{k=1}^{p}a[k]z^{-k}=\Pi_{m=1}^{r}(1-z_{m}z^{-1})\Pi_{m=r+1}^{r+l}(1-z_{m}z^{-1})(1-\bar{z}_{m}z^{-1})$

where $p=r+2l$ and $z_{1},...,z_{r+l}$ are all within the unit circle with
$z_{1},...,z_{r}$ being real and $z_{r+1},...,z_{r+l}$ being complex with non-
zero imaginary parts. Assume that the measured process is

$y[n]=x[n]+v[n]$

where $v[.]$ is white, independent of $w[.]$ and with autocorrelation
$\sigma_{v}^{2}\delta[n]$. Assume that

$\sigma_{w}^{2}+\sigma_{v}^{2}A(z)A(z^{-1})=B(z)B(z^{-1})$

with

$B(z)=K\Pi_{m=1}^{p}(1-p_{m}z^{-1})$

where $p_{1},...,p_{p}$ are complex numbers within the unit circle. Find the
value of $K$ in terms of $\sigma_{w}^{2},\sigma_{v}^{2},a[1],...,a[p]$.
Determine the transfer function of the optimal non-causal and causal Wiener
filters for estimating the process $x[.]$ based on $y[.]$ and cast the causal
Wiener filter in time recursive form. Finally, determine the mean square
estimation errors for both the filters.

[2] Consider the state model

${\bf X}[n+1]={\bf A}{\bf X}[n]+{\bf b}w[n+1]$

and an associated measurement model

${\bf z}[n]={\bf Cx}[n]+{\bf v}[n]$

Given an AR(2)process

$x[n]=-a[1]x[n-1]-a[2]x[n-2]+w[n]$

and a measurement model

$z[n]=x[n]+v[n]$

with $w[.]$ and $v[.]$ being white and mutually uncorrelated processes with
variances of $\sigma_{w}^{2}$ and $\sigma_{v}^{2}$ respectively. Cast this
time series model in the above state variable form with ${\bf A}$ being a
$2\times 2$ matrix, ${\bf X}[n]$ a $2\times 1$ state vector, ${\bf b}$ a
$2\times 1$ vector and ${\bf C}$ a $1\times 2$ matrix. Write down explicitly
the Kalman filter equations for this model and prove that in the steady state,
ie, as $n\rightarrow\infty$, the Kalman filter converges to the optimal causal
Wiener filter. For proving this, you may assume an appropriate factorization
of $\sigma_{v}^{2}A(z)A(z^{-1})+\sigma_{w}^{2}$ into a minimum phase and a
non-minimum phase part where $A(z)=1+a[1]z^{-1}+a[2]z^{-2}$ is assumed to be
minimum phase. Note that the steady state Kalman filter is obtained by setting
the state estimation error conditional covariance matrices for prediction and
filtering, namely ${\bf P}[n+1|n]$, ${\bf P}[n|n]$ as well as the Kalman gain
$K[n]$ equal to constant matrices and thus express the Kalman filter in steady
state form as

$\hat{\bf X}[n+1|n+1]={\bf D}\hat{\bf X}[n|n]+{\bf f}z[n+1]$

where ${\bf D,f}$ are functions of the steady state values of ${\bf P}[n|n]$
and ${\bf P}[n+1|n]$ which satisfy the steady state Riccati equation.

[3] Let ${\bf X}_{k},k=1,2,...,N$ and $\theta$ be random vectors defined on
the same probability space with the conditional distribution of ${\bf
X}_{1},...,{\bf X}_{N}$ given $\theta$ being that of iid $N(\theta,{\bf
R}_{1})$ random vectors. Assume further that the distribution of $\theta$ is
$N(\mu,{\bf R}_{2})$. Calculate (a) $p({\bf X}_{1},...,{\bf
X}_{N}|\theta)$,(b) $p(\theta|{\bf X}_{1},...,{\bf X}_{N})$, (3) the ML
estimate of $\theta$ given ${\bf X}_{1},...,{\bf X}_{N}$ and (4) the MAP
estimate of $\theta$ given ${\bf X}_{1},...,{\bf X}_{N}$. Also evaluate

$Cov(\hat{\theta}_{ML}-\theta),Cov(\theta_{MAP}-\theta)$

Which one has a larger trace ? Explain why this must be so.

[4] Let ${\bf X}_{1},...,{\bf X}_{N}$ be iid $N(\mu,{\bf R})$. Evaluate the
Fisher information matrix elements

$-\mathbb{E}[\frac{\partial^{2}}{\partial\mu_{a}\partial\mu_{b}}ln(p({\bf
X}_{1},...,{\bf X}_{N}|\mu,{\bf R})]$
$-\mathbb{E}[\frac{\partial^{2}}{\partial\mu_{a}\partial R_{bc}}ln(p({\bf
X}_{1},...,{\bf X}_{N}|\mu,{\bf R})]$
$-\mathbb{E}[\frac{\partial^{2}}{\partial R_{ab}\partial R_{cd}}ln(p({\bf
X}_{1},...,{\bf X}_{N}|\mu,{\bf R})]$

Taking the special case when the ${\bf X}_{k}^{\prime}s$ are 2-dimensional
random vectors, evaluate the Cramer-Rao lower bound for
$Var(\hat{\mu}_{a}),a=1,2$ and $Var(\hat{R}_{ab}),a,b=1,2$ where $\hat{\mu}$
and $\hat{\bf R}$ are respectively unbiased estimates of $\mu$ and ${\bf R}$.

[5] Let $\rho(\theta)$ denote a mixed quantum state in $\mathbb{C}^{N}$
dependent upon a classical parameter vector
$\theta=(\theta_{1},...\theta_{p})^{T}$. Let $X,Y$ be two observables with
eigendecompositions

$X=\sum_{a=1}^{N}|e_{a}>x(a)<e_{a}|,Y=\sum_{a=1}^{N}|f_{a}>y(a)<f_{a}|,<e_{a}|e_{b}>=<f_{a}|f_{b}>=\delta_{ab}$

First measure $X$ and note its outcome $x(a)$. Then allowing for state
collapse, allow the state of the system to evolve from this collapsed state
$\rho_{c}$ to the new state

$T(\rho_{c})=\sum_{k=1}^{m}E_{k}\rho_{c}E_{k}^{*},\sum_{k=1}^{m}E_{k}^{*}E_{k}=I$

Then measure $Y$ in this evolved state and note the outcome $y(b)$. Calculate
the joint probability distribution $P(x(a),y(b)|\theta)$. Explain by
linearizing about some $\theta_{0}$, how you will calculate the maximum
likelihood estimate of $\theta$ as a function of the measurements $x(a),y(b)$.
Also explain how you will calculate the CRLB for $\theta$. Finally, explain
how you will select the observables $X,Y$ to be measured in such a way that
the CRLB is minimum when $\theta$ is a scalar parameter.

[6] For calculating time recursive updates of parameters in a linear AR model,
we require to apply the matrix inversion lemma to a matrix of the form

$(\lambda.{\bf R}+{\bf bb}^{T})^{-1}$

while for calculating order recursive updates, we require to use the
projection operator update formula in the form

${\bf P}_{span(,\xi)}={\bf P}_{W}+{\bf P}_{{\bf P}_{W}^{\perp}\xi}$

and also invert a block structured matrices of the form

$\left(\begin{array}[]{cc}{\bf A}&{\bf b}\\\ {\bf b}^{T}&c\end{array}\right)$

and

$\left(\begin{array}[]{cc}c&{\bf b}^{T}\\\ {\bf b}&{\bf A}\end{array}\right)$

in terms of ${\bf A}^{-1}$. Explain these statements in the light of the
recurisive least squares lattice algorithm for simultaneous time and order
recursive updates of the prediction filter parameters.

###  10.4 Question paper on statistical signal processing, long test

####  10.4.1 Atom excited by a quantum field

[1] Let the state at time $t=0$ of an atom be $\rho_{A}(0)$ and that of the
field be the coherent state $|\phi(u)><\phi(u)|$ so that the state of both the
atom and the field at time $t=0$ is

$\rho(0)=\rho_{A}(0)\otimes|\phi(u)><\phi(u)|$

Let the Hamiltonian of the atom and field in interaction be

$H(t)=H_{A}+\sum_{k}\omega(k)a(k)^{*}a(k)+\delta.V(t)$

where $H_{A}$ acts in the atom Hilbert space,
$H_{F}=\sum_{k}\omega(k)a(k)^{*}a(k)$ acts in the field Boson Fock space and
the interaction Hamiltonian $V(t)$ acts in the tensor product of both the
spaces. Assume that $V(t)$ has the form

$V(t)=\sum_{k}(X_{k}(t)\otimes a_{k}+X_{k}(t)^{*}\otimes a(k)^{*})$

where $X_{k}(t)$ acts in the atom Hilbert space. Note that writing

$exp(itH_{A})X_{k}(t)exp(-itH_{A})=Y_{k}(t),$
$exp(itH_{F})a_{k}.exp(-itH_{F})=a_{k}.exp(-i\omega(k)t)=a_{k}(t)$

we have on defining the unperturbed (ie, in the absence of interaction) system
plus field Hamiltonian

$H_{0}=H_{A}+H_{F}=H_{A}+\sum_{k}\omega(k)a(k)^{*}a(k)$

that

$W(t)=exp(itH_{0})V(t).exp(-itH_{0})=\sum_{k}(Y_{k}(t)\otimes
a_{k}(t)+Y_{k}(t)^{*}\otimes a_{k}(t)^{*})$

show that upto $O(\delta^{2})$ terms, the total system $\otimes$ field
evolution operator $U(t)$ can be expressed as

$U(t)=U_{0}(t).S(t)$

where

$U_{0}(t)=exp(-itH_{0})=exp(-itH_{A}).exp(-itH_{F})$

is the unperturbed evolution operator and

$S(t)=1-i\delta\int_{0}^{t}W(s)ds-\delta^{2}\int_{0<s_{2}<s_{1}<t}W(s_{1})W(s_{2})ds_{1}ds_{2}$
$=1-i\delta S_{1}(t)-\delta^{2}S_{2}(t)$

Show that with this approximation, the state of the system and bath at time
$T$ is given by

$\rho(T)=U_{0}(T)(1-i\delta S_{1}(T)-\delta^{2}S_{2}(T))\rho(0).(1+i\delta
S_{1}(T)-\delta^{2}S_{2}(T)^{*})U_{0}(T)^{*}$

Evaluate using this expression, the atomic state at time $T$, as

$\rho_{A}(T)=Tr_{2}\rho(T)$

Use the fact that $\rho(0)=\rho_{A}(0)\otimes|\phi(u)><\phi(u)|$ and that

$Tr(a(k)|\phi(u)><\phi(u)|)=<\phi(u)|a(k)|\phi(u)>=u(k),$
$Tr(a(k)^{*}|\phi(u)><\phi(u)|)=<\phi(u)|a(k)^{*}|\phi(u)>=\bar{u}(k),$
$Tr(a(k)a(m)|\phi(u)><\phi(u)|)=<\phi(u)|a(k)a(m)|\phi(u)>=u(k)u(m),$
$Tr(a(k)a(m)^{*}|\phi(u)><\phi(u)|)=<\phi(u)|a(k)a(m)^{*}|\phi(u)>$
$=<\phi(u)|[a(k),a(m)^{*}]|\phi(u)>+<\phi(u)|a(m)^{*}a(k)|\phi(u)>$
$=\delta(k,m)+u(k)\bar{u}(m)$

and

$Tr(a(k)^{*}a(m)|\phi(u)><\phi(u)|)=<\phi(u)|a(k)^{*}a(m)|\phi(u)>=$
$=\bar{u}(k)u(m)$

Remark: More generally, we can expand any general initial state of the system
(atom) and bath (field) as

$\rho(0)=\int F_{A}(u,\bar{u})\otimes|\phi(u)><\phi(u)|dud\bar{u}$ $=\int
F_{A}(u,\bar{u})exp(-|u|^{2})\otimes|e(u)><e(u)|du.d\bar{u}$

In this case also, we can calculate assuming that $\rho(0)$ is the initial
state, final state at time $T$ using the formula

$\rho(T)=U_{0}(T)(1-i\delta S_{1}(T)-\delta^{2}S_{2}(T))\rho(0).(1+i\delta
S_{1}(T)-\delta^{2}S_{2}(T)^{*})U_{0}(T)^{*}$

by evaluating this using the fact that if $Y_{1},Y_{2}$ are system operators,
then

$(Y_{1}\otimes a_{k})(\int
F(u,\bar{u})\otimes|\phi(u)><\phi(u)|dud\bar{u})(Y_{2}\otimes a_{m})$ $=\int
Y_{1}F(u,\bar{u})Y_{2}exp(-|u|^{2})\otimes
a_{k}|e(u)><e(u)|a_{m}dud\bar{u}---(1)$

Now

$a_{k}|e(u)>=u(k)|e(u)>$ $a_{k}^{*}|e(u)>=\partial|e(u)>/\partial u(k)$

so (1) evaluates to after using integration by parts,

$\int
Y_{1}F_{1}(u,\bar{u})Y_{2}u(k)exp(-|u|^{2})\otimes(\partial/\partial\bar{u}(m))(|e(u)><e(u)|)du.d\bar{u}$
$=\int(-\partial/\partial\bar{u}(m))(Y_{1}F(u,\bar{u})Y_{2}).exp(-|u|^{2}u(k)))\otimes|e(u)><e(u)|dud\bar{u}$
$=\int(-\partial/\partial\bar{u}(m)+u(m))(Y_{1}F(u,\bar{u})Y_{2}u(k))|\phi(u)><\phi(u)|du.d\bar{u}$

Likewise,

$(Y_{1}\otimes a_{k}^{*})(\int
F(u,\bar{u})\otimes|\phi(u)><\phi(u)|dud\bar{u})(Y_{2}\otimes a_{m})$ $=\int
Y_{1}F(u,\bar{u})Y_{2}\otimes a_{k}^{*}|\phi(u)><\phi(u)|a_{m}dud\bar{u}$
$=\int Y_{1}F(u,\bar{u})Y_{2}exp(-|u|^{2})\otimes(\partial/\partial
u(k)).(\partial/\partial\bar{u}(m))(|e(u)><e(u)|)du.d\bar{u}$
$=\int(\partial^{2}/\partial
u(k)\partial\bar{u}(m))(Y_{1}F(u,\bar{u})Y_{2}.exp(-|u|^{2}))\otimes|e(u)><e(u)|du.d\bar{u}$
$=\int(exp(|u|^{2})(\partial^{2}/\partial
u(k)\partial\bar{u}(m))(Y_{1}F(u,\bar{u})Y_{2}.exp(-|u|^{2})))\otimes|\phi(u)><\phi(u)|du.d\bar{u}$
$=\int(\partial/\partial
u(k)+\bar{u}(k))(\partial/\partial\bar{u}(m)-u(m))(Y_{1}F(u,\bar{u})Y_{2})\otimes|\phi(u)<\phi(u)|dud\bar{u}$

and likewise terms like

$(Y_{1}\otimes a_{k}^{*})(\int
F(u,\bar{u})\otimes|\phi(u)><\phi(u)|dud\bar{u})(Y_{2}\otimes a_{m}^{*})$

and

$(Y_{1}\otimes a_{k})(\int
F(u,\bar{u})\otimes|\phi(u)><\phi(u)|dud\bar{u})(Y_{2}\otimes a_{m}^{*})$

We leave it as an exercise to the reader to evaluate these terms and hence
derive upto the second order Dyson series approximation for the evolution
operator the approximate final state of the atom and field and then by partial
tracing over the field variables, to derive the approximate final atomic state
under this interaction of atom and field (ie, system and bath).

[2] If the state of the system is given by $\rho=\rho_{0}+\delta.\rho_{1}$,
then evaluate the Von-Neumann entropy of $\rho$ as a power series in $\delta$.

Hint:

####  10.4.2 Estimating the power spectral density of a stationary Gaussian
process using an AR model:Performance analysis of the algorithm

The process $x[n]$ has mean zero and autocorrelation $R(\tau)$. It’s $p^{th}$
order prediction error process is given by

$e[n]=x[n]+\sum_{k=1}^{p}a[k]x[n-k]$

where the $a[k]^{\prime}s$ are chosen to minimize $\mathbb{E}e[n]^{2}$. The
optimal normal equations are

$\mathbb{E}(e[n]x[n-m])=0,m=1,2,...,p$

and these yield the Yule-Walker equations

$\sum_{k=1}^{p}R[m-k]a[k]=-R[m],m=1,2,...,p$

or equivalently in matrix notation,

${\bf a}=-{\bf R}_{p}^{-1}{\bf r}_{p}$

where

${\bf R}_{p}=((R[m-k]))_{1\leq m,k\leq p},{\bf r}_{p}=((R[m]))_{m=1}^{p}$

The minimum mean square prediction error is then

$\sigma^{2}=\mathbb{E}(e[n]^{2}=\mathbb{E}(e[n]x[n])=R[0]+\sum_{k=1}^{p}a[k]R[k]=R[0]+{\bf
a}^{T}{\bf r}_{p}$ $=R[0]-{\bf r}_{p}^{T}{\bf R}_{p}^{-1}{\bf r}_{p}$

If the order of the AR model $p$ is large, then $e[n]$ is nearly a white
process, in fact in the limit $p\rightarrow\infty$, $e[n]$ becomes white, ie,
$\mathbb{E}(e[n]x[n-m])$ becomes zero for all $m\geq 1$ and hence so does
$\mathbb{E}(e[n]e[n-m])$. The ideal known statistics AR spectral estimate is
given by

$S_{AR,p}(\omega)=S(\omega)=\frac{\sigma^{2}}{|A(\omega)|^{2}},$
$A(\omega)=1+{\bf a}^{T}{\bf e}(\omega),$ ${\bf e}(\omega)=((exp(-j\omega
m)))_{m=1}^{p}$

Since in practice, we do not have the exact correlations available with us, we
estimate the AR parameters using the least squares method, ie, by minimizing

$\sum_{n=1}^{N}(x[n]+{\bf a}^{T}{\bf x}_{n})^{2}$

where

${\bf x}_{n}=((x(n-m)))_{m=1}^{p}$

The finite data estimate of ${\bf R}_{p}$ is

$\hat{R}_{p}=N^{-1}\sum_{n=1}^{N}{\bf x}_{n}{}_{n}^{T}$

and that of ${\bf r}_{p}$ is

$\hat{r}_{p}=N^{-1}\sum_{n=1}^{N}x[n]{\bf x}_{n}$

We write $R$ for $R_{p}$ and $\hat{R}=R+\delta R$ for $\hat{R}_{p}$. Likewise,
we write $r$ for $r_{p}$ and $r+\delta r$ for $\hat{r}_{p}$ for notational
convenience. Note that $\hat{R}$ and $\hat{r}$ are unbiased estimates of $R$
and $r$ respectively and hence $\delta R$ and $\delta r$ have zero mean. Now
the least squares estimate $\hat{a}=a+\delta a$ of $a$ is given by

$\hat{a}=a+\delta a=-\hat{R}^{-1}\hat{r}=-(R+\delta R)^{-1}(r+\delta r)$
$=-(R^{-1}-R^{-1}\delta RR^{-1}+R^{-1}\delta R.R^{-1}\delta R.R^{-1})(r+\delta
r)$ $=a-R^{-1}\delta r-R^{-1}\delta Ra+R^{-1}\delta R.R^{-1}\delta
r+R^{-1}\delta R.R^{-1}\delta R.a$

upto quadratic orders in the data correlation estimate errors. Thus, we can
write

$\delta a=-R^{-1}\delta r-R^{-1}\delta Ra+R^{-1}\delta R.R^{-1}\delta
r+R^{-1}\delta R.R^{-1}\delta R.a$

Likewise, the estimate of the prediction error energy $\sigma^{2}$ is given by

$\hat{\sigma}^{2}=\sigma^{2}+\delta\sigma^{2}=\hat{R}[0]+\hat{a}^{T}\hat{r}$

and hence

$\delta\sigma^{2}=\delta R[0]+a^{T}\delta r+\delta a^{T}r+\delta a^{T}\delta
r$ $=\delta R[0]+a^{T}\delta r+(-R^{-1}\delta r-R^{-1}\delta Ra+R^{-1}\delta
R.R^{-1}\delta r+R^{-1}\delta R.R^{-1}\delta r)^{T}\delta r$ $+(-R^{-1}\delta
r-R^{-1}\delta Ra)^{T}\delta r$

again upto quadratic orders in the data correlation estimate errors. In terms
of the fourth moments of the process $x[n]$, we can now evaluate the bias and
mean square fluctuations of $\hat{a}$ and $\hat{\sigma}^{2}$ w.r.t their known
statistics values $a,\sigma^{2}$ respectively. First we observe that the bias
of $\hat{a}$ is

$\mathbb{E}(\delta a)=\mathbb{E}(-R^{-1}\delta r-R^{-1}\delta Ra+R^{-1}\delta
R.R^{-1}\delta r+R^{-1}\delta R.R^{-1}\delta R.a)$ $=\mathbb{E}(R^{-1}\delta
R.R^{-1}\delta r)+\mathbb{E}(R^{-1}\delta R.R^{-1}\delta R.a)$

Specifically, for the $i^{th}$ component of this, we find that with
$P=R^{-1}$, and using Einstein’s summation convention over repeated indices,

$\mathbb{E}(\delta a_{i})=$ $\mathbb{E}(P_{ij}\delta R_{jk}P_{km}\delta
r_{m})+\mathbb{E}(P_{ij}\delta R_{jk}P_{km}\delta R_{ml}a_{l})$
$=P_{ij}P_{km}\mathbb{E}(\delta R_{jk}\delta
r_{m})+P_{ij}P_{km}a_{l}\mathbb{E}(\delta R_{jk}\delta R_{ml})$

Now,

$\mathbb{E}(\delta R_{jk}.\delta
r_{m})=N^{-2}\sum_{n,l=1}^{N}\mathbb{E}[(x[n-k]x[n-j]-R[j-k])(x[l]x[l-m]-R[m])]$
$=N^{-2}\sum_{n,l=1}^{N}(\mathbb{E}(x[n-k]x[n-j]x[l]x[l-m])-R[j-k]R[m])$

upto fourth degree terms in the data. Note that here we have not yet made any
assumption that $x[n]$ is a Gaussian process, although we are assuming that
the process is stationary at least upto fourth order. Writing

$\mathbb{E}(x[n_{1}]x[n_{2}]x[n_{3}]x[n_{4}])=C(n_{2}-n_{1},n_{3}-n_{1},n_{4}-n_{1})$

we get

$\mathbb{E}(\delta R_{jk}.\delta r_{m})=$
$N^{-2}\sum_{n,l=1}^{N}C(k-j,l-n+k,l-n+k-m)$

####  10.4.3 Statistical performance analysis of the MUSIC and ESPRIT
algorithms

Let

$\hat{R}=R+\delta R$

be an estimate of the signal correlation matrix $R$. Let $v_{1},...,v_{p}$
denote the signal eigenvectors and $v_{p+1},...,v_{N}$ the noise eigenvectors.
The signal subspace eigen-matrix is

$V_{S}=[v_{1},...,v_{p}]\in\mathbb{C}^{N\times p}$

and the noise subspace eigen-matrix is

$V_{N}=[v_{p+1},...,v_{N}]$

so that

$V=[V_{S}|V_{N}]$

is an $N\times N$ unitary matrix. The signal correlation matrix model is

$R=EDE^{*}+\sigma^{2}I$

$\sigma^{2}$ is estimated as $\hat{\sigma}^{2}$ which is the average of the
$N-p$ smallest eigenvalues of $R$. In this expression,

$E=[e(\omega_{1}),...,e(\omega_{p})]$

is the signal manifold matrix where

$e(\omega)=[1,exp(j\omega),...,exp(j(N-1)\omega)]^{T}$

is the steering vector along the frequency $\omega$. Note that since $N>p$,
$E$ has full column rank $p$. The signal eigenvalues of $R$ are

$\lambda_{i}=\mu_{i}+\sigma^{2},i=1,2,...,p$

while the noise eigenvalues are

$\lambda_{i}=\sigma^{2},i=p+1,...,N$

Thus,

$Rv_{i}=\lambda_{i}v_{i},i=1,2,...,N$

The known statistics exact MUSIC spectral estimator is

$P(\omega)=(1-N^{-1}\parallel V_{S}^{*}e(\omega)\parallel^{2})^{-1}$

which is infinite when and only when
$\omega\in\\{\omega_{1},...,\omega_{p}\\}$. Alternately, the zeros of the
function

$Q(\omega)=1/P(\omega)=1-N^{-1}\parallel V_{S}^{*}e(\omega)\parallel^{2}$

are precisely $\\{\omega_{1},...,\omega_{p}\\}$ because, the
$V_{N}^{*}e(\omega)=0$ iff $e(\omega)$ belongs to the signal subspace
$span\\{e(\omega_{i}):1\leq i\leq p\\}$ iff
$\\{e(\omega),e(\omega_{i}),i=1,2,...,p\\}$ is a linearly dependent set iff
$\omega\in\\{\omega_{1},...,\omega_{p}\\}$, the last statement being a
consequence of the fact that $N\geq p+1$ and the Van-Der-Monde determinant
theorem. Now, owing to finite data effects, $v_{i}$ gets perturbed to
$v_{i}+\delta v_{i}$ for each $i=1,2,...,p$ so that correspondingly, $V_{S}$
gets perturbed to $\hat{V}_{S}=V_{S}+\delta V_{S}$. Then $Q$ gets perturbed to
(upto first orders of smallness, ie, upto $O(\delta R)$

$Q(\omega)+\delta Q(\omega)=1-N^{-1}\parallel(V_{S}+\delta
V_{S})^{*}e(\omega)\parallel^{2}$

so that

$\delta Q(\omega)=(-2/N)Re(e(\omega)^{*}\delta V_{S}V_{S}^{*}e(\omega))$

Let $\omega$ be one of the zeros of $Q(\omega)$, ie, one of the frequencies of
the signal. Then, owing to finite data effects, it gets perturbed to
$\omega+\delta\omega$ where

$Q^{\prime}(\omega)\delta\omega+\delta Q(\omega)=0$

ie,

$\delta\omega=-\delta Q(\omega)/Q^{\prime}(\omega)$

and

$Q^{\prime}(\omega)=(-2/N)Re(e(\omega)^{*}P_{S}e^{\prime}(\omega))=(-2/N)Re<e(\omega)|P_{S}|e^{\prime}(\omega)>$

where

$P_{S}=V_{S}V_{S}^{*}$

is the orthogonal projection onto the signal subspace. Note that by first
order perturbation theory

$|\delta v_{i}>=\sum_{j\neq i}\frac{|v_{j}><v_{j}|\delta
R|v_{i}>}{\lambda_{i}-\lambda_{j}},i=1,2,...,N$

and

$\delta\lambda_{i}=<v_{i}|\delta R|v_{i}>$

So

$\delta V_{S}=column(\sum_{j\neq i}|v_{j}><v_{j}|\delta
R|v_{i}>/(\lambda_{i}-\lambda_{j}):i=1,2,...,p]$ $=[Q_{1}\delta
R|v_{1}>,...,Q_{p}\delta R|v_{p}>]$

where

$Q_{i}=\sum_{j\neq i}|v_{j}><v_{j}|/(\lambda_{i}-\lambda_{j})$

We can express this as

$\delta V_{S}=[Q_{1},...,Q_{p}](I_{p}\otimes\delta R.V_{S})$

and hence

$\delta Q(\omega)=(-2/N)Re(<e(\omega)|\delta V_{S}V_{S}^{*}|e(\omega))>)$
$=(-2/N)Re(<e(\omega)S(I_{p}\otimes\delta R).V_{S}V_{S}^{*}|e(\omega)>)$
$=(-2/N)Re(<e(\omega)|S(I_{p}\otimes\delta R)P_{S}|e(\omega)>)$

where as before $P_{S}=V_{S}V_{S}^{*}$ is the orthogonal projection onto the
signal subspace and

$S=[Q_{1},Q_{2},...,Q_{p}]$

The evaluation of $\mathbb{E}((\delta\omega)^{2})$ can therefore be easily
done once we evaluate $\mathbb{E}(\delta R(i,j)\delta R(k,m))$ and this is
what we now propose to do.

Statistical performance analysis of the ESPRIT algorithm. Here, we have two
correlation matrices $R,R_{1}$ having the structures

$R=EDE^{*}+\sigma^{2}I,R_{1}=ED\Phi^{*}E^{*}+\sigma^{2}Z$

where $E$ has full column rank, $D$ is a square, non-singular positive
definite matrix and $\Phi$ is a diagonal unitary matrix. The structure of $Z$
is known. We have to estimate the diagonal entries of $\Phi$ which give us the
signal frequencies. Let as before $V_{S}$ denote the matrix formed from the
signal eigenvectors of $R$ and $V_{N}$ that formed from the noise eigenvectors
of $R$. $V=[V_{S}|V_{N}]$ is unitary. When we form finite data estimates, $R$
is replaced by $\hat{R}=R+\delta R$ and $R_{1}$ by $\hat{R}_{1}=R_{1}+\delta
R_{1}$. The estimate of $\sigma^{2}$ is
$\hat{\sigma}^{2}=\sigma^{2}+\delta\sigma^{2}$ obtained by averaging the $N-p$
smallest eigenvalues of $\hat{R}$. The estimate of $R_{S}=EDE^{*}$ is then

$\hat{R}_{S}=R_{S}+\delta R_{S}=\hat{R}-\hat{\sigma}^{2}I=R-\sigma^{2}I+\delta
R-\delta\sigma^{2}$

Thus,

$\delta R_{S}=\delta R-\delta\sigma^{2}I$

Likewise, the estimate of $R_{S1}=ED\Phi^{*}E^{*}$ is

$\hat{R}_{S1}=R_{S1}+\delta
R_{S1}=\hat{R}_{1}-\hat{\sigma}^{2}Z=R_{1}-\sigma^{2}Z+\delta
R_{1}-\delta\sigma^{2}Z$

so that

$\delta R_{S1}=\delta R_{1}-\delta\sigma^{2}Z$

The rank reducing numbers of the matrix pencil

$R_{S}-\gamma R_{S1}=ED(I-\gamma\Phi^{*})E^{*}$

are clearly the diagonal entries of $\Phi$. Since $R(E)=R(V_{S})$, it follows
that these rank reducing numbers are the zeroes of the polynomial equation

$det(V_{S}^{*}(R_{S}-\gamma R_{S1})V_{S})=0$

or equivalently of

$det(M-\gamma V_{S}^{*}R_{S1}V_{S})=0$

where

$M=V_{S}^{*}R_{S}V_{S}=diag[\mu_{1},...,\mu_{p}]$

with

$\mu_{k}=\lambda_{k}-\sigma^{2},kj=1,2,...,p$

being the nonzero eigenvalues of $R_{S}$. When finite data effects are taken
into consideration, these become the rank reducing numbers of

$det(M+\delta M-\hat{\gamma}(V_{S}+\delta V_{S})^{*}(R_{S1}+\delta
R_{S1})(V_{S}+\delta V_{S}))=0$

or equivalently, upto first order perturbation theory, the rank reducing
number $\gamma$ gets perturbed to $\hat{\gamma}=\gamma+\delta\gamma$, where

$det(M-\gamma X+\delta M-\gamma.\delta X-\delta\gamma X)=0$

where

$X=V_{S}^{*}R_{S1}V_{S},\delta X=\delta
V_{S}^{*}R_{S1}V_{S}+V_{S}^{*}R_{S1}\delta V_{S}+V_{S}^{*}\delta R_{S1}V_{S}$

Let $\xi$ be a right eigenvector of $M-\gamma X$ and $\eta^{*}$ a left
eigenvector of the same corresponding to the zero eigenvalue. Note that
$M,X,\delta M,\delta X$ are all Hermitian but $M-\gamma X$ is not because
$\gamma=exp(j\omega_{i})$ is generally not real. Then, let $\xi+\delta\xi$ be
a right eigenvector of

$M-\gamma X+\delta M-\gamma\delta X-\delta\gamma X$

corresponding to the zero eigenvalue. In terms of first order perturbation
theory,

$0=(M-\gamma X+\delta M-\gamma\delta X-\delta\gamma X)(\xi+\delta\xi)$
$=(M-\gamma X)\delta\xi+(\delta M-\gamma\delta X)\xi-\delta\gamma X\xi$

Premultiplying by $\eta^{*}$ gives us

$\delta\gamma\eta^{*}X\xi=\eta^{*}(\delta M-\gamma\delta X)\xi$

or equivalently,

$\delta\gamma=\frac{<\eta|\delta M-\gamma\delta X\|\xi>}{<\eta|X|\xi>}$

which is the desired formula.

###  10.5 Quantum neural networks for estimating EEG pdf from speech pdf based
on Schrodinger’s equation

Reference:Paper by Vijay Upreti, Harish Parthasarathy, Vijayant Agarwal and
Guguloth Sagar.

Quantum neural networks for estimating the marginal pdf of one signal from the
marginal pdf of another signal. The idea is that the essential characteristics
of either speech or EEG signals are contained in their empirical pdf’s, ie,
the proportion of time that the signal amplitude spends within any Borel
subset of the real line. This hypothesis leads one to believe that by
estimating the pdf of the EEG data from that of speech data alone, we can
firstly do away with the elaborate apparatus required for measuring EEG
signals and secondly we can extract from the EEG pdf, the essential
characteristics of the EEG signal parameters which determine the condition of
a diseased brain. It is a fact that since the Hamiltonian operator $H(t)$ in
the time dependent Schrodinger equation is Hermitian, the resulting evolution
operator $U(t,s),t>s$ is unitary causing thereby the wave function modulus
square to integrate to unity at any time $t$ if at time $t=0$ it does
integrate to unity. Thus, the Schrodinger equation naturally generates a
family of pdf’s. Further, based on the assumption that the speech pdf and the
EEG pdf for any patient are correlated in the same way (a particular kind of
brain disease causes a certain distortion in the EEG signal and a correlated
distortion in the speech process with the correlation being patient
independent), we include a potential term in Schrodinger’s equation involving
speech pdf and hence while tracking a given EEG pdf owing to the adaptation
algorithm of the QNN, the QNN also ensures that the output pdf coming from
Schrodinger’s equation will also maintain a certain degree of correlation with
the speech pdf. The basic idea behind the quantum neural network is to
modulate the potential by a ”weight process” with the weight increasing
whenever the desired pdf exceeds the pdf generated by Schrodinger’s equation
using the modulus of the wave function squared. A hieuristic proof of why the
pdf generated by the Schrodinger equation should increase when the weight
process and hence th modulated potential increases is provided at the end of
the paper.

First consider the following model for estimating the pdf $p(t,x)$ of one
signal via Schrodinger’s wave equation which naturally generates a continuous
family of pdf’s via the modulus square of the wave function:

$ih\partial_{t}\psi(t,x)=(-h^{2}/2)\partial_{x}^{2}\psi(t,x)+W(t,x)V(x)\psi(t,x)$

The weight process $W(t,x)$ is generated in such a way that $|\psi(t,x)|^{2}$
tracks the given pdf $p(t,x)$. This adaptation scheme is

$\partial_{t}W(t,x)=-\beta W(t,x)+\alpha(p(t,x)-|\psi(t,x)|^{2})$

where $\alpha,\beta$ are positive constants. This latter equation guarantees
that if $W(t,x)$ is very large and positive (negative), then the term $-\beta
W(t,x)$ is very large and negative (positive) causing $W$ to decrease
(increase). In other words, $-\beta.W$ is a threshold term that prevents the
weight process from getting too large in magnitude which could cause the
break-down of the quantum neural network. We are assuming that $V(x)$ is a
non-negative potential. Thus, increase in $W$ will cause an increase in $WV$
which as we shall see by an argument given below, will cause a corresponding
increase in $|\psi|^{2}$ and likewise a decrease in $W$ will cause a
corresponding decrease in $|\psi|^{2}$. Now we add another potential term
$p_{0}(t,x)V_{0}(x)$ to the above Schrodinger dynamics where $p_{0}$ is the
pdf of the input signal. This term is added because we have the understanding
that the output pdf $p$ is in some way correlated with the input pdf. Then the
dynamics becomes

$ih\partial_{t}\psi(t,x)=(-h^{2}/2)\partial_{x}^{2}\psi(t,x)+W(t,x)V(x)\psi(t,x)+p_{0}(t,x)V_{0}(x)\psi(t,x)$

Statistical performance analysis of the QNN: Suppose $p_{0}(t,x)$ fluctuates
to $p_{0}(t,x)+\delta p_{0}(t,x)$ and $p(t,x)$ to $p(t,x)+\delta p(t,x)$. Then
the corresponding QNN output wave function $\psi(t,x)$ will fluctuate to
$\psi(t,x)+\delta\psi(t,x)$ where $\delta\psi$ satisfies the linearized
Schrodinger equation and likewise the weight process $W(t,x)$ will fluctuate
to $W(t,x)+\delta W(t,x)$. The linearized dynamics for these fluctuations are

$ih\partial_{t}\delta\psi(t,x)=(-h^{2}/2)\partial_{x}^{2}\delta\psi(t,x)+V(x)(W\delta\psi+\psi\delta
W)+V_{0}(x)(p_{0}\delta\psi+\psi\delta p_{0})$ $\partial_{t}\delta
W=-\beta\delta W+\alpha(\delta p-\bar{\psi}.\delta\psi-\psi.\delta\bar{\psi})$

Writing

$\psi(t,x)=A(t,x).exp(iS(t,x)/h)$

we get

$\delta\psi=(\delta A+iA\delta S/h)exp(iS/h)$

so that with prime denoting spatial derivative,

$\delta\psi^{\prime}=(\delta A^{\prime}+iA^{\prime}\delta S/h+iA\delta
S^{\prime}/h+iS^{\prime}\delta A/h-AS^{\prime}\delta S/h^{2})exp(iS/h)$
$\delta\psi^{\prime\prime}=(\delta A^{\prime\prime}+iA^{\prime\prime}\delta
S/h+2iA^{\prime}\delta S^{\prime}/h+iA\delta
S^{\prime\prime}/h+2iS^{\prime}\delta A^{\prime}/h+iS^{\prime\prime}\delta
A/h-AS^{\prime}\delta S^{\prime}/h^{2}$ $-A^{\prime}S^{\prime}\delta
S/h^{2}-AS^{\prime\prime}\delta S/h^{2}-AS^{\prime}\delta
S^{\prime}/h^{2}-S^{{}^{\prime}2}\delta A/h^{2}-iAS^{{}^{\prime}2}\delta
S/h^{3}).exp(iS/h)$

Thus we obtain the following differential equation for the wave function
fluctuations caused by random fluctuations in the EEG and speech pdf’s:

$ih\partial_{t}\delta A-\partial_{t}(A\delta S)-\delta A.\partial_{t}S-A\delta
S\partial_{t}S/h$ $=(-h^{2}/2)\delta A^{\prime\prime}-ihA^{\prime\prime}\delta
S/2-ihA^{\prime}\delta S^{\prime}-ihA.\delta
S^{\prime\prime}/2-ihS^{\prime}\delta A^{\prime}$ $-ihS^{\prime\prime}\delta
A/2+AS^{\prime}\delta S^{\prime}/2+A^{\prime}S^{\prime}\delta
S/2+AS^{\prime\prime}\delta S/2+AS^{\prime}\delta
S^{\prime}/2+S^{{}^{\prime}2}\delta A/2+iAS^{{}^{\prime}2}\delta S/2h$
$+V_{0}(x)p_{0}(t,x)(\delta A+iA\delta S/h)+V_{0}(x)A\delta p_{0}(t,x)$
$+V(x)W(t,x)(\delta A+iA\delta S/h)+V(x)A\delta W)$

On equating real and imaginary parts in this equation, we get two pde’s for
the two functions $A$ and $S$. However, it is easier to derive these equations
by first deriving the unperturbed pde’s for $A,S$ and then forming the
perturbations. The unperturbed equations are (See Appendix A.1)

$h\partial_{t}A=-h\partial_{x}A.\partial_{x}S-(1/2)A\partial_{x}^{2}S$
$-A\partial_{t}S=-((h^{2}/2)\partial_{x}^{2}A-(1/2)A(\partial_{x}S)^{2})+WVA+p_{0}V_{0}A$
$\partial_{t}W(t,x)=-\beta(W(t,x)-W_{0}(x))+\alpha(p(t,x)-A^{2})$

Forming the variation of this equation around a fixed operating point
$p_{0}(x),p(x)=|\psi(t,x)|^{2}$ and $W(x)$ with
$\psi(t,x)=A(x).exp(-iEt+iS_{0}(x))$ gives us

$h\partial_{t}\delta A=-h\delta A^{\prime}.S_{0}^{\prime}-hA^{\prime}\delta
S^{\prime}-(1/2)S_{0}^{\prime\prime}\delta A-(1/2)A\delta S^{\prime\prime},$
$-A\partial_{t}\delta S+E\delta A=(-h^{2}/2)\delta
A^{\prime\prime}-(1/2)S_{0}^{{}^{\prime}2}\delta A-AS_{0}^{\prime}\delta
S^{\prime\prime}$ $+VW\delta A+VA\delta W+V_{0}A\delta p_{0}+V_{0}p_{0}\delta
A,$ $\partial_{t}\delta W=-\beta\delta W+\alpha.\delta p-2\alpha A\delta A$

Appendix

A.1

Proof that increase in potential causes increase in probability density
determined as modulus square of the wave function.

Consider Schrodinger’s equation in the form

$ih\partial_{t}\psi(t,x)=(-h^{2}/2)\partial_{x}^{2}\psi(t,x)+V(x)\psi(t,x)$

We write

$\psi(t,x)=A(t,x).exp(iS(t,x)/h)$

where $A,S$ are real functions with $A$ non-negative. Thus,
$|\psi|^{2}=A^{2}$. Substituting this expression into Schrodinger’s equation
and equating real and imaginary parts gives

$h\partial_{t}A=-h\partial_{x}A.\partial_{x}S-(1/2)A\partial_{x}^{2}S)$
$-A\partial_{t}S=-((h^{2}/2)\partial_{x}^{2}A-(1/2)A(\partial_{x}S)^{2})+VA$

We write

$S(t,x)=-Et+S_{0}(x)$

so that $E$ is identified with the total energy of the quantum particle. Then,
the second equation gives on neglecting $O(h^{2})$ terms,

$(\partial_{x}S_{0})=\pm\sqrt{2(E-V(x))}$

and then substituting this into the first equation with $\partial_{t}A=0$
gives

$S_{0}^{\prime\prime}/S_{0}^{\prime}=-2hA^{\prime}/A$

or

$ln(S_{0}^{\prime})=-2h.ln(A)+C$

so

$A=K_{1}(S_{0}^{\prime})^{-1/2}=K_{2}(E-V(x))^{-1/4}$

which shows that as long as $V(x)<E$, $A(x)$ and hence
$A(x)^{2}=|\psi(x)|^{2}$ will increase with increasing $V(x)$ confirming the
fact that by causing the driving potential to increase when $p_{E}-|\psi|^{2}$
is large positive, we can ensure that $|\psi|^{2}$ will increase and viceversa
thereby confirming the a validity of our adaptive algorithm for tracking the
EEG pdf.

A.2 Calculating the empirical distribution of a stationary stochastic process.

Let $x(t)$ be a stationary stochastic process. The empirical density estimate
of $N$ samples of this process spaced $\tau_{k},k=1,2,...,n-1$ seconds apart
from the first sample, ie, an estimate of the joint density of
$x(t),x(t+\tau_{k}),k=1,2,...,N-1$ is given bylog(

$\hat{p}_{T}(x_{1},...,x_{N};\tau_{1},...,\tau_{N-1})=T^{-1}\int_{0}^{T}\Pi_{k=1}^{N}\delta(x(t+\tau_{k})-x_{k})dt$

where $\tau_{0}=0$. The Gartner-Ellis theorem in the theory of large
deviations provides a reliability estimate of how this empirical density
performs. Specifically, if $f$ is a function of $N$ variables, then the moment
generating functional of $\hat{p}_{T}(.,\tau_{1},...,\tau_{N-1})$ is given by

$\mathbb{E}[exp(\int\hat{p}_{T}(x_{1},...,x_{N};\tau_{1},...,\tau_{N-1})f(x_{1},...,x_{N})dx_{1}...dx_{N})]$
$=\mathbb{E}[exp(T^{-1}\int_{0}^{T}f(x(t),x(t+\tau_{1}),...,x(t+\tau_{N-1}))dt)]$
$=M_{T}(f|\tau_{1},...,\tau_{N-1})$

say. Then, the rate functional for the family of random fields
$\hat{p}_{T}(.,\tau_{1},...\tau_{N-1}),T\rightarrow\infty$ is given by

$I(p)=sup_{f}(\int
f(x_{1},...,x_{N})p(x_{1},...,x_{N})dx_{1}...dx_{N}-\Lambda(f))$

where

$\Lambda(f)=lim_{T\rightarrow\infty}T^{-1}log(M_{T}(T.f))$

assuming that this limit exists. It follows that as $T\rightarrow\infty$, the
asymptotic probability of $\hat{p}_{T}$ taking values in a set $E$ is given by

$T^{-1}log(Pr(\hat{p}_{T}(.,\tau_{1},...,\tau_{N-1})\in E)\approx-
inf(I(p):p\in E)$

or equivalently,

$Pr(\hat{p}_{T}(.,\tau_{1},...,\tau_{N-1})=p)\approx
exp(-TI(p)),T\rightarrow\infty$

Appendix A.3

We can also use the electroweak theory or more specially, the Dirac-Yang-Mills
non-Abelian gauge theory for matter and gauge fields to track a given pdf.
Such an algorithm enables us to make use of elementary particles in a particle
accelerator to simulate probability densities for learning about the
statistics of signals.

Let $A_{\mu}^{a}(x)\tau_{a}$ be a non-Abelian gauge field assuming values in a
Lie algebra $\mathfrak{g}$ which is a Lie sub-algebra of $\mathfrak{u}(N)$.
Here, $\tau_{a},a=1,2,...,N$ are Hermitian generators of $\mathfrak{g}$.
Denote their structure constants by $\\{C(abc)\\}$, ie,

$[\tau_{a},\tau_{b}]=iC(abc)\tau_{c}$

Then the gauge covariant derivative is

$\nabla_{\mu}=\partial_{\mu}+ieA_{\mu}^{a}\tau_{a}$

and the corresponding Yang-Mills antisymmetric gauge field tensor, also called
the curvature of the gauge covariant derivative is given by

$ieF_{\mu\nu}=[\nabla_{mu},\nabla_{\nu}]=$
$=ie(A_{\nu,\mu}^{a}-A_{\mu,\nu}^{a})-eC(abc)A_{\mu}^{b}A_{\nu}^{c})\tau_{a}$

so that

$F_{\mu\nu}=F_{\mu\nu}^{a}\tau_{a}$

where

$F_{\mu\nu}^{a}=A_{\nu,\mu}^{a}-A_{\mu,\nu}^{a})-eC(abc)A_{\mu}^{b}A_{\nu}^{c}$

The action functional for the matter field $\psi(x)$ having $4N$ components
and the gauge potentials $A_{\mu}^{a}(x)$ is then given by (S.Weinberg, The
quantum theory of fields, vol.II)

$S[A,\psi]=\int Ld^{4}x,$
$L=\bar{\psi}(\gamma^{\mu}(i\partial_{\mu}+eA_{\mu}^{a}\tau_{a})-m)\psi-(1/4)F_{\mu\nu}^{a}F^{a\mu\nu}$

where

$\bar{\psi}=\psi^{*}\gamma^{0}$

The equations of motion obtained by setting the variation of $S$ w.r.t the
matter wave function $\psi$ and the gauge potentials $A$ are given by

$[\gamma^{\mu}(i\partial_{\mu}+eA_{\mu}^{a}\tau_{a})-m]\psi=0,$
$D_{\nu}F^{\mu\nu}=J^{\mu}$

where $D_{\mu}$ is the gauge covariant derivative in the adjoint
representation, ie,

$[\partial_{\nu}+ieA_{\nu},F^{\mu\nu}]=J^{\mu}$

or equivalently in terms of components,

$\partial_{\nu}F^{a\mu\nu}-eC(abc)A_{\nu}^{b}F^{c\mu\nu}=J^{a\mu}$

with $J^{a\mu}$ being the Dirac-Yang-Mills current:

$J^{a\mu}=-e\bar{\psi}(\gamma^{\mu}\otimes\tau_{a})\psi$

it is clear that $\psi$ satisfies a Dirac-Yang-mills matter wave equation and
it can be expressed in the form of Dirac’s equation with a self-adjoint
Hamiltonian owing to which $J^{a\mu}$ is a conserved current and in
particular, the $|\psi|^{2}$ integrates over space to give a constant at any
time which can be set to unity by the initial condition. In other words,
$|\psi(x)|^{2}$ is a probability density at any time $t$ where $x=(t,{\bf
r})$. Now it is easy to make the potential in this wave equation adaptive, ie,
depend on the error $p(x)-|\psi(x)|^{2}$. Simply add an interaction term
$W(x)(p(x)-|\psi(x)|^{2})V(x)$ to to the Hamiltonian. This is more easily
accomplished using the electroweak theory where there is in addition to the
Yang-Mills gauge potential terms a photon four potential $B_{\mu}$ with
$B_{0}(x)=W(x)(p(x)-|\psi(x)|^{2})V(x)$. This results in the following wave
equation

$[\gamma^{\mu}(i\partial_{\mu}+eA_{\mu}^{a}\tau_{a}+eB_{\mu})-m]\psi=0$

The problem with this formalism is that a photon Lagrangian term
$(-1/4)(B_{\nu,\mu}-B_{\mu,\nu})(B^{\nu,\mu}-B^{\mu,\nu})$ also has to be
added to the total Lagrangian which results in the Maxwell field equation

$F^{ph\mu\nu}_{,\nu}=-e\bar{\psi}\gamma^{\mu}\psi$

where

$F^{ph}_{\mu\nu}=B_{\nu,\mu}-B_{\mu,\nu}$

is the electromagnetic field. In order not to violate these Maxwell equations,
we must add another external charge source term to the rhs so that the Maxwell
equation become

$F^{ph\mu\nu}_{,\nu}=-e\bar{\psi}\gamma^{\mu}\psi+\rho(x)\delta^{\mu}_{0}$

$\rho(x)$ is an external charge source term that is given by

$-e\psi^{*}\psi+\rho(x)=-\nabla^{2}(W(x)(p(x)-|\psi(x)|^{2})V(x))=-\nabla^{2}B_{0}$

so that

$B_{0}=W(p-|\psi|^{2})V$

The change in the Lagrangian produced by this source term is
$-\rho(x)B_{0}(x)$. This control charge term can be adjusted for appropriate
adaptation. Thus in effect our Lagrangian acquires two new terms, the first is

$e\bar{\psi}\gamma^{0}B_{0}\psi=e\psi^{*}\psi.B_{0}$

and the second is

$-\rho.B_{0}$

where

$\rho=-\nabla^{2}(W(p-\psi^{*}\psi))+e\psi^{*}\psi$

The dynamics of the wave function $\psi$ then acquires two new terms, the
first coming the variation w.r.t $\bar{\psi}$ of the term
$e\psi^{*}\psi.B_{0}=e\bar{\psi}\gamma^{0}\psi.B_{0}$, this term is
$eB_{0}\gamma^{0}\psi$, the second is the term coming from the variation w.r.t
$\bar{\psi}$ of the term
$-\rho.B_{0}=B_{0}\nabla^{2}(W(p-\bar{\psi}\gamma^{0}\psi))$. This variation
gives

$(\nabla^{2}B_{0})(W\gamma^{0}\psi)$

The resulting equations of motion for the wave function get modified to

$[\gamma^{\mu}(i\partial_{\mu}+eA^{a}_{\mu}\tau_{a}+eB_{\mu})-m+\nabla^{2}B_{0}.W\gamma^{0}]\psi=0$

This equation is to be supplemented with the Maxwell and Yang-Mills gauge
field equations and of course the weight update equation for $W(x)$.

###  10.6 Statistical Signal Processing, M.Tech, Long TestMax marks:50

[1] [a] Prove that the quantum relative entropy is convex, ie, if
$(\rho_{i},\sigma_{i}),i=1,2$ are pairs of mixed states in $\mathbb{C}^{d}$,
then

$D(p_{1}\rho_{1}+p_{2}\rho_{2}|p_{1}\sigma_{1}+p_{2}\sigma_{2})\leq
p_{1}D(\rho_{1}|\sigma_{1})+p_{2}D(\rho_{2}|\sigma_{2})$

for $p_{1},p_{2}\geq 0,p_{1}+p_{2}=1$.

hint: First prove using the following steps that if $K$ is a quantum
operation, ie, TPCP map, then for two states $\rho,\sigma$, we have

$D(K(\rho)|K(\sigma))\leq D(\rho|\sigma)---(a)$

Then define

$\rho=\left(\begin{array}[]{cc}p_{1}\rho_{1}&0\\\
0&p_{2}\rho_{2}\end{array}\right),$
$\sigma=\left(\begin{array}[]{cc}p_{1}\sigma_{1}&0\\\
0&p_{2}\sigma_{2}\end{array}\right),$

and $K$ to be the operation given by

$K(\rho)=E_{1}\rho.E_{1}^{*}+E_{2}\rho.E_{2}^{*}$

where

$E_{1}=[I|0],E_{2}=[0|I]\in\mathbb{C}^{d\times 2d}$

Note that

$E_{1}^{*}E_{1}+E_{2}^{*}E_{2}=I_{2d}$

To prove (a), make use of the result on the asymptotic error rate in
hypothesis testing on tensor product states: Given that the first kind of
error, ie, probability of detecting $\rho^{\otimes n}$ given that the true
state is $\sigma^{\otimes n}$ goes to zero, the minimum (Neyman-Pearson test)
probability of error of the second kind goes to zero at the rate of
$-D(\rho|\sigma)$ (this result is known as quantum Stein’s lemma). Now
consider any sequence of tests $T_{n}$, ie $0\leq T_{n}\leq I$ on
$(\mathbb{C}^{d})^{\otimes n}$. Then consider using this test for testing
between the pair of states $K^{\otimes n}(\rho^{\otimes n})$ and $K^{\otimes
n}(\sigma^{\otimes n})$. Observe that the optimum performance for this
sequential hypothesis testing problem is the same as that of the dual problem
of testing $\rho^{\otimes n}$ versus $\sigma^{\otimes n}$ using the dual test
$K^{*\otimes n}(T_{n})$. This is because

$Tr(K^{\otimes n}(\rho^{\otimes n})T_{n})=Tr(\rho^{\otimes n}K^{*\otimes
n}(T_{n}))$

etc. It follows that the optimum Neyman-Pearson error probability of the first
kind for testing $\rho^{\otimes n}$ versus $\sigma^{\otimes n}$ will be much
smaller than that of testing $K^{\otimes n}(\rho^{\otimes n})$ versus
$K^{\otimes n}(\sigma^{\otimes n})$ since $K^{*\otimes n}(T_{n})$ varies only
over a subset of tests as $T_{n}$ varies over all tests in
$(\mathbb{C}^{d})^{\otimes n}$. Hence going to the limit gives the result

$exp(-nD(\rho|\sigma))\leq exp(-nD(K(\rho)|K(\sigma))$

asymptotically as $n\rightarrow\infty$ which actually means that

$D(\rho|\sigma)\geq D(K(\rho)|K(\sigma))$

[b] Prove using [a] that the Von-Neumann entropy $H(\rho)=-Tr(\rho.log(\rho))$
is a concave function, ie if $\rho,\sigma$ are two states, then

$H(p_{1}\rho+p_{2}\sigma)\geq p_{1}H(\rho)+p_{2}H(\sigma),p_{1},p_{2}\geq
0,p_{1}+p_{2}=1$

hint: Let $K$ be any TPCP map. Let

$\rho=\left(\begin{array}[]{cc}p_{1}\rho_{1}&0\\\
0&p_{2}\rho_{2}\end{array}\right)\in\mathbb{C}^{2d\times 2d},$

and

$\sigma=I_{2d}/2d$

where $\rho_{k}\in\mathbb{C}^{d\times d},k=1,2$. Then choose

$E_{1}=[I|0],E_{2}=[0|I]\in\mathbb{C}^{d\times 2d}$

and

$K(X)=E_{1}XE_{1}^{*}+E_{2}XE_{2}^{*}\in\mathbb{C}^{d\times
d},X\in\mathbb{C}^{2d\times 2d}$

Note that

$E_{1}^{*}E_{1}+E_{2}^{*}E_{2}=I_{d}$

Show that

$K(\sigma)=K(I_{2d}/2d)=I_{d}/d$

Then we get using [a],

$D(K(\rho)|K(\sigma))\leq D(\rho|sigma)$

Show that this yields

$H(p_{1}\rho_{1}+p_{2}\rho_{2})+log(d)\geq
p_{1}H(\rho_{1})+p_{2}H(\rho_{2})-H(p)+log(2d)$

where

$H(p)=-p_{1}log(p_{1})-p_{2}log(p_{2})$

Show that

$H(p)\leq log(2)$

and hence deduce the result.

[2] [a] Let ${\bf X}[n],n\geq 0$ be a $M$-vector valued AR process, ie, it
satisfies the first order vector difference equation

${\bf X}[n]={\bf A}{\bf X}[n-1]+{\bf W}[n],n\geq 1$

Assume that ${\bf W}[n],N\geq 1$ is an iid $N({\bf 0},{\bf Q})$ process
independent of ${\bf X}[0]$ and that ${\bf X}[0]$ is a Gaussian random vector.
Calculate the value of

${\bf R}[0]=\mathbb{E}({\bf X}[0]{\bf X}[0]^{T})$

in terms of ${\bf A}$ and ${\bf Q}$ so that ${\bf X}[n],n\geq 0$ is wide-sense
stationary in the sense that

$\mathbb{E}({\bf X}[n]{\bf X}[n-m]^{T})={\bf R}[m]$

does not depend on $n$. In such a case, evaluate the ACF ${\bf R}[m]$ of
$\\{{\bf X}[n]\\}$ in terms of ${\bf A}$ and ${\bf Q}$. Assume that ${\bf A}$
is diagonable with all eigenvalues being strictly inside the unit circle.
Express your answer in terms of the eigenvalues and eigen-projections of ${\bf
A}$.

[b] Assuming ${\bf X}[0]$ known, evaluate the maximum likelihood estimator of
${\bf A}$ given the data $\\{{\bf X}[n]:1\leq n\leq N\\}$. Also evaluate the
approximate mean and covariance of the estimates of the matrix elements of
${\bf A}$ upto quadratic orders in the correlations $\\{{\bf R}[m]\\}$ based
on the data ${\bf X}[n],1\leq n\leq N$ and upto $O(1/N)$. In other words,
calculate using perturbation theory, $\mathbb{E}(\delta A_{ij})$ and
$\mathbb{E}(\delta A_{ij}\delta A_{km})$ upto $O(1/N)$ where

$\hat{A}-{\bf A}=\delta{\bf A}$

where $\hat{A}$ is the maximum likelihood estimator of ${\bf A}$.

hint: Write

$N^{-1}\sum_{n=1}^{N}{\bf X}[n-i]{\bf X}[n-j]^{T}-{\bf R}[j-i]=\delta R(j,i)$

and then evaluate $\delta{\bf A}$ upto quadratic orders in $\delta R(j,i)$.
Finally, use the fact that since ${\bf X}[n],n\geq 0$ is a Gaussian process,

$\mathbb{E}(X_{a}[n_{1}]X_{b}[n_{2}]X_{c}[n_{3}]X_{d}[n_{4}])=R_{ab}[n_{1}-n_{2}]R_{cd}[n_{3}-n_{4}]+R_{ac}[n_{1}-n_{3}]R_{bd}[n_{2}-n_{4}]+R_{ad}[n_{1}-n_{4}]R_{bc}[n_{2}-n_{3}]$

where

$((R_{ab}(\tau)))_{1\leq a,b\leq M}={\bf R}(\tau)=\mathbb{E}({\bf X}[n]{\bf
X}[n-\tau]^{T})$

[c] Calculate the Fisher information matrix for the parameters $\\{A_{ij}\\}$
and hence the corresponding Cramer-Rao lower bound.

[3] Let ${\bf X}[n]$ be a stationary $M$-vector valued zero mean Gaussian
process with ACF ${\bf R}(\tau)=\mathbb{E}({\bf X}[n]{\bf
X}[n-\tau]^{T})\in\mathbb{R}^{M\times M}$. Define its power spectral density
matrix by

${\bf S}(\omega)=\sum_{\tau=-\infty}^{\infty}{\bf
R}(\tau)exp(-j\omega\tau)\in\mathbb{C}^{M\times M}$

Prove that ${\bf S}(\omega)$ is a positive definite matrix. Define its
windowed periodogram estimate by

$\hat{S}_{N}(\omega)=\hat{X}_{N}(\omega)(\hat{X}_{N}(\omega))^{*}$

where

$\hat{X}_{N}(\omega)=N^{-1/2}\sum_{n=0}^{N-1}{\bf W}(n){\bf X}(n)exp(-j\omega
n)$

where ${\bf W}(n)$ is an $M\times M$ matrix valued window function. Calculate

$\mathbb{E}\hat{S}_{N}(\omega),Cov((S_{N}(\omega_{1}))_{ab},(S_{N}(\omega_{2}))_{cd})$

and discuss the limits of these as $N\rightarrow\infty$. Express these means
and covariances in terms of

$\hat{W}(\omega)=\sum_{\tau=-\infty}^{\infty}W(\tau)exp(-j\omega\tau)$

[4] This problem deals with causal Wiener filtering of vector valued
stationary processes. Let ${\bf X}(n),{\bf S}(n),n\in\mathbb{Z}$ be two
jointly WSS processes with correlations

${\bf R}_{XX}(\tau)=\mathbb{E}({\bf X}(n){\bf X}(n-\tau)^{T}),{\bf
R}_{SX}(\tau)=\mathbb{E}({\bf S}(n){\bf X}(n-\tau)^{T})$

Let

${\bf S}_{XX}(z)=\sum_{\tau}{\bf R}_{XX}(\tau)z^{-\tau}$

Prove that

${\bf R}_{XX}(-\tau)={\bf R}_{XX}(\tau)^{T}$

and hence

${\bf S}_{XX}(z^{-1})={\bf S}_{XX}(z)^{T}$

Assume that this power spectral density matrix can be factorized as

${\bf S}_{XX}(z)={\bf L}(z){\bf L}(z^{-1})^{T}$

where ${\bf L}(z)$ and ${\bf G}(z)={\bf L}(z)^{-1}$ admit expansions of the
form

${\bf L}(z)=\sum_{n\geq 0}{\bf l}[n]z^{-n},|z|\geq 1$ ${\bf G}(z)=\sum_{n\geq
0}{\bf g}[n]z^{-n},|z|\geq 1$

with

$\sum_{n\geq 0}\parallel{\bf l}[n]\parallel<\infty,$ $\sum_{n\geq
0}\parallel{\bf g}[n]\parallel<\infty$

Then suppose

${\bf H}(z)=\sum_{n\geq 0}{\bf h}[n]z^{-n}$

is the optimum causal matrix Wiener filter that takes ${\bf X}[.]$ as input
and outputs an estimate $\hat{S}[.]$ of ${\bf S}[.]$ in the sense that

$\hat{S}[n]=\sum_{k\geq 0}{\bf h}[k]{\bf X}[n-k]$

with $\\{{\bf h}[k]\\}$ chosen so that

$\mathbb{E}[\parallel{\bf S}[n]-\hat{S}[n]\parallel^{2}]$

is a minimum, then prove that ${\bf h}[.]$ satisfies the matrix Wiener-Hopf
equation

$\sum_{k\geq 0}{\bf h}[k]{\bf R}_{XX}[m-k]={\bf R}_{SX}[m],m\geq 0$

Show that this equation is equivalent to

$[{\bf H}(z){\bf S}_{XX}(z)-{\bf S}_{SX}(z)]_{+}=0$

Prove that the solution to this equation is given by

${\bf H}(z)=[{\bf S}_{SX}(z){\bf L}(z^{-1})^{-T}]_{+}{\bf L}(z)^{-1}$ $=[{\bf
S}_{SX}(z){\bf G}(z^{-1})^{T}]_{+}{\bf G}(z)$

where

$[{\bf S}_{SX}(z){\bf G}(z^{-1})^{T}]_{+}$ $=[\sum{\bf R}_{SX}[n]{\bf
g}[m]^{T}z^{-n+m}]_{+}=$ $\sum_{n\geq 0}z^{-n}\sum_{m\geq 0}{\bf
R}_{SX}[n+m]{\bf g}[m]^{T}$

Prove that this is a valid solution for a stable Causal matrix Wiener filter,
it is sufficient that

$\sum_{n\geq 0}\parallel{\bf R}_{SX}[n]\parallel<\infty$

[5] Write short notes on any two of the following:

[a] Recursive least squares lattice filter for prediction with time and order
updates of the prediction filter coefficients.

[b] Kolmogorov’s formula for the entropy rate of a stationary Gaussian process
in discrete time in terms of the power spectral density and its relationship
to the infinite order one step prediction error variance.

[c] The Cq Shannon coding theorem and its converse and a derivation of the
classical Shannon noisy coding theorem and its converse from it.

[d] Cramer-Rao lower bound for vector parameters and vector observations.

[e] The optimum detector in a binary quantum hypothesis testing problem
between two mixed states.

###  10.7 Monotonocity of quantum relative Renyi entropy

Let $A,B$ be positive definite matrices and let $0\leq t\leq 1$, then prove
that

$(tA+(1-t)B)^{-1}\leq t.A^{-1}+(1-t).B^{-1}---(1)$

ie, the map $A\rightarrow A^{-1}$ is operator convex on the space of positive
matrices.

Solution: Proving (1) is equivalent to proving that

$A^{1/2}(tA+(1-t)B)^{-1}.A^{1/2}\leq tI+(1-t)A^{1/2}B^{-1}A^{1/2}$

which is the same as proving that

$(tI+(1-t)X)^{-1}\leq tI+(1-t)X^{-1}---(2)$

where

$X=A^{-1/2}BA^{-1/2}$

It is enough to prove (2) for any positive matrix $X$. By the spectral theorem
for positive matrices, to prove (2), it is enough to prove

$(t+(1-t)x)^{-1}\leq t+(1-t)/x---(3)$

for all positive real numbers $x$. Proving (3) is equivalent to proving that

$(t+(1-t)x)(tx+1-t)\geq x$

for all $x>0$, or equivalently,

$(t^{2}+(1-t)^{2})x+t(1-t)x^{2}+t(1-t)\geq x$

or equivalently, adding and subtracting $2t(1-t)x$ on both sides to complete
the square,

$-2t(1-t)x+t(1-t)x^{2}+t(1-t)\geq 0$

or equivalently,

$1+x^{2}-2x\geq 0$

which is true.

Problem: For $p\in(0,1)$, show that $x\rightarrow-x^{p}$, $x\rightarrow
x^{p-1}$ and $x\rightarrow x^{p+1}$ are operator convex on the space of
positive matrices.

Solution: Consider the integral

$I(a,x)=\int_{0}^{\infty}u^{a-1}du/(u+x)$

It is clear that this integral is convergent when $x\geq 0$ and $0<a<1$. Now

$I(a,x)=x^{-1}\int_{0}^{\infty}u^{a-1}(1+u/x)^{-1}du$
$=x^{a-1}\int_{0}^{\infty}y^{a-1}(1+y)^{-1}dy$

on making the change of variables $u=xy$. Again make the change of variables

$1-1/(1+y)=y/(1+y)=z,y=1/(1-z)-1=z/(1-z)$

Then

$I(a,x)=x^{a-1}\int_{0}^{1}z^{a-1}(1-z)^{-a+1}(1-z)(1-z)^{-2}dz=x^{a-1}\int_{0}^{1}z^{a-1}(1-z)^{-a}dz$
$=x^{a-1}\beta(a,1-a)=x^{1-a}\Gamma(a)\Gamma(1-a)=x^{a-1}\pi/sin(\pi a)$

Thus,

$x^{a-1}=(sin(\pi a)/\pi)I(a,x)=(sin(\pi
a)/\pi)\int_{0}^{\infty}u^{a-1}du/(u+x)$

and since for $u>0$, as we just saw, $x\rightarrow(u+x)^{-1}$ is operator
convex, the proof that $f(x)=x^{a-1}$ is operator convex when $a\in(0,1)$.
Now, Now,

$u^{a-1}x/(u+x)=u^{a-1}(1-u/(u+x))=u^{a-1}-u^{a}/(u+x)$

Integrating from $u=0$ to $u=\infty$ and making use of the previous result
gives

$x^{a}=(sin(\pi a)/\pi)\int_{0}^{\infty}(u^{a-1}-u^{a}/(u+x))du$

from which it immediately becomes clear that $x\rightarrow-x^{a}$ is operator
convex on the space of positive matrices. Now consider $x^{a+1}$. We have

$u^{a-1}x^{2}/(u+x)=u^{a-1}((u+x)^{2}-u^{2}-2ux)/(u+x)=$
$u^{a-1}(u+x)-u^{a+1}/(u+x)-2u^{a}x/(u+x)=u^{a-1}(u+x)-u^{a+1}/(u+x)-2u^{a}(1-u/(u+x))$

and integration thus gives

$x^{a+1}=(sin(\pi a)/\pi)\int_{0}^{\infty}(u^{a-1}x+u^{a+1}/(u+x)-u^{a})du$

from which it becomes immediately clear that $x^{a+1}$ is operator convex.

Monotonicity of Renyi entropy: Let $s\in\mathbb{R}$. For two positive matrices
$A,B$, consider the function

$D_{s}(A|B)=log(Tr(A^{1-s}B^{s}))$

Our aim is to show that for $s\in(0,1)$ and a TPCP map $K$, we have for two
states $A,B$,

$D_{s}(K(A)|K(B))\geq D_{s}(A|B)$

while for $s\in(-1,0)$, we have

$D_{s}(K(A)|K(B))\leq D_{s}(A|B)$

For $a\in(0,1)$, and $x>0$ we’ve seen that

$x^{a+1}=sin(\pi a)/\pi)\int_{0}^{\infty}(u^{a-1}x+u^{a+1}/(u+x)-u^{a})du$

and hence for $s\in(-1,0)$, replacing $a$ by $-s$ gives

$x^{1-s}=-(sin(\pi s)/\pi)\int_{0}^{\infty}(u^{a-1}x+u^{a+1}/(u+x)-u^{a})du$

Let $s\in(-1,0)$. Then the above formula shows that $x^{1-s}$ is operator
convex. Hence, if $A,B$ are any two density matrices and $K$ a TPCP map, then

$K(A)^{1-s}\leq K(A^{1-s})$

whence

$B^{s/2}K(A)^{1-s}B^{s/2}\leq B^{s/2}K(A^{1-s})B^{s/2}$

for any positive $B$ and taking taking trace gives

$Tr(K(A)^{1-s}B^{s})\leq Tr(K(A^{1-s})B^{s})$

Replacing $B$ by $K(B)$ here gives

$Tr(K(A)^{1-s}K(B)^{s})\leq Tr(K(A^{1-s})K(B)^{s})$

Further since $s\in(-1,0)$ it follows that $x^{s}$ is operator convex. Hence

$K(B)^{s}\leq K(B^{s})$

and so

$A^{1/2}K(B)^{s}A^{1/2}\leq A^{1/2}K(B^{s})A^{1/2}$

for any positive $A$ and replacing $A$ by $K(A^{1-s})$ and then taking trace
gives

$Tr(K(A^{1-s})K(B)^{s})\leq Tr(K(A^{1-s})K(B^{s}))$

More generally, we have the following:

If $X$ is any matrix, and $s\in(0,1)$ and $A,B$ positive, and $K$ any TPCP map

$Tr(X^{*}K(A)^{1-s}XK(B)^{s})\geq Tr(X^{*}K(A)^{1-s}XK(B^{s}))$

(since $K(B)^{s}\geq K(B^{s})$ because $B\rightarrow B^{s}$ is operator
concave and since $K(A)^{1-s}X$ is positive)

$=Tr(K(A)^{1-s}XK(B^{s})X^{*})\geq Tr(K(A^{1-s})XK(B^{s})X^{*})$

(since $K(A)^{1-s}\geq K(A^{1-s})$ because $A\rightarrow A^{1-s}$ is operator
concave and since $XK(B^{s})X^{*}$ is positive)

$=Tr(X^{*}K(A^{1-s})XK(B^{s}))$

Since $X$ is an arbitary complex square matrix, this equation immediately can
be written as

$K(A)^{1-s}\otimes K(B)^{s}\geq K(A^{1-s})\otimes K(B^{s})$

for all positive $A,B$ and $s\in(0,1)$. We can express this equation as

$K(A)^{1-s}\otimes K(B)^{s}\geq(K\otimes K)(A^{1-s}\otimes B^{s})$

and hence if $I$ denotes the identity matrix of the same size as $A,B$, then

$Tr(K(A)^{1-s}K(B)^{s})\geq Tr((A^{1-s}\otimes B^{s})(K\otimes
K)(Vec(I).Vec(I)^{*}))$

Note that if $\\{e_{k}:k=1,2,...,n\\}$ is an onb for $\mathbb{C}^{n}$, the
Hilbert space on which $A,B$ act, then

$I=\sum_{k}|e_{k}><e_{k}|,Vec(I)=\sum_{k}|e_{k}\otimes e_{k}>=|\xi>$
$Vec(I).Vec(I)^{*}=|\xi><\xi|=\sum_{ij}E_{ij}\otimes E_{ij}$

where

$E_{ij}=|e_{i}><e_{j}|$

Thus, for $s\in(0,1)$,

$Tr(K(A)^{1-s}K(B)^{s})\geq\sum_{i,j}Tr(K_{ij}A^{1-s}K_{ji}B^{s})$

where

$K_{ij}=K(E_{ij})$

Likewise, if $s\in(0,1)$ so that $s-1\in(-1,0),2-s\in(1,2)$, then both the
maps $A\rightarrow A^{s-1}$ and $A\rightarrow A^{2-s}$ are operator convex and
hence for any TPCP map $K$ , $K(A)^{s-1}\leq K(A)^{s-1},K(A)^{2-s}\leq
K(A^{2-s})$, from which we immediately deduce that

$Tr(X^{*}K(A)^{2-s}XK(B)^{s-1})\leq Tr(X^{*}K(A^{2-s})XK(B^{s-1}))$

or equivalently,

$K(A)^{2-s}\otimes K(B)^{s-1}\leq K(A^{2-s})\otimes K(B^{s-1})$

which result in the inequality

$Tr(K(A)^{2-s}K(B)^{s-1})\leq\sum_{i,j}Tr(K_{ij}A^{2-s}K_{ji}B^{s-1})$

Remark: Lieb’s inequality states that for $s\in(0,1)$, the map

$(A,B)\rightarrow A^{1-t}\otimes B^{t}=f_{t}(A,B)$

is joint operator concave, ie, if $A_{1},A_{2},B_{1},B_{2}$ are positive and
$p,q>0,p+q=1$ then with

$A=pA_{1}+qA_{2},B=pB_{1}+qB_{2}$

we have

$f_{t}(A,B)\geq pf_{t}(A_{1},B_{1})+qf_{t}(A_{2},B_{2})$

From this, we get the result that

$(A,B)\rightarrow Tr(X^{*}A^{1-t}XB^{t})=H_{t}(A,B|X)$

is jointly concave. Now let $K$ denote the pinching

$K(A)=(A+UAU^{*})/2$

where $U$ is a unitary matrix. Then, we get

$H_{t}(K(A),K(B)|X)\geq(1/2)H_{t}(A,B|X)+(1/2)H_{t}(UAU^{*},UBU^{*}|X)$

and replacing $X$ by $I$ in this inequality gives

$Tr(K(A)^{1-t}K(B)^{t})\geq Tr(A^{1-t}B^{t})$

which proves the monotonicity of the Renyi entropy when $t\in(0,1)$ under
pinching operations $K$ of the above kind. Now, this inequality will also hold
for any pinching $K$ since any pinching can be expressed as a composition of
such elementary pinchings of the above kind. We now also wish to show the
reverse inequality when $t\in(-1,0)$, ie show that if $t\in(0,1)$, then

$Tr(K(A)^{2-t}K(B)^{t-1})\leq Tr(A^{2-t}B^{t-1})$

##  Chapter 11 My commentary on the Ashtavakra Gita

A couple of years ago, my colleague and friend presented me with a verse to
verse translation of the celebrated Ashtavakra Gita in English along with a
commentary on each verse. I was struck by the grandeur and the simplicity of
the Sanskrit used by Ashtavakra for explaining to King Janaka, the true
meaning of self-realization, how to attain it immediately without any
difficulty and also King Janaka’s reply to Ashtavakra about how he has
attained immeasurable joy and peace by understanding this profound truth.
Ashtavakra was a sage who lived during the time in which the incidents of the
Raamayana took place and hence his Gita dates much much before even the
Bhagavad Gita spoken by Lord Krisnna to Arjuna was written down by Sage Vyasa
through Ganapati the scribe. One can see in fact many of the slokas in the
Bhagavad Gita have been adapted from the Ashtavakra Gita and hence I feel that
it is imperative for any seeker of the absolute truth and everlasting peace,
joy and tranquility to read the Ashtavakra Gita, so relevant in our modern
times where there are so many disturbances. Unlike the Bhagavad Gita,
Ashtavkra does not bring in the concept of a personal God to explain true
knowledge and the process of self-realization. He simply says that you attain
self-realization immediately at this moment when you understand that the
physical universe that you see around you is illusory and the fact that your
soul is the entire universe. Every living being’s soul is the same and it
fills the entire universe and that all the physical events that take place
within our universe as perceived by our senses are false, ie, they do not
occur. In the language of quantum mechanics, these events occur only because
of the presence of the observer, namely the living individuals with their
senses. The absolute truth is that each individual is a soul and not a body
with his/her senses and this soul neither acts nor causes to act. Knowledge is
obtained immediately at the moment when we understand the truth that the
impressions we have from our physical experiences are illusory, ie false and
that the truth is that our soul fills the entire universe and that this soul
is indestructible and imperishable as even the Bhagavad Gita states. The
Bhagavad Gita however distinguishes between two kinds of souls, the jivatma
and the paramatma, the former belonging to the living entities in our universe
and the latter to the supreme personality of Godhead. The Bhagavad Gita tells
us that within each individual entity, both the jivatma and the paramatma
reside and when the individual becomes arrogant, he believes that his jivatma
is superior to the paramatma and this then results in self conflict owing to
which the individual suffers. By serving the paramatma with devotion as
outlined in the chapter on Bhakti yoga, the jivatma is able to befriend the
paramatma and then the individual lives in peace and tranquility. To explain
self-knowledge more clearlh without bringing in the notion of a God,
Ashtavakra tells Raja Janaka that there are two persons who are doing exactly
the same kind of work every day both in their offices as well as in their
family home. However, one of them is very happy and tranquil while the other
is depressed and is always suffering. What is the reason for this ? The
reason, Ashtavakra says is that the second person believes that his self is
doing these works and consequently the reactions generated by his works affect
him and make him either very happy or very sad. He is confused and does not
know how to escape from this knot of physical experiences. The first person,
on the other hand, knows very well that he is compelled to act in the way he
is acting owing to effects of his past life’s karma’s and that in fact his
soul is pure and has nothing at all to do with the results and effects of his
actions. He therefore neither rejoices nor does he get depressed. The physical
experiences that he goes through do not leave any imprints on his self because
he knows that the physical experiences are illusory and that his soul which
pure and without any blemish fills the entire universe and is the truth.
Ashtavarka further clarifies his point by saying that the person in ignorance
will practice meditation and try to cultivate detachment from the objects of
this world thinking that this will give him peace. Momentarily such a person
will get joy but when he comes of the meditation, immediately all the desires
of this illusory physical world will come back to him causing even more
misery. He will try to run to a secluded place like a forest and meditate to
forget about his experiences, but again when his meditation is over, misery
will come back to him. However, once he realizes that he is actually not his
body but his soul and that the soul is free of any physical quality and that
it is the immeasurable truth, then he will neither try forcefully to cultivate
detachment nor will he try forcefully to acquire possessions. Both the
extremes of attachment and detachment will not affect him. In short, he
attains peace even living in this world immediately at this moment when true
knowledge comes to him. According to Ashtavkara, the seeker of attachement as
well as the seeker of detachment are both unhappy because they have not
realized that the mind must not be forced to perform an activity. Knowledge is
already present within the individual, he just has to remain in his worldly
position as it is, performing his duties with the full knowledge that this
compulsion is due to his past karmas and then simultaneously be conscious of
the fact that his soul is the perfect Brahman and it pervades the entire
universe. At one point, Ashtavakra tells Janaka that let even Hari (Vishnu) or
Hara (Shiva) be your guru, even then you cannot be happy unless you have
realized at this moment the falsity of the physical world around you and the
reality of your soul as that which pervades the entire universe. Such a person
who has understood this, will according to Ashtavakra, be unmoved if either if
death in some form approaches him or if temptations of various kinds appear
before him. He will remain neutral to both of these extremes. Such a realized
soul has eliminated all the thought processes that occur within his mind. When
true knowledge dawns upon a person, his mind will become empty and like a dry
leaf in a strong wind, he will appear to be blown from one place to other
doing different kinds of karma but always with the full understanding that
these karmas have nothing at all to do with his soul which is ever at bliss,
imperturbable, imperishable and omnisicient. In the quantum theory of matter
and fields, we learn about how the observer perturbs a system in order to
measure it and gain knowledge about the system. When the system is thus
perturbed, one measures not the true system but only the perturbed system.
This is the celebrated Heisenberg uncertainty principle. Likewise, in
Ashtavakra’s philosophy, if we attempt to gain true knowledge about the
universe and its living entities from our sensory experience, we are in effect
perturbing the universe and hence we would gain only perturbed knowledge, not
true knowledge. True knowledge comes only with the instant realization that we
are not these bodies, we are all permanent souls. Even from a scientific
standpoint, as we grow older, all our body cells eventually get replaced by
new ones, then what it is that in our new body after ten years that we are
able to retain our identity ? What scientific theory of the DNA or RNA can
explain this fact ? What is the difference between the chemical composition of
our body just before it dies and just after it dies ? Absolutely no
difference. Then what is it that disinguishes a dead body from a living body ?
That something which distinguishes the two must be the soul which is not
composed of any chemical. It must be an energy, the life force which our
modern science is incapable of explaining.

##  Chapter 12 LDP-UKF based controller

State equations:

$x(n+1)=f(x(n),u(n))+w(n+1)$

Measurement equations:

$z(n)=h(x(n),u(n))+v(n)$

$w,v$ are independent white Gaussian processes with mean zero and covariances
$Q,R$ respectively. $x_{d}(n)$ is the desired process to be tracked and it
satisfies

$x_{d}(n+1)=f(x_{d}(n),u(n))$

UKF equations: Let $\\{\xi(k):1\leq k\leq K\\}$ and $\\{\eta(k):1\leq k\leq
K\\}$ be independent iid $N(0,I)$ random vectors. The UKF equations are

$\hat{x}(n+1|n)=K^{-1}\sum_{k=1}^{K}f(\hat{x}(n|n)+\sqrt{P(n|n)}\xi(k),u(n))$
$\hat{x}(n+1|n+1)=\hat{x}(n+1|n)+K(n)(z(n+1)-\hat{z}(n+1|n))$

where

$K(n+1)=\Sigma_{XY}\Sigma_{YY}^{-1}$

Where

$\Sigma_{XY}=K^{-1}\sum_{k=1}^{K}[\sqrt{P(n+1|n)}\xi(k)(h(\hat{x}(n+1|n)+\sqrt{P(n+1|n)}\eta(k))-\hat{z}(n+1|n))^{T}],$
$\hat{z}(n+1|n)=K^{-1}\sum_{k=1}^{K}h(\hat{x}(n+1|n)+\sqrt{P(n+1|n)}\eta(k),u(n+1))$
$\Sigma_{YY}=K^{-1}.\sum_{k=1}^{K}(h(\hat{x}(n+1|n)+\sqrt{P(n+1|n)}\eta(k))(,,)^{T}$
$P(n+1|n)=K^{-1}\sum_{k=1}^{K}[(f(\hat{x}(n|n)+\sqrt{P(n|n)}\xi(k),u(n))-\hat{x}(n+1|n))(,,)^{T}]+Q$
$P(n+1|n+1)=\Sigma_{XX}-\Sigma_{XY}.\Sigma_{YY}^{-1}\Sigma_{XY}^{T}$

where

$\Sigma_{XX}=P(n+1|n)$

Tracking error feedback controller: Let

$f(n)=x_{d}(n)-\hat{x}(n|n),f(n+1|n)=x_{d}(n+1)-\hat{x}(n+1|n)$

Controlled dynamics:

$x(n+1)=f(x(n),u(n))+w(n+1)+K_{c}f(n)$

Aim: To choose the controller coefficient $K_{c}$ so that the probability

$P(max_{1\leq n\leq N}(|e(n)|^{2}+|f(n)|^{2})>\epsilon)$

is minimized where

$e(n)=x(n)-\hat{x}(n),e(n+1|n)=x(n+1)-\hat{x}(n+1|n),\hat{x}(n)=\hat{x}(n|n)$

Linearized difference equations for $e(n),f(n)$: (Approximate)

$e(n+1|n)=x(n+1)-\hat{x}(n+1|n)=f(x(n),u(n))+w(n+1)+K_{c}f(n+1|n)-K^{-1}.\sum_{k}f(\hat{x}(n)+\sqrt{P(n|n)}\xi(k),u(n))$
$=f(x(n),u(n))-f(\hat{x}(n),u(n))-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)+w(n+1)+K_{c}f(n)$
$=f^{\prime}(\hat{x}(n),u(n))e(n)-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)+w(n+1)+K_{c}f(n)---(1)$
$f(n+1|n)=x_{d}(n+1)-\hat{x}(n+1|n)=f_{(}x_{d}(n),u(n))-K^{-1}.\sum_{k}f(\hat{x}(n)+\sqrt{P(n|n)}\xi(k),u(n))$
$=f^{\prime}(\hat{x}(n),u(n))f(n)-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)---(2)$
$e(n+1)=x(n+1)-\hat{x}(n+1|n+1)=$
$e(n+1|n)+K_{c}f(n+1|n)-K(n)(z(n+1)-\hat{z}(n+1|n))$
$=e(n+1|n)+K_{c}f(n+1|n)-K(n)(h(x(n+1),u(n+1))+v(n+1)-K^{-1}\sum_{k}h(\hat{x}(n+1|n)+\sqrt{P(n+1|n)}\eta(k),u(n+1)))$
$=e(n+1|n)+K_{c}f(n+1|n)-K(n)(h^{\prime}(\hat{x}(n+1|n),u(n+1))e(n+1|n)+v(n+1)$
$-h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}.K^{-1}.\sum_{k}\eta(k))$
$=(I-K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)))e(n+1|n)+K_{c}f(n)-K(n)v(n+1)$
$+K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}K^{-1}\sum_{k}\eta(k)---(3)$

Finally,

$f(n+1)=x_{d}(n+1)-\hat{x}(n+1|n+1)=$ $f(n+1|n)-K(n)(z(n+1)-\hat{z}(n+1|n))$
$=f(n+1|n)-K(n)(h(x(n+1),u(n+1))+v(n+1)-K^{-1}\sum_{k}h(\hat{x}(n+1|n)+\sqrt{P(n+1|n)}\eta(k),u(n+1))$
$=f(n+1|n)-K(n)(h^{\prime}(\hat{x}(n+1|n),u(n+1))e(n+1|n)$
$-h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}K^{-1}\sum_{k}\eta(k))-K(n)v(n+1)$
$=f(n+1|n)-K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)))e(n+1|n)$
$+K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}.K^{-1}.\sum_{k}\eta(k)-K(n)v(n+1)---(4)$

Substituting for $e(n+1|n)$ and $f(n+1|n)$ from (1) and (2) respectively into
(3) and (4) results in the following recursion for $e(n),f(n)$:

$e(n+1)=$
$f^{\prime}(\hat{x}(n),u(n))f(n)-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)-K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)))(f^{\prime}(\hat{x}(n),u(n))e(n)$
$-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)+w(n+1)+K_{c}f(n))$
$+K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}.K^{-1}.\sum_{k}\eta(k)-K(n)v(n+1)---(5)$

and

$f(n+1)=$
$f^{\prime}(\hat{x}(n),u(n))f(n)-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)$
$-K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1))).$
$.(f^{\prime}(\hat{x}(n),u(n))e(n)-f^{\prime}(\hat{x}(n),u(n))\sqrt{P(n|n)}K^{-1}\sum_{k}\xi(k)+w(n+1)+K_{c}f(n))$
$+K(n)h^{\prime}(\hat{x}(n+1|n),u(n+1)).\sqrt{P(n+1|n)}.K^{-1}.\sum_{k}\eta(k)-K(n)v(n+1)---(6)$

If we regard the error processes $e(n),f(n)$ and the noise processes
w(n),v(n), $\xi_{1}(n)=K^{-1}.\sum_{k=1}^{K}\xi(k)$ and
$\eta_{1}(n)=K^{-1}\sum_{k}\eta(k)$ to be all of the first order of smallness
in perturbation theory, then to the same order, we can replace in (4) and (5),
$\hat{x}(n)$ appearing on the rhs by $x_{d}(n)$ and then the error thus
incurred will be only of the second order of smallness. Doing so, we can
express (4) and (5) in the form after equilibrium has been obtained as

$e(n+1)=A_{11}e(n)+(A_{12}+A_{18}K_{c})f(n)+A_{13}w(n+1)+A_{14}A_{15}v(n+1)+A_{16}\xi_{1}(n+1)+A_{17}\eta_{1}(n+1)$
$f(n+1)=A_{21}e(n)+(A_{22}+A_{28}K_{c})f(n)+A_{23}w(n+1)+A_{24}A_{25}v(n+1)+A_{26}\xi_{1}(n+1)+A_{27}\eta_{1}(n+1)$

Define the aggregate noise process

$W[n]=[w(n)^{T},v(n)^{T},\xi_{1}(n)^{T},\eta_{1}(n)^{T}]^{T}$

It is a zero mean vector valued white Gaussian process and its covariance
matrix is block diagonal:

$\mathbb{E}(W(n)W(m)^{T})=R_{W}(n)\delta(n-m)$

where

$R_{W}(n)=\mathbb{E}(W(n)W(n)^{T})=diag[Q,R,P[n|n)/K,P(n+1|n)/K]$

Let $P$ denote the limiting equilibrium value of $P[n|n]$ as
$n\rightarrow\infty$. The limiting equilibrium value of $P[n+1|n]$ is then
given approximately by

$P_{+}=f^{\prime}(\hat{x},u)Pf^{\prime}(\hat{x},u)^{T}+Q$

where $\hat{x}$ may be replaced by the limiting value of $x_{d}$, assuming it
to be asymptotically d.c and $f^{\prime}(x,u)$ is the Jacobian matrix of
$f(x,u)$ w.r.t $x$. We write $f^{\prime}(x_{d},u)=F$ and then

$P_{+}=FPF^{T}+Q$

If we desire to use a more accurate value of $P_{+}$, then we must replace it
by

$P_{+}=Cov(f(x_{d}+P.\xi_{1}))+Q$

where $\xi_{1}$ is an $N(0,K^{-1}P)$ random vector. Now, let $R_{W}$ denote
the limiting equilibrium value of $R_{W}(n)$, ie,

$R_{W}=diag[Q,R,P/K,P_{+}/K]$

Then, the above equations for

$\chi[n]=[e[n]^{T},f[n]^{T}]^{T}$

can be expressed as

$\chi[n+1]=(G_{0}+G_{1}K_{c}G_{2})\chi[n]+G_{3}W[n+1]$

and its solution is given by

$\chi[n]=\sum_{k=0}^{n-1}(G_{0}+G_{1}K_{c}G_{2})^{n-1-k}G_{3}W[k+1]$

This process is zero mean Gaussian with an autocorrelation given by

$\mathbb{E}(\chi[n]\chi[m]^{T})=\sum_{k=0}^{min(n,m)-1}(G_{0}+G_{1}K_{c}G_{2})^{n-1-k}.R_{W}.(G_{0}+G_{1}K_{c}G_{2})^{Tm-1-k}$
$=R_{\chi}[n,m|K_{c}]$

say. The LDP rate function of the process $\chi[.]$ over the time interval
$[0,N]$ is then given by

$I[\chi[n]:0\leq n\leq
N]=(1/2)\sum_{n,m=0}^{N}\chi[n]^{T}Q_{\chi}[n,m|K_{c}]\chi[m]$

where

$Q_{\chi}=((R_{\chi}[n,m|K_{c}]))_{0\leq n,m\leq N}^{-1}$

According to the LDP for Gaussian processes, the probability that
$\sum_{n=0}^{N}\parallel\chi[n]\parallel^{2}$ will exceed a threshold $\delta$
after we scale $\chi$ by a small number $\sqrt{\epsilon}|$ is given
approximately by

$exp(-\epsilon^{-1}min(I(\chi):\sum_{n=0}^{N}\parallel\chi[n]\parallel^{2}>\delta))$

and in order to minimize this deviation probability, we must choose $K_{c}$ so
that the maximum eigenvalue $\lambda_{max}(K_{c})$ of
$((R_{\chi}[n,m|K_{c}]))_{0\leq n,m\leq N}^{-1}$ is a minimum.

##  Chapter 13 Abstracts for a book on quantum signal processing using quantum
field theory

Chapter 1

An introduction to quantum electrodynamics

[1] The Maxwell and Dirac equations.

[2] Quantization of the electromagnetic field using Bosonic creation and
annihilation operators.

[3] Second Quantization of the Dirac field using Fermionic creation and
annihilation operators.

[4] The interaction term between the Dirac field and the Maxwell field.

[5] Propagators for the photons and electron-positron fields.

[7] The Dyson series and its application to calculation of amplitudes for
absorption, emission and scattering processes between electrons and positrons.

[8] Representing the terms in the Dyson series for interactions between
photons, electrons and positrons using Feynman diagrams.

[9] The role of propagators in Feynman diagrammatic calculations.

[10] Quantum electrodynamics using the Feynman path integral method.

[11] Calculating the propagator of fields using Feynman path integrals.

[12] Electrons self energy, vacuum polarization, Compton scattering and
anomalous magnetic moment of the electron.

Chapter 2

The design of quantum gates using quantum field theory

[1] Design of quantum gates using first quantized quantum mechanics by
controlled modulation of potentials.

[2] Design of quantum gates using harmonic oscillators with charges perturbed
by and electric field.

[3] Design of quantum gates using 3-D harmonic oscillators perturbed by
control electric and magnetic field.

[4] Representing the Klein-Gordon Hamiltonian with perturbation using
truncated spatial Fourier modes within a box.

[5] Design of the perturbation potential in Klein-Gordon quantization for
realizing very large sized quantum gates.

[6] Representing the Hamiltonian of the Dirac field plus electromagnetic field
in terms of photon and electron-positron creation and annihilation operators.

[7] Design of control classical electromagnetic fields and control classical
current sources in quantum electrodynamics for approximating very large sized
quantum unitary gates.

[8] The statistical performance of quantum gate design methods in the presence
of classical noise.

Chapter 3

The quantum Boltzmann equation with applications to the design of quantum
noisy gates, ie, TPCP channels.

[1] The classical Boltzmann equation for a plasma of charged particles
interacting with an external electromagnetic field.

[2] Calculating the conductivity and permeability of a plasma interacting with
an electromagnetic field using first order perturbation theory.

[3] The quantum Boltzmann equation for the one particle density operator for a
system of $N$ identically charged particles interacting with a classical
electromagnetic field. The effect of quantum charge and current densities on
the dynamics of the Maxwell field and the effect of the Maxwell field on the
dynamics of the one particle density operator.

[4] The quantum nonlinear channel generated by the evolving the one particle
density operator under an external field in the quantum Boltzmann equation.

Chapter 4

The Hudson-Parthasarathy quantum stochastic calculus with application to
quantum gate and quantum TPCP channel design.

[1] Boson and Fermion Fock spaces.

[2] Creation, annihilation and conservation operator fields in Boson and
Fermion Fock spaces.

[3] Creation, annihilation and conservation processes in Boson Fock space.

[4] The Hudson-Parthasarathy quantum Ito formulae.

[5] Hudson-Parthasarathy-Schrodinger equation for describing the evolution of
a quantum system in the presence of bath noise.

[6] The GKSL (Lindlbad) master equation for open quantum systems.

[7] Design of the Lindlbad operators for an open quantum system for generation
of a given TPCP map, ie, design of noisy quantum channels.

[8] Schrodinger’s equation with Lindlbad noisy operators in the presence of an
external electromagnetic field. Design of the electromagnetic field so that
the evolution matches a given TPCP map in the least squares sense.

[9] Quantum Boltzmann equation in the presence of an external electromagnetic
field with Lindblad noise terms. Design of the Lindblad operators and the
electromagnetic field so that the evolution matches a given nonlinear TPCP
map.

Chapter 5

Non-Abelian matter and gauge fields applied to the design of quantum gates.

[1] Lie algebra of a Lie group.

[2] Gauge covariant derivatives.

[3] The non-Abelian matter and gauge field Lagrangian that is invariant under
local gauge transformations.

[4] Application of non-Abelian matter and gauge field theory to the design of
large sized quantum gates in the first quantized formalism, gauge fields are
assumed to be classical and the control fields and current are also assumed to
be classical.

Chapter 6

The performance of quantum gates in the presence of quantum noise.

Quantum plasma physics: For an $N$ particle system with identitcal particles

$i\rho_{,t}=[\sum_{k}H_{k},\rho]+[\sum_{i<j}V_{ij},\rho]+Lindlbadnoiseterms.$

where

$H_{k}=-(\nabla_{k}-ieA(t,r_{k}))^{2}/2m+e\Phi(t,r_{k})$ $=H_{0k}+V_{k}$

where

$H_{0k}=-\nabla_{k}^{2}/2m,V_{k}=(ie/2m)(2.divA_{k}+(A_{k},\nabla_{k}))+e^{2}A_{k}^{2}/2m+e\Phi_{k}$

Approximate one particle Boltzmann equation obtained by partial tracing over
the remaining $N$ particles.

$i\rho_{1,t}=[H_{1},\rho_{1}]+(N-1)Tr_{2}[V_{12},\rho_{12}]+Lindbladnoiseterms$
$i\rho_{12,t}=[H_{1}+H_{2}+V_{12},\rho_{12}]+(N-2)Tr_{3}[V_{13}+V_{23},\rho_{123}]+Lindbladnoiseterms$

Writing

$\rho_{12}=\rho_{1}\otimes\rho_{1}+g_{12}$

we get from the above two equations,

$i(\rho_{1,t}\otimes\rho_{1}+\rho_{1}\otimes\rho_{1,t})+ig_{12,t}=$
$[H_{1},\rho_{1}]\otimes\rho_{1}+\rho_{1}\otimes[H_{1},\rho_{1}]+ig_{12,t}=$
$[H_{1},\rho_{1}]\otimes\rho_{1}+\rho_{1}\otimes[H_{1},\rho_{1}]+[H_{1}+H_{2}+V_{12},g_{12}]$
$+Lindbladnoise$

Here, we have neglected the cubic terms, ie, the terms involving $\rho_{123}$.
After making the appropriate cancellations, we get

$ig_{12,t}=[H_{1}+H_{2}+V_{12},g_{12}]+Lindbladnoise$

Thus we get for the one particle density,

$i\rho_{1,t}=[H_{1},\rho_{1}]+(N-1)Tr_{2}[V_{12},\rho_{1}\otimes\rho_{1}]+(N-1)Tr_{2}[V_{12},g_{12}]$
$+Lindbladnoise$

We can express this approximate equation as

$i\rho_{1,t}=[H_{1},\rho_{1}]+(N-1)Tr_{2}[V_{12},\rho_{1}\otimes\rho_{1}]$
$+(N-1)Tr_{2}[V_{12},exp(-itad(H_{1}+H_{2}+V_{12}))(g_{12}(0))]$

Since the nonlinear term is small, we can approximate the $\rho_{1}(t)$
occurring in it by $exp(-itad(H_{1}))(\rho_{1}(0))$. Thus,

$i\rho_{1,t}(t)=[H_{1},\rho_{1}(t)]+(N-1)Tr_{2}[V_{12},exp(-it.ad(H_{1}+H_{2}))(\rho_{1}(0)\otimes\rho_{1}(0))]$
$+(N-1)Tr_{2}[V_{12},exp(-itad(H_{1}+H_{2}+V_{12}))(g_{12}(0))]+\theta(\rho_{1}(t))$

where $\theta(\rho_{1})$ is the Lindblad noise term. The solution is

$\rho_{1}(t)=exp(-itad(H_{1}))(\rho_{1}(0))+$
$(N-1)\int_{0}^{t}exp(-i(t-s)ad(H_{1}))(Tr_{2}[V_{12},exp(-is.ad(H_{1}+H_{2}))(\rho_{1}(0)\otimes\rho_{1}(0))])ds$
$+(N-1)\int_{0}^{t}exp(-i(t-s)ad(H_{1}))(Tr_{2}[V_{12},exp(-isad(H_{1}+H_{2}+V_{12}))(g_{12}(0))])ds$
$+Lindbladnoiseterms.$

The first term is the initial condition term, the second term is the collision
term and the third term is the scattering term. Note that taking the trace
gives us

$i\partial_{t}(Tr(\rho_{1}(t))=0$

which means that the above quantum Boltzmann equation as it evolves the one
particle state, determines a nonlinear trace preserving mapping. This suggests
that the quantum Boltzmann equation can be used to design nonlinear TPCP maps
by controlling the electromagnetic potentials $A,\Phi$ as well as the Lindblad
terms.

Now let $\rho_{1}(t,r,r^{\prime})$ denote the position space representation of
the one particle density operator. Then the charge density corresponding to
this operator is given by $e\rho_{1}(t,r,r)$ and the current density is

$J(t,r)=(e/m)\nabla_{r^{\prime}}Im(\rho(t,r,r^{\prime}))|_{r^{\prime}=r}+(e/m)A(t,r)\rho_{1}(t,r,r)$

This is based on the fact that if $\psi(t,r)$ denotes a pure state in the
position representation, ie, a wave function, then the mixed state is an
average of $\psi(t,r)\psi(t,r^{\prime})^{*}$ over ensembles of such pure
states, ie,

$\rho_{1}(t,r,r^{\prime})=<\psi(t,r)\psi(t,r^{\prime})^{*}>$

In particular, the charge density is

$e<|\psi(t,r)|^{2}>=e\rho_{1}(t,r,r)$

Likewise the current density for a pure state is

$J(t,r)=(ie/2m)(\psi(t,r)^{*}(\nabla-
ieA(t,r))\psi(t,r)-\psi(t,r)(\nabla+ieA(t,r))\psi(t,r)^{*})$
$=(e/m)A(t,r)|\psi(t,r)|^{2}-(e/m)Im(\psi(t,r)^{*}\nabla\psi(t,r))$

which on taking average gives

$J(t,r)=(e/m)A(t,r)\rho_{1}(t,r,r)-(e/m)\nabla_{r^{\prime}}Im(\rho_{1}(t,r^{\prime},r)|_{r^{\prime}=r})$
$=(e/m)A(t,r)\rho_{1}(t,r,r)+(e/m)\nabla_{r^{\prime}}Im(\rho_{1}(t,^{\prime}r,r)|_{r^{\prime}=r})$

The classical em field then satisfies the Maxwell equations

$\square
A(t,r)=\mu_{0}J(t,r)=\mu_{0}((e/m)A(t,r)\rho_{1}(t,r,r)-(e/m)\nabla_{r^{\prime}}Im(\rho_{1}(t,r^{\prime},r)|_{r^{\prime}=r}))$
$\square\Phi(t,r)=\rho(t,r)/\epsilon_{0}=\epsilon_{0}^{-1}(e\rho_{1}(t,r,r))$

where

$\square=(c^{-2}\partial_{t}^{2}-\nabla^{2})$

is the wave operator. The complete system of the quantum Boltzmann or rather
quantum Vlasov (quantum for the density operator and classical for the em
field) is therefore summarized below:

$H_{1}=-(\nabla_{1}+ieA(t,r_{1}))^{2}/2m+e\Phi(t,r_{1})$
$V_{12}=V(|r_{1}-r_{2}|)$ $\rho_{1}(t)=exp(-itad(H_{1}))(\rho_{1}(0))+$
$(N-1)\int_{0}^{t}exp(-i(t-s)ad(H_{1}))(Tr_{2}[V_{12},exp(-is.ad(H_{1}+H_{2}))(\rho_{1}(0)\otimes\rho_{1}(0))])ds$
$+(N-1)\int_{0}^{t}exp(-i(t-s)ad(H_{1}))(Tr_{2}[V_{12},exp(-isad(H_{1}+H_{2}+V_{12}))(g_{12}(0))])ds$
$+Lindbladnoiseterms.$ $\square
A(t,r)=\mu_{0}((e/m)A(t,r)\rho_{1}(t,r,r)-(e/m)\nabla_{r^{\prime}}Im(\rho_{1}(t,r^{\prime},r)|_{r^{\prime}=r}))$
$\square\Phi(t,r)=\epsilon_{0}^{-1}(e\rho_{1}(t,r,r))$

Second quantization of the quantum plasma equations.

Consider the Dirac field of electrons and positrons. The wave function
$\psi(t,r)$ is a Fermionic field operator satisfying the Dirac equation

$[\gamma^{\mu}(i\partial_{\mu}+eA_{\mu})-m]\psi(t,r)=0$

The unperturbed field $\psi_{0}(t,r)$, ie, when $A_{\mu}=0$ can be expanded as
a superposition of annihilation and creation operator fields in momentum space
of electrons and positrons. This is the case when we second quantize the
motion of a single electron-positron pair. When however we second quantize the
motion of $N$ identical electron-positron pairs and then form the marginal
density operator of a single electron-positron pair, we obtain the quantum
Boltzmann equation for the one particle density operator field
$\rho_{1}(t,r^{\prime},r^{\prime})$ which in the absence of mutual
interactions $V(r_{1}-r_{2})$, is simply
$\psi_{0}(t,r)\psi_{0}(t,r^{\prime})^{*}$ which is expressible as a quadratic
combination of Fermionic creation and annihilation operator fields in momentum
space. The single particle Dirac Hamiltonian is

$H_{D}=(\alpha,-i\nabla+eA)+\beta m-e\Phi$

which in the absence of electromagnetic interactions reads

$H_{D0}=(\alpha,-i\nabla)+\beta m$

The second quantized density operator field in position space representation
that corresponds to this unperturbed Dirac Hamiltonian is given by

$\rho_{0}(t,r,r^{\prime})=\psi_{0}(t,r)\psi_{0}(t,r^{\prime})$

where

$\psi_{0}(t,r)=\int[a(P,\sigma)u(P,\sigma)exp(-i(E(P)t-P.r))+b(P,\sigma)^{*}v(P,\sigma)exp(i(E(P)t-P.r))]d^{3}P$

with $u(P,\sigma)$ and $v(-P,\sigma)$ being eigenvectors for
$H_{D0}(P)=(\alpha,P)+\beta m$ with eigenvalues $\pm E(P)$ respectively where
$E(P)=\sqrt{P^{2}+m^{2}}$. Thus, we can write after discretization in the
momentum domain,

$\rho_{100}(t,r,r^{\prime})=\sum_{k,m}[a(k)a(m)f_{1}(t,r,r^{\prime}|k,m)+a(k)a(m)^{*}f_{2}(t,r,r^{\prime})]+H.c$

where $a(k),a(k)^{*}$ are respectively Fermionic annihilation and creation
operators satisfying the canonical anticommutation relations

$[a(k),a(m)^{*}]_{+}=\delta(k,m),[a(k),a(m)]_{+}=0=[a(k)^{*},a(m)^{*}]_{+}$

After interaction with the quantum em field, the density operator
$\rho_{10}(t,r,r^{\prime})$ satisfies the equation

$i\partial_{t}\rho_{10}(t)=[H_{D0}+H_{int}(t),\rho_{10}(t)]$

where

$H_{int}(t)=e(\alpha,A(t,r))-e\Phi(t,r)$

where $A(t,r),\Phi(t,r)$ are now both Bosonic quantum operator fields
expressible upto zeroth order approximations as linear combinations of Bosonic
annihilation and creation operators $c(k),c(k)^{*},k=1,2,...$ that satisfy the
canonical commutation relations:

$[c(k),c(m)^{*}]=\delta(k,m),[c(k),c(m)]=[c(k)^{*},c(m)^{*}]=0$

We have upto first order perturbation theory,

$\rho_{10}=\rho_{100}+\rho_{101}$

where

$i\partial_{t}\rho_{101}(t)=[H_{D0},\rho_{101}(t)]+[H_{int}(t),\rho_{100}(t)]$

This has the solution

$\rho_{101}(t)=\int_{0}^{t}exp(-i(t-s)ad(H_{D0})([H_{int}(s),\rho_{100}(s)])ds$

It is clear that this expression for $\rho_{101}(t)$ is a trilinear
combination of the photonic creation and annihilation operators and the
electron-positron creation and annihilation operators such that it is linear
in the former and quadratic in the latter. Thus, we can write approximately

$\rho_{101}(t,r,r^{\prime})=\sum_{klm}[c(k)a(l)a(m)g_{1}(t,r,r^{\prime}|klm)+c(k)^{*}a(l)a(m)g_{2}(t,r,r^{\prime}|klm)+c(k)a(l)^{*}a(m)g_{3}(t,r,r^{\prime}|klm)+c(k)^{*}a(l)^{*}a(m)g_{4}(t,r,r^{\prime}|klm)]+H.c$

After taking into account self interactions of the plasma of electrons and
positrons, the one particle density operator field $\rho_{1}(t,r,r^{\prime})$
satisfies the quantum Boltzmann equation

$i\rho_{1,t}(t,r_{1},r_{1}^{\prime})=\int(H_{D}(t,r_{1},r_{1}^{\prime\prime})\rho_{1}(t,r^{\prime\prime},r^{\prime})-\rho_{1}(t,r,r^{\prime\prime})H_{D}(t,r^{\prime\prime},r^{\prime}))d^{3}r^{\prime\prime}$
$+(N-1)\rho_{10}(t,r_{1},r_{1}^{\prime})\int(V(r_{1},r_{2})-V(r_{1}`^{\prime},r_{2}))\rho_{10}(t,r_{2},r_{2})d^{3}r_{2}$
$+(N-1)\int(V(r_{1},r_{2})-V(r_{1}^{\prime},r_{2}))g_{12}(t,r_{1},r_{2}|r_{1}^{\prime},r_{2})d^{3}r_{2}---(a)$

where

$g_{12}(t,r_{1},r_{2}|r_{1}^{\prime},r_{2}^{\prime})$

Note that in perturbation theory,

$\rho_{1}=\rho_{10}+\rho_{11}$

where $\rho_{10}$ as calculated above, satisfies (a) with $V=0$.

##  Chapter 14 Quantum neural networks and learning using mixed state dynamics
for open quantum systems

Let $\rho(t)$ be the state of an evolving quantum system according to the
dynamics

$i\rho^{\prime}(t)=[H_{0}+V(t),\rho(t)]+\theta(\rho(t))$

where $V(t)$ is the control potential and $\theta(\rho)$ is the Lindblad noise
term. Our aim is to design this control potential $V(t)$ so that the
probability density $\rho(t,r,r)$ in the position representation tracks a
given probability density function $p(t,r)$ and the quantum current
$(e/m)\nabla_{r^{\prime}}Im(\rho(t,r,r^{\prime})|_{r^{\prime}=r})$ tracks a
given current density $J(t,r)$. By analogy with pure state dynamics and fluid
dynamical systems, the corresponding problem is formulated as follows:
Consider the wave function $\psi(t,r)$ that satisfies Schrodinger’s equation
for a charged particle in an electromagnetic field:

$i\partial_{t}\psi(t,r)=(-(\nabla+ieA(t,r))^{2}/2-e\Phi(t,r)+W(t,r)V(t,r))\psi(t,r)$

The aim is to control the neural weight $W(t,r)$, so that
$p_{q}(t,r)=|\psi(t,r)|^{2}$ tracks the fluid density $\rho(t,r)$ that is
normalized so that $\int\rho(t,r)d^{3}r=1$ and further that the probability
current density

$J_{q}(t,r)=Im(\psi(t,r)\nabla\psi(t,r)^{*})+A(t,r)|\psi(t,r)|^{2}$

tracks a given current density $J(t,r)$ of the fluid. Note that we are
assuming that $p,J$ satisfy the equation of continuity:

$\partial_{t}\rho+div(J)=0$

This is an effective QNN algorithm for simulating the motion of a fluid
because the corresponding quantum mechanical probability and current density
pairs $(p_{q},J_{q})$ automatically by virtue of Schrodinger’s equation
satisfy the equation of continuity and simultaneously guarantee that $p_{q}$
is a probability density, ie, it is non-negative and integrates over space to
unity. Thus rather than using a QNN for tracking only the density of a fluid
or a probability density, we can in addition use it to track also the fluid
current density by adjusting the control weights $W(t,r)$. Note that for open
quantum systems, ie, in the presence of Lindblad noise terms, the equation of
continuity for the QNN does not hold. This means that when bath noise
interacts with our QNN, we would not get an accurate simulation of the fluid
mass and current density. However, if there are sources of mass generation in
the fluid so that the equation of continuity reads

$\partial_{t}\rho(t,r)+divJ(t,r)=g(t,r)$

then we can use the Lindblad operator coefficient terms as additional control
parameters to simulate $\rho,J$ for such a system with mass generation $g>0$
or mass destruction $g<0$.

$|J(t,r)-J_{q}(t,r)|^{2}=|Im(\psi(t,r)\nabla\psi(t,r)^{*})+A(t,r)|\psi(t,r)|^{2}-J(t,r)|^{2}$

is the space-time instantaneous norm error square between the true fluid
current density and that predicted by the QNN. Its gradient w.r.t $A(t,r)$ is

$\delta|J(t,r)-J_{q}(t,r)|^{2}/\delta A(t,r)=$
$2|\psi(t,r)|^{2}(J(t,r)-J_{q}(t,r))$

Note that this is a vector. This means that for tracking the current density,
the update equation for the control vector potential must be

$\partial_{t}A(t,r)=-\mu\delta|J(t,r)-J_{q}(t,r)|^{2}/\delta A(t,r)=$
$-2\mu|\psi(t,r)|^{2}(J(t,r)-J_{q}(t,r))$

We can also derive a corresponding update equation for the scalar potential in
Schrodinger’s equation based on matching the true mass density of the fluid to
that predicted by the QNN as follows.

$|p(t,r)-p_{q}(t,r)|^{2}=|p(t,r)-|\psi(t,r)|^{2}|^{2}$

so

$\partial_{t}\Phi(t,r)=-\mu\delta|p(t,r)-|\psi(t,r)|^{2}|^{2}/\delta\Phi(t,r)=$
$2\mu[\bar{\psi}(t,r)\delta\psi(t,r)/\delta\Phi(t,r)+\psi(t,r)\delta\bar{\psi}(t,r)/\delta\Phi(t,r)][p(t,r)-|\psi(t,r)|^{2}]$

Now using the Dyson series expansion,

$\psi(t,r)=U_{0}(t)\psi(0,r)+\sum_{n\geq
1}(-i)^{n}\int_{0<t_{n}<...<t_{1}<t}U_{0}(t-t_{1})V(t_{1})U_{0}(t_{1}-t_{2})....U_{0}(t_{n-1}-t_{n})\psi(0,r)dt_{1}...dt_{n-1}$

giving

$\delta\psi(t,r)/\delta V(t,r)=V(t,r)\sum_{n\geq
1}(-i)^{n}\int_{0<t_{n}<...<t_{2}<t}U_{0}(t-t_{2})V(t_{2})U_{0}(t_{2}-t_{3})...U_{0}(t_{n-1}-t_{n})\psi(0,r)dt_{2}...dt_{n}$
$=-iV(t,r)(\psi(t,r)-\psi(0,r))$

Taking the conjugate of this equation gives us

$\delta\bar{\psi}(t,r)/\delta V(t,r)=iV(t,r)(\bar{\psi}(t,r)-\bar{\psi}(0,r))$

Thus the update equation for $\Phi(t,r)=V(t,r)/e$ becomes

$\partial_{t}\Phi(t,r)=2\mu.\Phi(t,r)Im(\psi(t,r)-\psi(0,r)).(p(t,r)-|\psi(t,r)|^{2})$

Note that here, we can more generally consider the error energy to be a
weighted linear combination of the error energies between the fluid mass
density and its QNN estimate and that between the fluid current density and
its QNN estimate:

$E=w_{1}|p(t,r)-|\psi(t,r)|^{2}|^{2}+w_{2}|J(t,r)-Im(\psi(t,r)\nabla\psi(t,r)^{*})+A(t,r)|\psi(t,r)|^{2}|^{2}$

while calculating the updates for the vector and scalar potentials using this
energy function, we must take into account the dependence of $\psi(t,r)$ and
$\bar{\psi}(t,r)$ on both $A(t,r)$ and $\Phi(t,r)$. For this we write the
Hamiltonian as

$H(t)=H_{0}+eV_{1}(t)+e^{2}V_{2}(t)$

where

$H_{0}=-\nabla^{2}/2,V_{1}(t)=\Phi(t,r)+(1/2)(divA(t,r)+2(A(t,r),\nabla)),V_{2}(t)=A(t,r)^{2}/2$

Then in terms of variational derivatives, from the Dyson series expansion,

$\delta\psi(t,r)/\delta A(t,r)=e((i/2)\nabla\psi(t,r)-i\nabla\psi(t,r))$
$=(-i/2)\nabla\psi(t,r)$

and taking conjugate

$\delta\bar{\psi}(t,r)/\delta A(t,r)=(i/2)\nabla\bar{\psi}(t,r)$

so we get

$\delta|\psi(t,r)|^{2}/\delta A(t,r)=$
$(-i/2)(\bar{\psi}(t,r)\nabla\psi(t,r)-\psi(t,r)\nabla\bar{\psi}(t,r))$
$=Im(\bar{\psi}(t,r)\nabla\psi(t,r))$

Exercise: Using the above formulas, evaluate

$\delta E/\delta A(t,r),\delta E/\delta\Phi(t,r)$

QNN in the presence of Hudson-Parthasarathy noise:

Let $|\psi(t)>$ be a pure state in the Hilbert space
$\mathfrak{h}\otimes\Gamma_{s}(L^{2}(\mathbb{R}_{+}))$ evolving according to
the Hudson-Parthasarathy noisy Schrodinger equation where
$\mathfrak{h}=L^{2}(\mathbb{R}_{+})$:

$d|\psi(t)>=(-(iH+LL^{*}/2)dt+LdA-L^{*}dA^{*})|\psi(t)>$

where $H,L$ are system operators. Specifically, choose

$H=-\nabla^{2}/2+V(t,r),L=(a,\nabla)+(b,r)$

where $a,b$ are constant complex 3-vectors. For a specific representation of
the creation and annihilation processes $A(t)^{*},A(t)$, we choose an onb
$|e_{n}>$ for $L^{2}(\mathbb{R}_{+})$ and then write

$A(t)=\sum_{n}a_{n}<\chi_{[0,t]}>|e_{n}>,A(t)^{*}=\sum_{n}a_{n}^{*}<e_{n}|\chi_{[0,t]}>$

where $a_{n}$ are operators in a Hilbert space satisfying the CCR

$[a_{n},a_{m}]=0,[a_{n},a_{m}^{*}]=\delta[n-m],[a_{n}^{*},a_{m}^{*}]=0$

Then an easy calculation shows that

$[A(t),A(s)^{*}]=\sum_{n}<\chi_{[0,t]}|e_{n}><e_{n}|\chi_{[0,s]}>=<\chi_{[0,t]}|\chi_{[0,s]}>=min(t,s)$

from which we easily deduce the quantum Ito formula of Hudson and
Parthasarathy. We can also introduce a conservation process $\Lambda(t)$

$\Lambda(t)=\sum_{n,m}a_{n}^{*}a_{m}<e_{n}|\chi_{[0,t]}|e_{m}>$

and deduce that

$[A(t),\Lambda(s)]=\sum_{n,m}a_{m}<\chi_{[0,t]}|e_{n}><e_{n}|\chi_{[0,s]}|e_{m}>$
$=\sum_{m}a_{m}<\chi_{[0,min(t,s)]}|e_{m}>=A(min(t,s))$

and taking the adjoint of this equation then gives us

$[\Lambda(s),A(t)^{*}]=A(min(t,s))^{*}$

These equations yield the other quantum Ito formulas involving the products of
the creation and annihilation operator differentials with the conservation
operator differentials. Now suppose that we assume that our wave function on
the tensor product of the system and bath space has the form

$|\psi(t)>=|\psi_{s}(t)>\otimes|\phi(u)>$

where $|\phi(u)>$ is the bath coherent state and $|\psi_{s}(t)>$ is the system
state. Then substituting this into the Hudson-Parthasarathy noisy Schrodinger
equation gives us using

$dA(t)|\phi(u)>=dt.u(t)|\phi(u)>$ $dA(t)^{*}|\phi(u)>=(dB(t)-u(t)dt)|\phi(u)>$

where

$B(t)=A(t)+A(t)^{*}$

is a classical Brownian motion process and since the bath is in the coherent
state, it is should be regarded as a Brownian motion with drift. Thus, the HPS
equation assumes the form for such a special class of pure states on system
and bath space assumes the form of a stochastic Schrodinger equatiion

$i\partial_{t}|\psi_{s}(t)>=(-(iH+LL^{*}/2)dt+u(t)dtL-(dB(t)-u(t)dt)L^{*})|\psi_{s}(t)>$
$=(((-iH-LL^{*}/2+u(t)(L+L^{*}))dt-L^{*}dB(t))|\psi_{s}(t)>$

This stochastic Schrodinger equation is parametrized by the coherent state
parameter function $u\in L^{2}(\mathbb{R}_{+})$. We can now use this
stochastic Schrodinger equation to track a given probability density $p(t,r)$
using control on all $H,L,u(t)$. This would thus define a QNN in the Hudson-
Parthasarathy-Stochastic-Schrodinger framework. The idea is that Brownian
noise is undesired but one cannot help it enter into the picture and hence one
must track the given pdf in spite of it being present.

Using quantum electrodynamics (QED) to track joint probability densities in a
QNN.

The Lagrangian of (QED) is given by the sum of a Maxwell term, a Dirac term
and an interaction term:

$L=L_{M}+L_{D}+L_{int}=$
$(-1/4)F_{\mu\nu}F^{\mu\nu}+\bar{\psi}(\gamma^{\mu}(i\partial_{\mu}+eA_{\mu})-m)\psi$
$=(-1/4)F_{\mu\nu}F^{\mu\nu}+\bar{\psi}(i\gamma^{\mu}(\partial_{\mu}-m)\psi+e\bar{\psi}\gamma^{\mu}\psi.A_{\mu}$

Thus we identify

$L_{M}=(-1/4)F_{\mu\nu}F^{\mu\nu},L_{D}=\bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi,$
$L_{int}=e\bar{\psi}\gamma^{\mu}\psi.A_{\mu}$

To proceed further, we write down the Schrodinger equation corresponding to
this second quantized system as

$iU^{\prime}(t)=H(t)U(t)$

where $H(t)$ is the second quantized Hamiltonian of the system expressed in
terms of the creation and annihilation operator fields of the photons and
Fermions. To get at this Hamiltonian, we must first calculate to a given order
of perturbation, the Dirac wave field and the Maxwell wave field in terms of
these Bosonic and Fermionic creation and annihilation operators. Let us as an
example, derive these expressions upto second order of perturbation. The
Maxwell and Dirac equations derived from the Lagrangian are

$\partial_{\alpha}\partial^{\alpha}A^{\mu}=e\psi^{*}\alpha^{\mu}\psi,$
$(i\gamma^{\mu}\partial_{\mu}-m)\psi=-e\gamma^{\mu}\psi.A_{\mu}$

where we have applied the Lorentz gauge condition

$\partial_{\mu}A^{\mu}=0$

in deriving the former. In the absence of interactions, we can express the
solutions as

$\psi_{0}(x)=\sum_{k}(a(k)f_{k}(x)+a(k)^{*}g_{k}(x)),$
$A^{\mu}(x)=\sum_{k}c(k)\phi_{k}^{\mu}(x)+c(k)^{*}\bar{\phi}_{k}^{\mu}(x)$

where $c(k),c(k)^{*}$ satisfy the Bosonic CCR while $a(k),a(k)^{*}$ satisfy
the Fermionic CAR. Writing

$\psi=\psi_{0}+\psi_{1}+\psi_{2},A_{\mu}=A_{\mu 0}+A_{\mu 1}+A_{\mu 2},$

we get

a.Zeroth order perturbation equations.

$(i\gamma^{\mu}\partial_{\mu}-m)\psi_{0}=0,\partial_{\alpha}\partial^{\alpha}A_{\mu
0}=0,$

b.First order perturbation equations.

$(i\gamma^{\mu}\partial_{\mu}-m)\psi_{1}=-e\gamma^{\mu}\psi_{0}A_{\mu 0},$
$\partial_{\alpha}\partial^{\alpha}A_{\mu
1}=e\psi_{0}^{*}\alpha_{\mu}\psi_{0}$

c.Second order perturbation equations.

$(i\gamma^{\mu}\partial_{\mu}-m)\psi_{1}=-e\gamma^{\mu}(\psi_{0}A_{\mu
1}+\psi_{1}A_{\mu 0})$ $\partial_{\alpha}\partial^{\alpha}A_{\mu
2}=e(\psi_{0}^{*}\alpha_{\mu}\psi_{1}+\psi_{1}^{*}\alpha_{\mu}\psi_{0})$

In general, writing the $n^{th}$ order perturbation series

$A_{\mu}=\sum_{k=0}^{n}A_{\mu k},\psi=\sum_{k=0}^{n}\psi_{k},$

we obtain the $n^{th}$ order perturbation equations as

$(i\gamma^{\mu}\partial_{\mu}-m)\psi_{n}=-e\gamma^{\mu}\sum_{k=0}^{n-1}\psi_{k}A_{\mu
n-1-k},$ $\partial_{\alpha}\partial^{\alpha}A_{\mu
n}=e\sum_{k=0}^{n-1}\psi_{k}^{*}\alpha_{\mu}\psi_{n-1-k}$

##  Chapter 15 Quantum gate design, magnet motion,quantum filtering,Wigner
distributions

###  15.1 Designing a quantum gate using QED with a control c-number four
potential

Consider the interaction Lagrangian between the em four potential and the
Dirac current:

$L_{int}=e\bar{\psi}\gamma^{\mu}\psi.A_{\mu}$

The corresponding contribution of this interaction Lagrangian to the
interaction Hamiltonian is its negative (in view of the Legendre
transformation) integrated over space:

$H_{int}=-L_{int}=-e\int\bar{\psi}\gamma^{\mu}\psi.A_{\mu}d^{3}x$

The contribution of this term upto second order in the em potential to the
scattering amplitude between two states $|p_{1},\sigma_{1},p_{2},\sigma_{2}>$
and $|p_{1}^{\prime},\sigma_{1}^{\prime},p_{2}^{\prime},\sigma_{2}^{\prime}>$
in which there are two electrons with the specified four momenta and
z-component of the spins or equivalently two positrons or one electron and one
positron is given by

$A_{2}(p_{1}^{\prime},\sigma_{1}^{\prime},p_{2}^{\prime},\sigma_{2}^{\prime}|p_{1},\sigma_{1},p_{2},\sigma_{2})=-e^{2}<p_{1}^{\prime},\sigma_{1}^{\prime},p_{2}^{\prime},\sigma_{2}^{\prime}|T(\int
H_{int}(x_{1})H_{int}(x_{2}))d^{4}x_{1}d^{4}x_{2}|p_{1},\sigma_{1},p_{2},\sigma_{2}>$

Let $u(p,\sigma)$ denote the free particle Dirac wave function in the
momentum-spin domain. Thus it satisfies

$(\gamma^{\mu}p_{\mu}-m)u(p,\sigma)=0,\sigma=1,2$

Taking the adjoint of this equation after premultiplying by $\gamma^{0}$ and
making use of the self-adjointness of the matrices $\gamma^{0}\gamma^{\mu}$
gives us

$\bar{u}(p,\sigma)(\gamma^{\mu}p_{\mu}-m)=0$

where

$\bar{u}(p,\sigma)=u(p,\sigma)^{*}\gamma^{0}$

Now observe that for one particle momentum-spin states of an electron,

$a(p^{\prime},\sigma^{\prime})|p,\sigma>=\delta(\sigma^{\prime},\sigma)\delta^{3}(p^{\prime}-p)|0>$

and therefore,

$\psi(x)|p,\sigma>_{e}=exp(-ip.x)u(p,\sigma)|0>$

and for a one particle positron state

$\psi(x)^{*}|p,\sigma>_{p}=exp(ip.x)v(p,\sigma)^{*}|0>$

or equivalently,

$\bar{\psi}(x)|p,\sigma>=\psi(x)^{*}\gamma^{0}|p,\sigma>_{p}=exp(ip.x)\bar{v}(p,\sigma)|0>$

Thus one of the terms in
$A_{2}(p_{1}^{\prime},\sigma_{1}^{\prime},p_{2}^{\prime},\sigma_{2}^{\prime}|p_{1},\sigma_{1},p_{2},\sigma_{2})$
for two electron momentum state transitions is given by

$-e^{2}\int<p_{1}^{\prime},\sigma_{1}^{\prime},p_{2}^{\prime},\sigma_{2}^{\prime}|(\bar{\psi}(x_{1})\otimes\bar{\psi}(x_{2}))\gamma^{\mu}\otimes\gamma^{\nu})(\psi(x_{1})\otimes\psi(x_{2}))|p_{1},\sigma_{1},p_{2},\sigma_{2}>D_{\mu\nu}(x_{1}-x_{2})d^{4}x_{1}d^{4}x_{2}$
$=\int
D_{\mu\nu}(x_{1}-x_{2})(\bar{u}(p_{1}^{\prime},\sigma_{1}^{\prime})\otimes\bar{u}(p_{2}^{\prime},\sigma_{2}^{\prime}))(\gamma^{\mu}\otimes\gamma^{\nu})(u(p_{1},\sigma_{1})\otimes
u(p_{2},\sigma_{2}))exp(-i(p_{1}-p_{1}^{\prime}).x_{1}-i(p_{2}-p_{2}^{\prime}).x_{2})d^{4}x_{1}d^{4}x_{2}$
$=\hat{D}_{\mu\nu}(p_{1}-p_{1}^{\prime})\delta^{4}(p_{1}+p_{2}-p_{1}^{\prime}-p_{2}^{\prime}).$
$.(\bar{u}(p_{1}^{\prime},\sigma_{1}^{\prime})\otimes\bar{u}(p_{2}^{\prime},\sigma_{2}^{\prime}))(\gamma^{\mu}\otimes\gamma^{\nu})(u(p_{1},\sigma_{1})\otimes
u(p_{2},\sigma_{2}))$

This evaluates to

$[(\bar{u}(p_{1}^{\prime},\sigma_{1}^{\prime})\gamma^{\mu}u(p_{1},\sigma_{1})).(\bar{u}(p_{2}^{\prime},\sigma_{2}^{\prime})\gamma_{\mu}u(p_{2},\sigma_{2}))/(p_{1}^{\prime}-p_{1})^{2}]\delta^{4}(p_{1}+p_{2}-p_{1}-p_{2}^{\prime})$

Likewise another term having a different structure also comes from the same
expression for $A_{2}$ and that is

$[(\bar{u}(p_{2}^{\prime},\sigma_{2}^{\prime})\gamma^{\mu}u(p_{1},\sigma_{1})).(\bar{u}(p_{1}^{\prime},\sigma_{1}^{\prime})\gamma_{\mu}u(p_{2},\sigma_{2}))/(p_{2}^{\prime}-p_{1})^{2}]\delta^{4}(p_{1}+p_{2}-p_{1}-p_{2}^{\prime})$

The total amplitude is a superposition of these two amplitudes. Here,
$D_{\mu\nu}(x)$ is the photon propagator and $\hat{D}_{\mu\nu}(p)$ is its four
dimensional space-time Fourier transform and is given by $\eta_{\mu\nu}/p^{2}$
where $p^{2}=p^{\mu}p_{\mu}$. This propagator is defined by

$D_{\mu\nu}(x_{1}-x_{2})=<0|T(A_{\mu}(x_{1})A_{\nu}(x_{2}))|0>$

Now we consider a situation when apart from the quantum em field, there is
also an external classical control em field described by the four potential
$A_{\mu}^{c}$. Then in the $n^{th}$ order term in the Dyson series for the
scattering of $M$ electrons, there will be a term of the form

$\int<p_{k}^{\prime},\sigma_{k}^{\prime},k=1,2,...,M|T(H_{int}(x_{1})...H_{int}(x_{n}))|p_{k},\sigma_{k},k=1,2,...,M>d^{4}x_{1}...d^{4}x_{n}$

where

$H_{int}(x)=e\bar{\psi}(x)\gamma^{\mu}\psi(x)(A_{\mu}(x)+A_{\mu}^{c}(x))$

Here, $A_{\mu}$ is the four potential of the quantum electromagnetic field
while $A_{\mu}^{c}(x)$ is the four potential of the classical c-number control
electromagnetic four potential. Expanding these, it is clear that this term
will be a sum of terms of the form

$\int<p_{k}^{\prime},\sigma_{k}^{\prime},k=1,2,...,MT(\Pi_{k=1}^{n}\bar{\psi}(x_{k})\gamma^{\mu_{k}}\psi(x_{k}))|p_{k},\sigma_{k},k=1,2,...,M>$
$.<0|T(A_{\mu_{1}}(x_{1})...A_{\mu_{r}}(x_{r}))|0>\Pi_{m=r+1}^{n}A_{\mu_{m}}^{c}(x_{m})d^{4}x_{1}...d^{4}x_{n}$

###  15.2 On the motion of a magnet through a pipe with a coil under gravity
and the reaction force exerted by the induced current through the coil on the
magnet

Problem formulation

Assume that the coil is wound around a tube with $n_{0}$ turns per unit length
extending from a height of $h$ to $h+b$. The tube is cylindrical and extends
from $z=0$ to $z=d$. Taking relativity into account via the retarded
potentials, if $m(t)$ denotes the magnetic moment of the magnet, then the
vector potential generated by it is approximately obtained from the retarded
potential formula as

$A(t,r)=\mu m(t-r/c)\times r/4\pi r^{3}---(1)$

where we assume that the magnet is located near the origin. This approximates
to

$A(t,r)=\mu m(t)\times r/4\pi r^{3}-\mu m^{\prime}(t)\times r/4\pi
cr^{2}---(2)$

Now Assume that the magnet falls through the central axis with its axis always
being oriented along the $-\hat{z}$ direction. Then, after time $t$, let its
$z$ coordinate be $\xi(t)$. We have the equation of motion

$M\xi^{\prime\prime}(t)=F(t)-Mg---(3)$

where $F(t)$ is the $z$ component of the force exerted by the magnetic field
generated by the current induced in the coil upon the magnet. Now, the flux of
the magnetic field generated by the magnet through the coil is given by

$\Phi(t)=n_{0}\int_{X_{2}+Y^{2}\leq R_{T}^{2},h\leq z\leq
h+b}B_{z}(t,X,Y,Z)dXdYdZ---(4)$

where

$B_{z}(t,X,Y,Z)=\hat{z}.B(t,X,Y,Z)---(5a)$ $B(t,X,Y,Z)=curlA(t,{\bf
R}-\xi(t)\hat{z})---(5b)$

where

${\bf R}=(X,Y,Z)$

and where $R_{T}$ is the radius of the tube. The current through the coil is
then

$I(t)=\Phi^{\prime}(t)/R_{0}---(6)$

where $R_{0}$ is the coil resistance. The force of the magnetic field
generated by this current upon the falling magnet can now be computed easily.
To do this, we first compute the magnetic field $B_{c}(t,X,Y,Z)$ generated by
the coil using the usual retarded potential formula

$B_{c}(t,X,Y,Z)=curlA_{c}(t,X,Y,Z)---(7a)$

where

$A_{c}(t,X,Y,Z)=(\mu/4\pi)\int_{0}^{2n_{0}\pi}(I(t-|{\bf
R}-R_{T}(\hat{x}.cos(\phi)+\hat{y}.sin(\phi))-\hat{z}(h+b/2n_{0}\pi)\phi|/c).$
$.(-\hat{x}.sin(\phi)+\hat{y}.cos(\phi))/|{\bf
R}-R_{T}(\hat{x}.cos(\phi)+\hat{y}.sin(\phi))-\hat{z}(b/2n_{0}\pi)\phi|)R_{T}d\phi---(7a)$

Note that $A_{c}(t,X,Y,Z)$ is a function of $\xi(t)$ and also of $m(t)$ and
$m^{\prime}(t)$. It follows that $I(t)$ is a function of
$\xi(t),\xi^{\prime}(t),m(t),m^{\prime}(t),m^{\prime\prime}(t)$. Then we
calculate the interaction energy between the magnetic fields generated by the
magnet and by the coil as

$U_{int}(\xi(t),\xi^{\prime}(t),t)=(2\mu)^{-1}\int(B_{c}(t,X,Y,Z),B(t,X,Y,Z))dXdYdZ---(8)$

and then we get for the force exerted by the coil’s magnetic field upon the
falling magnet as

$F(t)=-\partial U_{int}(\xi(t),\xi^{\prime}(t),t)/\partial\xi(t)---(9)$

Remark: Ideally since the interaction energy between the two magnetic fields
appears in the Lagrangian $\int(E^{2}/c^{2}-B^{2})d^{3}R/2\mu$, we should
write down the Euler-Lagrange equations using the Lagrangian

$L(t,\xi^{\prime}(t),\xi^{\prime}(t))=M\xi^{\prime}(t)^{2}/2-U_{int}(\xi(t),\xi^{\prime}(t),t)-Mg\xi(t)--(10)$

in the form

$\frac{d}{dt}\partial L/\partial\xi^{\prime}-\partial L/\partial\xi=0---(11)$

which yields

$M\xi^{\prime\prime}(t)-\frac{d}{dt}\partial U/\partial\xi^{\prime}+\partial
U_{int}/\partial\xi+Mg=0--(12)$

3\. Numerical simulations

Define the force function as in (9):

$F(\xi(t),\xi^{\prime}(t)=-\partial
U_{int}(\xi(t),\xi^{\prime}(t))/\partial\xi(t)$

Neglecting the term involving the gradient of $U_{int}$ w.r.t $\xi^{\prime}$
(this is justified because the dominant reaction force of the coil field on
the magnet is due to the spatial gradient of the dipole interaction energy),
we can discretize (2) with a time step of $\$\delta$ to get

$M(\xi[n+1]-2\xi[n]+\xi[n-1])/\delta^{2}+F_{int}(\xi[n],(\xi[n]-\xi[n-1])/\delta)+Mg=0$

or equivalently in recursive form,

$\xi[n+1]==2\xi[n]-\xi[n-1]-(\delta^{2}/M)F_{int}(\xi[n],(\xi[n]-\xi[n-1])/\delta)-\delta^{2}g$

Fig.1 shows plot of $\xi[n]$ obtained using a ”for loop”. The current through
the coil as a function of time has also been plotted in fig.2 using the
formula

$I(t)=R_{0}^{-1}\frac{d}{dt}(n_{0}\int_{X_{2}+Y^{2}\leq R_{T}^{2},h\leq z\leq
h+b}B_{z}(t,X,Y,Z)dXdYdZ)$ $=I(t|\xi(t),\xi^{\prime}(t))$

where

$B_{z}(t,X,Y,z)=\hat{z}.curlA(t,X,Y,Z-\xi(t))=\partial A_{y}(X,Y,Z)\partial
X-\partial A_{x}(X,Y,Z)/\partial Y$

with

$A(X,Y,Z)=\mu_{0}.{\bf m}\times{\bf R}/4\pi R^{3},{\bf R}=(X,Y,Z)$

Here relativistic corrections have been neglected. The formula for the
interaction energy between the magnet and the field generated by the coil
discussed in the previous section can be simplified after making appropriate
approximations to $-{\bf m}.{\bf B}_{c}=mB_{cz}$ where $B_{cz}$, the
$z$-component of the reaction magnetic field produced by the coil is evaluated
at the site of the magnet, ie, at $(0,0,\xi(t))$. Thus, the formula for the
force of the field on the coil is approximated as

$F_{int}(\xi,\xi^{\prime})=F_{int}(\xi)=-m\partial
B_{cz}(t,0,0,\xi|\xi,\xi^{\prime})/\partial\xi$

The coil magnetic field is evaluated using the approximate non-relativistic
Biot-Savart formula

$B_{c}(t,X,Y,Z|\xi,\xi^{\prime})=(\mu_{0}I(t|\xi,\xi^{\prime})/4\pi))\int_{0}^{2n_{0}\pi}R_{T}d\phi(-\hat{x}.sin(\phi)+\hat{y}.cos(\phi))\times((X-R_{T}.cos(\phi))\hat{x}+(Y-R_{T}.sin(\phi))$
$.\hat{y}+(Z-h-b\phi/2n_{0}\pi)\hat{z})/[(X-R_{T}.cos(\phi))^{2}+(Y-R_{T}.sin(\phi))^{2}+(Z-h-b\phi/2n_{0}\pi)^{2}]^{3/2}$

These approximations are shown to produce excellent agreement with hardware
experiments as a comparison of figs.() shows.

Taking into account relativistic corrections: The magnetic field generated by
the coil has the form $B_{c}(t,X,Y,Z|\xi(t),\xi^{\prime}(t))$. More precisely,
if $I_{0}(t)$ is the current of the electrons within the magnet, the magnetic
vector potential generated by this current is given by

$A(t,R)=(\mu/4\pi)\int_{magnet}I_{0}(t-|R-\xi(t)-R^{\prime}|/c)dR^{\prime}/|R-\xi(t)-R^{\prime}|$

In this expression, $\xi(t)$ is the position vector of the centre of the
magnet at time $t$ and $R^{\prime}$ is the position on the magnet electron
loop relative to its centre. Since the magnet current is a constant in the
case of a permanent magnet, the above simplifies to

$A(t,R)=(\mu I_{0}/4\pi)\int_{magnet}dR^{\prime}/|R-\xi(t)-R^{\prime}|$