# Quantum squeezing cannot beat the standard quantum limit

###### Abstract.

Quantum entanglement between particles is expected to allow one to perform
tasks that would otherwise be impossible [Bell1964, Feynman1982, Deutsch1985,
Shor1994, Giovannetti2001]. In quantum sensing and metrology, entanglement is
often claimed to enable a precision that cannot be attained with the same
number of particles and time, forgoing entanglement [Helstrom1969,
Holevo1973a, Giovannetti2004, Giovannetti2006, DemkowiczDobrzanski2012,
Zwierz2012, Pezze2018]. Two distinct approaches exist: creation of entangled
states that either _i)_ respond quicker to the signal, or _ii)_ are associated
with lower noise and uncertainty. The second class of states are generally
called squeezed states. Here we show that if our definition of success is – a
precision that is impossible to achieve without entanglement – then squeezed
states cannot succeed. In doing so we show that a single non-separable
squeezed state provides fundamentally no better precision, per unit time, than
a single particle.

Liam P. McGuinness

## Prelude

I have asked for and received a lot of feedback on this work from experts in
the field111If you have any feedback, please contact me.. Here I try to distil
the arguments presented in the main text as clearly as possible.

If one wants to compare the precision of two measurement devices, a good
method is to break each device down into its individual components and analyse
the precision each component provides. If you show that each component of the
first device has no better precision than each component of the second, it is
possible to conclude that the first device cannot outperform the second. For
rigour, two additional points are needed:

  1. (1)

A check that the second device has more components (as more components improve
precision).

  2. (2)

An assumption that the components are independent.

This paper performs such an analysis. The mathematical content is simply to
analyse the (optimum) precision per unit time, that quantum mechanics predicts
an individual component can achieve222To be precise, I analyse the maximum
information each individual component provides per unit time, for a single
measurement.. One proviso is that this analysis only applies when each
component is measured once. The physics is to associate an individual
component with a single non-separable state vector. As non-separable states
can be entangled, this allows quantum correlations to improve the measurement
precision, however, with assumption (2) above, these are the only form of
correlations present.

Using basic quantum mechanics, it is straightforward to bound the amount of
information per unit time, a single state vector can provide on a signal
$\theta$. The amount of information (and the measurement precision) just
depends on the vector response to $\theta$, i.e.
$\frac{\mathrm{d}}{\mathrm{d}\theta}$. This is a well-known and standard
result (see Refs [91–93]). In terms of the mathematics and physics there is
nothing even mildly controversial about what I am saying here. Note, we don’t
even need to worry about the length of the state vector – entangled or
unentangled – any pure state has unit length.

I then make the observation that in the quantum squeezing community, people
claim to improve the precision of their measurement device (beyond a
fundamental limit) without improving the state response to the signal. In
fact, in a squeezed device, the response to the signal of a single non-
separable squeezed state is often described as being the same as that of a
single particle. In case you think that I am misrepresenting the position of
the squeezing community, please look through the papers for yourself. Refs
[12–15] are review papers and a good starting point. See also the references
within [14] and Refs [33, 35–85, 96–101]. There is a mathematical and logical
contradiction between these two observations. They cannot both be true. It
turns out that the results of the squeezing community are in error.

To prove that a device constructed from an ensemble of $N$ squeezed particles
can not provide fundamentally more information per unit time than possible
from a device made from the same number of independent particles (this is
known as the standard quantum limit), I note:

  1. (1)

An entangled ensemble of $N$ particles contains less indivisible state vectors
than an unentangled ensemble of $N$ particles.

That’s it. Much of the main text discusses experiments with spins, but the
analysis is general and equally applies to measuring devices composed of
photons, i.e. optical interferometers such as LIGO.

Just to reiterate the logical arguments made here.

  1. (1)

Assuming the postulates of quantum mechanics are correct, the minimum
uncertainty per unit time, that any single non-separable state provides on an
unknown signal is bounded by the state response to the signal. This bound
takes into account the minimum noise when measuring a single state vector.

  2. (2)

In the literature, a squeezed state is commonly described as having the same
signal response as a single particle state. However, the measurement noise of
this state is claimed to be less than the measurement noise of a single
particle. I agree that this argument sounds reasonable. It is persuasive
because it appeals to our intuitive concept of signal-to-noise determining the
measurement precision. I refer you to preceding point. Based on our
understanding of quantum mechanics, this is impossible.

  3. (3)

An ensemble of entangled squeezed states contains less non-separable state
vectors than an ensemble of unentangled states. Therefore squeezed ensembles
cannot beat the standard quantum limit.

If I could pinpoint why people have such strong resistance to these results,
it seems to be because there is a vast body of theoretical and experimental
work claiming the contrary. If you are at that point, then I ask you first to
go through the arguments that I am making and look for an error. If none is
forthcoming, then I urge you to consider reassessing the analysis presented by
the rest of the scientific community (see the FAQ’s section and Ref. [31] for
a summary of my efforts to date).

I have lost a lot of confidence in the peer-review system (see Refs
[108–110]), but that is not to say that I do not want feedback. Of course I
would like other people to check my work for rigour. Not only that, I also
want to demonstrate to non-experts that this has been done. I do not think
that the peer-review system performs those two tasks very well. For those two
goals, a much more effective strategy is to financially incentivise people to
perform a thorough peer-review, i.e. offer a financial reward to anyone who
finds an error. Therefore I offer a prize of US$10,000 to the first person
that finds an error that invalidates the conclusions of this work. Please send
to me via email333Note, in Ref. [31] I have also offered a prize of US$10,000
for the first experimental demonstration of a precision beyond the standard
quantum limit, and an additional prize of US$10,000 for the first
demonstration using $N$-partite entanglement that surpasses the single
particle limit. At the least, the measurement time and number of particles
should include all unitary evolution..

## Introduction

Consider a sensing device composed of $N$ spin-$1/2$ particles,444Note, these
results apply to photon interferometry and are discussed in this context in
Appendix 1. which is used to measure some unknown signal $\theta$. Taking just
one spin and using the spin direction $\accentset{\rightharpoonup}{S}$ of this
particle as a meter, then by error propagation any estimate $\tilde{\theta}$
of the value of $\theta$ has an uncertainty555More fully, we can include the
possibility of a finite bias $\epsilon$ and average over $p(\theta=\Theta)$,
the _a priori_ probability distribution of $\theta$, to obtain the expected
uncertainty:
$\langle\Delta\tilde{\theta}\rangle\geq\sqrt{\int^{\Theta}p(\theta=\Theta)\left(\left|\frac{\partial\accentset{\rightharpoonup}{S}}{\partial\theta}\right|^{-2}\left(\Delta\accentset{\rightharpoonup}{S}\right)^{2}+\epsilon^{2}\right)d\theta}$.:

$\Delta\tilde{\theta}\geq\left|\frac{\partial\accentset{\rightharpoonup}{S}}{\partial\theta}\right|^{-1}\Delta\accentset{\rightharpoonup}{S},$
(1)

where $\Delta\accentset{\rightharpoonup}{S}$ is the uncertainty in determining
the spin direction. The derivative term is often called the measurement
signal, and $\Delta\accentset{\rightharpoonup}{S}$ the measurement
noise666Incorrectly so, it is the measurement uncertainty not noise., so that
the ‘signal-to-noise ratio’ determines the measurement precision. With $N$
identical and independent spins, in general one cannot do better than

$\Delta\tilde{\theta}\geq\Delta\tilde{\theta}_{1}/\sqrt{N},$ (2)

where $\Delta\tilde{\theta}_{1}$ bounds the uncertainty using a single
particle. Note, for Eq. (2) to hold, $\Delta\tilde{\theta}_{1}$ must represent
the best possible uncertainty that can be attained with a single particle,
otherwise one could beat the limit simply by improving
$\Delta\tilde{\theta}_{1}$. Furthermore, in general $\Delta\tilde{\theta}_{1}$
is a function of time since the spin response is given by unitary evolution
and with more time one can increase the spin response or perform more
measurements to improve the uncertainty. Under this definition, Eq. (2) is
called the standard quantum limit (SQL) and it sets an uncertainty bound per
unit time that is impossible to surpass with a given number of identical and
independent spins.

There are two approaches to overcoming the SQL using entanglement, either make
the spin response greater, or reduce the uncertainty in measuring the spin
direction [Gross2012, Ma2011, Ludlow2015, Pezze2018]. The first makes use of
entangled NOON, CAT or GHZ states so that, in theory:
$\frac{\partial\accentset{\rightharpoonup}{S}_{N}(t)}{\partial\theta}=N\frac{\partial\accentset{\rightharpoonup}{S}_{1}(t)}{\partial\theta}$,
whilst
$\Delta\accentset{\rightharpoonup}{S}_{N}=\Delta\accentset{\rightharpoonup}{S}_{1}$,
where the subscript denotes the number of spins in the sensor [Pezze2018,
Gross2012, Ludlow2015, Bollinger1996, Dowling1998, Childs2000, Campos2003,
Kok2004, Leibfried2004, Nagata2007, Resch2007, Berry2009, Dorner2009,
Escher2011, Giovannetti2011, Daryanoosh2018, Shaniv2018, Xie2021]. In effect a
single spin with greater magnetic moment is created such that the response of
the entangled device is $N$-fold greater than that of a single spin. It is
worth noting here, that $\Delta\accentset{\rightharpoonup}{S}$ is the result
of sampling from an unknown probability distribution and since a measurement
of both a single entangled state and a single spin provide only one sample,
they have the same measurement uncertainty. We will not discuss this approach
except to note that when the resources required to generate entanglement are
fully accounted for, no improvement over the SQL or $\Delta\tilde{\theta}_{1}$
has been demonstrated [McGuinness2021].

The second approach uses entangled squeezed states so that777Early proposals
did not require entanglement to overcome the SQL, resulting in confusion as to
whether entanglement is required (see Appendix 2). Conflicting definitions of
the SQL still remain, for example when photon states with non-Poissonian
statistics are called non-classical and claimed to surpass the SQL (I prefer
the descriptor ‘non-thermal’ since any single particle including a single
baseball is a non-classical state in this definition). We explicitly define
the SQL as whatever one can achieve without entanglement, thus it is a moot
point whether entanglement is required to surpass it. The definition of
squeezing is also not standardized see e.g. [Soerensen2001, Meyer2001,
Maccone2020].:
$\frac{\partial\accentset{\rightharpoonup}{S}_{N}(t)}{\partial\theta}=\frac{\partial\accentset{\rightharpoonup}{S}_{1}(t)}{\partial\theta}$,
whereas
$\Delta\accentset{\rightharpoonup}{S}_{N}=\Delta\accentset{\rightharpoonup}{S}_{1}/N$.
I.e. the spin response of the entangled device remains the same as that of a
single spin but the measurement noise reduces by a factor of $N$ [Caves1981,
Ma2011, Pezze2018, Gross2012, Ludlow2015, Walls1983, Wodkiewicz1985, Wu1986,
Slusher1987, Xiao1987, Polzik1992, Wineland1992, Kitagawa1993, Wineland1994,
Sanders1995, Kuzmich1998, Soerensen1998, Brif1999, Kuzmich2000, Meyer2001,
Esteve2008, Appel2009, Eberle2010, Gross2010, Koschorreck2010,
Koschorreck2010a, Leroux2010, Leroux2010a, LouchetChauvet2010,
SchleierSmith2010, Wasilewski2010, Luecke2011, Sewell2012, Aasi2013,
Taylor2013, Bohnet2014, Muessel2014, Strobel2014, Kruse2016, Polzik2016,
Cox2016, Davis2016, Bohnet2016, Hosten2016, Linnemann2016, Macri2016, Tse2019,
Braverman2019, Schulte2020, Malia2020, Bao2020, PedrozoPenafiel2020,
Casacio2021, Gilmore2021, Greve2022, Malia2022]. We can already see that there
is a conflict between our explanation of the origin of the measurement
uncertainty and what is observed with squeezed states. Shouldn’t the
uncertainty in estimating the direction of a single squeezed state be the same
as for a single spin? Where does this reduced uncertainty come from? Either
our layperson description is wrong or squeezed states do not perform as
described. We now show that, if entanglement provides any benefit over the
SQL, then it must come about from increasing the sensor response to the signal
and not through reduced noise888My apologies for changing between
‘uncertainty’ and ‘noise’, they do not mean the same thing. I would like to
use the correct terminology - uncertainty, but the papers in this field refer
to noise and I am quoting their claims. This conflation of terms is the crux
of the issue, I encourage the reader to take careful note of this point.
Returning to the start of this paragraph,
$\Delta\accentset{\rightharpoonup}{S}_{N}=\Delta\accentset{\rightharpoonup}{S}_{1}/N$
does not mean the measurement noise reduces by a factor of $N$, rather the
uncertainty in estimating the spin direction of
$\accentset{\rightharpoonup}{S}_{N}$, is $N$ times lower than estimating the
direction of a single particle $\accentset{\rightharpoonup}{S}_{1}$..

## Noise independent quantum precision bound

For rigour we make two adjustments to our language. First, rather than talk
about spin direction, we address the underlying mathematical object – the
state vector. To show that the uncertainty bound obtained from an entangled
ensemble containing squeezed states is worse than that of an unentangled
ensemble, we use a counting argument based on the number of indivisible state
vectors (i.e. a state vector that cannot be further factored into non-
separable states) in the ensemble. Henceforth, a state vector,
$\left|\psi\right>$ always refers to this basic unit we are dealing with – an
indivisible state. A single state vector is never used to describe the quantum
state of a separable ensemble, instead we keep note of the number of copies of
each state. Secondly, for technical reasons and conciseness we avoid
quantitatively defining uncertainty, we define instead the (Fisher)
information on $\theta$ denoted $\mathrm{I}\left[\theta,t\right]$, provided by
a given state, or more precisely, measurement of that state [Braunstein1994]:

$\mathrm{I}\left[\theta,t\right]\equiv\int^{X}\mathrm{d}X\frac{1}{\mathrm{Pr}\left[X|\theta,t\right]}\left(\frac{\partial\mathrm{Pr}\left[X|\theta,t\right]}{\partial\theta}\right)^{2},\quad\mathrm{I}\left[\theta\right]\equiv\sum_{i=1}^{R}\frac{1}{\mathrm{Pr}\left[X_{i}|\theta,t\right]}\left(\frac{\partial\mathrm{Pr}\left[X_{i}|\theta,t\right]}{\partial\theta}\right)^{2}$
(3)

where $\mathrm{Pr}\left[X|\theta,t\right]$ is the conditional probability to
obtain the measurement result $X$ in time $t$ given $\theta$ and the (LHS) RHS
assumes the measurement outcomes are (continuous) discrete with $R$
possibilities. We note that for any measurement, the estimation uncertainty
$\Delta\tilde{\theta}$ is a monotonically decreasing function of
$\mathrm{I}\left[\theta,t\right]$. This observation provides the necessary
tools to compare uncertainties, to say one is greater than another, and is
sufficient for our purposes.

_Key requirement:_ The central claim in squeezing enhanced metrology is that
squeezed states have an improved intrinsic noise (uncertainty) compared to a
single spin. It is clear that the response to $\theta$ of the squeezed state,
quantified by $\left|\frac{d\left|\psi(\theta,t)\right>}{d\theta}\right|$, is
not greater than that of a single spin. One reason this must be the case is
that if the noise can be reduced by a factor of $N$, then any further
improvement in the sensor response would violate the Heisenberg limit.

Finally, we only establish a bound on the information provided by a single
measurement of the ensemble. While this allows a direct like-for-like
comparison between measurement devices, it is important to note that this is
not how measurement sensitivity is usually reported. Often, when reporting
squeezing enhancements, comparison is made to a time-averaged limit with
$\mathrm{Hz}^{-1/2}$ uncertainty improvement. Further assumptions for the
proof are provided in Appendix 3.

_Counting approach:_ For a measurement device comprised of $N$ spins, assuming
we can prove

> Statement 1: A single squeezed state does not provide fundamentally more
> information on $\theta$, per unit time, than a single spin.

It then follows that the information bound on $\theta$ for any $N$ spin
ensemble containing $M$ squeezed states is lower than an ensemble of $N$
unentangled spins. The reason being that a squeezed ensemble can be separated
into a maximum of $M$ independent state vectors where $M<N$ (this follows from
the very definition of entanglement). Assuming these states are
independent999This assumption just ensures that any correlations come solely
from entanglement., the information provided by the squeezed ensemble is $M$
times that of a single squeezed state and is therefore less than the
information provided by the unentangled ensemble. In fact, this counting
argument shows that increasing squeezing leads to a worse uncertainty bound
since there are less states to average over. Meaning that, if the uncertainty
provided by an ensemble containing squeezed states ever surpasses the single
particle bound and if the degree of squeezing/entanglement is continuously
increased, then at some point the uncertainty must get worse. And the
converse: if the measurement uncertainty always improves for arbitrary amounts
of squeezing, then the uncertainty never surpasses the single particle bound.

Many mathematical statements equivalent to Statement 1 (excluding the counting
argument) have been published (see Appendix 4). For example, Wootters showed
in 1981 that $\mathrm{I}\left[\theta,t\right]$ can be interpreted as distance
metric over quantum states [Wootters1981], meaning that if the path mapped out
by unitary evolution is the same for two states then so is the information.
This is equivalent to saying that the uncertainty depends only on the sensor
response to $\theta$ [Ou1997, Childs2000, Giovannetti2004, Zwierz2012,
Pang2014, Yuan2015, Pang2017, Gorecki2020], therefore states with no enhanced
response provide no fundamental sensing advantage. Braunstein, Caves and
Milburn provided most of the mathematical content of the proof by showing that
for pure states, $\mathrm{I}\left[\theta,t\right]$ is given solely by the
gradient of the state vector with respect to $\theta$, and does not depend on
any intrinsic uncertainty of this state [Braunstein1996]. Here we detail the
arguments in full.

## Proof - Time-independent multiplicative Hamiltonian:
$\hat{H}(\theta)=\theta\cdot\hat{H}$

Denote a spin-squeezed state as $\left|\psi_{SS}(\theta,t)\right>$, and the
state of a single spin as $\left|\psi_{1}(\theta,t)\right>$. Denote the
maximum information on $\theta$ that can be provided by any measurement on
$\left|\psi_{SS}(\theta,t)\right>$, $\left|\psi_{1}(\theta,t)\right>$ as
$\mathrm{I}_{\mathrm{SS}}\left[\theta,t\right]$ and
$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$ respectively, then we have:

Claim: If
$\left(\frac{\mathrm{d}\left<\psi_{SS}(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi_{SS}(\theta,t)\right>}{\mathrm{d}\theta}\right)=\left(\frac{\mathrm{d}\left<\psi_{1}(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)$
then
$\mathrm{I}_{\mathrm{SS}}\left[\theta,t\right]\leq\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$.

Physical interpretation: Squeezed states claim to surpass the SQL by reducing
the uncertainty associated with the state, and not by increasing the response
of the state to the signal (i.e. the derivative). To refute this claim we need
to show that if the gradient of the squeezed state with respect to the signal
is the same as that of a single spin, then information bound on the squeezed
state is less than or equal to that of a single spin. As the derivative of a
state vector is also a state vector, we can’t say one state vector is greater
than another, i.e.
$\frac{d\left|\psi_{1}\right>}{d\theta}>\frac{d\left|\psi_{2}\right>}{d\theta}$
is not a mathematical statement (the vectors actually exist in different
Hilbert spaces). To obtain a non-negative real number, we take the inner-
product. Since state vectors are normalised, this operation returns the
magnitude of the derivative.

Proof: Braunstein, Caves and Milburn showed that for a pure state
[Braunstein1996]:

$\mathrm{I}\left[\theta,t\right]=4\left[\left(\frac{\mathrm{d}\left<\psi(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)-\left|\left<\psi(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}\right].$
(4)

Working in the Schrödinger picture where time-evolution is carried by quantum
states, an initial state $\left|\psi_{0}\right>$ evolves in response to the
Hamiltonian $\hat{H}(\theta,t)=\theta\cdot\hat{H}$, according to:
$\left|\psi_{0}\right>\rightarrow\hat{U}(\theta,t)\left|\psi_{0}\right>\equiv\left|\psi(\theta,t)\right>$,
where $\hat{U}(\theta,t)=\mathrm{Exp}\left[-{i\mkern 1.0mu}\theta
t\hat{H}/\hbar\right]$. Writing $\left|\psi_{0}\right>$ in terms of the $K$
eigenstates of $\hat{H}$, denoted by their eigenvalues
$\left|\psi_{E_{k}}\right>$ with complex amplitude $\alpha_{k}$:
$\left|\psi_{0}\right>=\sum_{k=1}^{m}\alpha_{i}\left|\psi_{E_{k}}\right>$, we
have:

$\left|\psi(\theta,t)\right>=\sum_{k=1}^{K}\mathrm{Exp}\left[-{i\mkern
1.0mu}\theta tE_{k}/\hbar\right]\alpha_{k}\left|\psi_{E_{k}}\right>,$

where the derivative of this state with respect to $\theta$ is:

$\frac{\mathrm{d}}{\mathrm{d}\theta}\left|\psi(\theta,t)\right>=\sum_{k=1}^{K}(-{i\mkern
1.0mu}tE_{k}/\hbar)\mathrm{Exp}\left[-{i\mkern 1.0mu}\theta
tE_{k}/\hbar\right]\alpha_{i}\left|\psi_{E_{k}}\right>.$

We first derive the maximum information a single spin can provide. For a
spin-1/2 with two eigenstates and denoting the eigenvalues of $\hat{H}$ as
$\pm\frac{\gamma}{2}$, we have:

$\left(\frac{\mathrm{d}\left<\psi_{1}(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)=\left(\frac{t\gamma}{2\hbar}\right)^{2}\left[|\alpha_{1}|^{2}+|\alpha_{2}|^{2}\right],$

and

$\left|\left<\psi_{1}(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}=\left(\frac{t\gamma}{2\hbar}\right)^{2}\left[|\alpha_{1}|^{2}-|\alpha_{2}|^{2}\right]^{2}.$

We can see that $\mathrm{I}\left[\theta,t\right]$ is maximised by initially
placing the spin in an equal superposition of eigenstates, so that
$|\alpha_{1}|^{2}-|\alpha_{2}|^{2}=0$. Then:

$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]=\left(\frac{t\gamma}{\hbar}\right)^{2}.$

We have shown that:
$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]=4\left(\frac{\mathrm{d}\left<\psi_{1}(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)$
and reproduced the well-known Heisenberg limit for a single spin
[Bollinger1996, Giovannetti2004, Pang2014]. Inserting into Eq. (4), the
equality stated in the claim, the maximum information of the squeezed state
is:

$\mathrm{I}_{\mathrm{SS}}\left[\theta,t\right]=\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]-4\left|\left<\psi_{SS}(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi_{SS}(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}.$

As the second term is a non-negative number,
$\mathrm{I}_{\mathrm{SS}}\left[\theta,t\right]\leq\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$
and the proof is complete.

In the above analysis, it may seem like we only consider quantum states and
not measurements, however Eq. (4) implicitly contains the Born rule. In
particular, $\mathrm{I}\left[\theta,t\right]$ is a projection of
$\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}$ meaning that
(for this Hamiltonian) the optimal measurement basis is a projection
orthogonal to the eigenstates101010Here orthogonal means at an angle of
$90^{\circ}$, not anti-parallel. [Braunstein1996, Childs2000, Giovannetti2004,
Pang2017]. Explicitly, for $\hat{H}=\gamma\hat{\sigma}_{z}/2$ where
$\hat{\sigma}_{z}$ is the Pauli-$z$ matrix, the measurement associated with
$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$ is a projective measurement in
the $x-y$ plane. Considering just a 2-dimensional space with discrete
measurement outcomes, we can denote the measurement results as ‘1’ and ‘0’,
thus allowing Eq. (3) to be expressed as111111We have used
$\mathrm{Pr}\left[0|\theta,t\right]=1-\mathrm{Pr}\left[1|\theta,t\right]$, for
a Bernoulli random variable taking only two values.:

$\mathrm{I}\left[\theta,t\right]=\left|\frac{\partial\mathrm{Pr}\left[1|\theta,t\right]}{\partial\theta}\right|^{2}\frac{1}{\mathrm{Pr}\left[1|\theta,t\right]\left(1-\mathrm{Pr}\left[1|\theta,t\right]\right)}.$
(5)

Here we can identify
$\left|\frac{\partial\mathrm{Pr}\left[1|\theta,t\right]}{\partial\theta}\right|$
and
$\sqrt{\mathrm{Pr}\left[1|\theta,t\right]\left(1-\mathrm{Pr}\left[1|\theta,t\right]\right)}$
with the measurement signal and noise in Eq. (1), where the latter is called
quantum projection noise [Itano1993]. Note that the description of some
quantum states as having intrinsically lower uncertainty is completely absent
in this analysis, and in Eq. (5) the noise and signal are not
independent121212We should be careful when equating terms in Eq. (1) and Eq.
(5) because they are not identical. In Eq. (1) the response of the meter is
independent of the measurement, whereas in Eq. (5),
$\mathrm{Pr}\left[1|\theta,t\right]$ depends on the measurement..

In Appendix 5, 6 we generalise the proof to the following situations. We show
that if we consider:

  * •

probability mixtures of squeezed states, then the information from these mixed
states cannot exceed a single spin.

  * •

the expected mean information, averaged over the prior probability
distribution of $\theta$, then squeezed states cannot outperform a single
spin.

  * •

a modified claim concerning the projection of the gradient
$\frac{\mathrm{d}\left|\psi\right>}{\mathrm{d}\theta}$ orthogonal to
$\left|\psi\right>$, then a squeezed state cannot cannot outperform a single
spin for estimation of signals in arbitrary time-dependent Hamiltonians
$\hat{H}(\theta,t)$.

## Discussion

We have proved a powerful and surprising theorem that allows us to immediately
exclude any of the methods proposed in [Caves1981, Walls1983, Wodkiewicz1985,
Slusher1985, Wu1986, Slusher1987, Xiao1987, Polzik1992, Wineland1992,
Kitagawa1993, Wineland1994, Sanders1995, Kuzmich1998, Soerensen1998, Brif1999,
Kuzmich2000, Meyer2001, Orzel2001, Andre2002, Esteve2008, Appel2009,
Eberle2010, Gross2010, Koschorreck2010, Koschorreck2010a, Leroux2010,
Leroux2010a, LouchetChauvet2010, Riedel2010, SchleierSmith2010,
Wasilewski2010, Abadie2011, Luecke2011, Ma2011, Sewell2012, Aasi2013,
Taylor2013, Bohnet2014, Muessel2014, Strobel2014, Kruse2016, Polzik2016,
Cox2016, Davis2016, Bohnet2016, Hosten2016, Hosten2016a, Linnemann2016,
Macri2016, Tse2019, Braverman2019, Schulte2020, Malia2020, Bao2020,
PedrozoPenafiel2020, Casacio2021, Gilmore2021, Colombo2022, Greve2022,
Malia2022, Braginsky1974, Braginsky1977, Unruh1979, Braginsky1980,
Braginsky1996] from achieving a measurement precision beyond the SQL. Of these
works, the following are experimental papers [Xiao1987, Slusher1985,
Slusher1987, Polzik1992, Soerensen1998, Kuzmich2000, Meyer2001, Orzel2001,
Esteve2008, Appel2009, Eberle2010, Gross2010, Koschorreck2010,
Koschorreck2010a, Leroux2010, Leroux2010a, LouchetChauvet2010, Riedel2010,
SchleierSmith2010, Wasilewski2010, Abadie2011, Luecke2011, Sewell2012,
Aasi2013, Taylor2013, Bohnet2014, Muessel2014, Strobel2014, Bohnet2016,
Cox2016, Hosten2016, Hosten2016a, Kruse2016, Linnemann2016, Braverman2019,
Tse2019, Bao2020, Malia2020, PedrozoPenafiel2020, Casacio2021, Gilmore2021,
Colombo2022, Greve2022, Malia2022], the majority of which claim to demonstrate
a measurement precision that cannot be obtained without squeezing. This begs
the question:

> Why are there so many experimental papers that seem to contradict this
> proof?

The answer is not that our proof is erroneous, but rather that in general,
analyses of experiments either do not compare the achieved precision to the
SQL or involve fundamental mistakes. Below we outline some common examples in
the literature.

_Reasons why quantum squeezing has never surpassed the SQL:_

A. Comparison to a different limit

  1. (1)

Comparison to an imperfect limit: An improvement in measurement precision for
a device operating far away from the SQL is demonstrated [Abadie2011,
Aasi2013, Tse2019, Casacio2021]. This is not the same as demonstrating a
measurement precision beyond the SQL.

  2. (2)

Tautological definition of the measurement precision: A different quantum
limit is introduced. This limit is defined as the precision of a measurement
device _as it is and without modifications_. Modifications that utilise
entanglement are allowed, whilst forbidding any other changes to the device
[Wu1986, Xiao1987, Taylor2013]. If the device was already operating at the
SQL, and squeezing implemented whilst keeping the number of particles and time
constant, then the sensitivity cannot improve.

B. Analytic errors

  1. (1)

Analysis of measurement noise not uncertainty: Conflation of noise and
uncertainty is one of the most common analytical mistakes in spin squeezing,
see e. g. [Orzel2001, Esteve2008, Appel2009, Gross2010, Koschorreck2010,
Koschorreck2010a, Leroux2010, LouchetChauvet2010, Riedel2010,
SchleierSmith2010, Sewell2012, Cox2016, Hosten2016, Kruse2016, Colombo2022,
Greve2022, Malia2022]. As an example, assume a projective measurement in the
$x$-basis is performed on a single spin pointing in an unknown direction in
the $x-y$ plane. Denote this state $\left|a\right>$ and label the outcomes of
the observable $\hat{S}_{x}=\hat{\sigma}_{x}/2$ as $\\{0,1\\}$. The
uncertainty in estimating the $x$-component of the spin direction is actually
independent of the measurement variance. If ten experiments produce the
following data-set: $\\{1,1,1,1,1,1,1,1,1,1\\}$, the SQL has not been
surpassed, even though the variance is less than 1/4. Compare to the data-set
obtained from a different state $\left|b\right>$: {0,1,1,1,0,0,1,1,1,0}.
Someone with knowledge of $\left|a\right>$, $\left|b\right>$ can prove this is
not a statistically anomaly by showing
$\left(\Delta\hat{S}_{x}\right)^{2}_{a}<\left(\Delta\hat{S}_{x}\right)^{2}_{b}$,
where the subscript indicates that the variance is calculated with respect to
these states. Despite this mathematical inequality, our uncertainty in
estimating the $x$-component of $\left|b\right>$ is the same as for
$\left|a\right>$. Ultimately we do not want measurements with less _noise_ ,
but measurements that provide lower _uncertainty_ on the spin
direction131313In the literature $(\Delta\accentset{\rightharpoonup}{S})^{2}$
is often incorrectly equated with the variance of some collective spin
observable $\hat{\bm{S}}_{z}=\sum_{i=1}^{N}\hat{S}_{z}$, such that a variance
$\left(\Delta\hat{\bm{S}}_{z}\right)^{2}<|S|/4$, where $|S|$ denotes the spin
vector length, is evidence of an enhanced precision. Note, in Eq. (1),
$\Delta\accentset{\rightharpoonup}{S}$ is the uncertainty in estimating the
spin direction, it is not the (square-root) variance of an operator. Nor is it
the square-root variance of some dataset.. Often prior information from state
preparation (and the value of $\theta$) is used to ensure that the measurement
statistics have low variance without reducing the uncertainty in estimating
$\theta$.

  2. (2)

Replace the signal gradient with the contrast: The measurement signal (c.f.
the derivative in Eq. (5)), is replaced with a constant term that does not
depend on $\theta$ or the measurement basis. A common error when the
measurement precision is defined, is to replace the gradient with the spin
vector length, characterised by the ‘contrast’, ‘visibility’, ‘coherence’ or
‘Wineland squeezing parameter’ [Meyer2001, Esteve2008, Gross2010, Leroux2010,
Leroux2010a, Riedel2010, Hosten2016, PedrozoPenafiel2020, Malia2020,
Malia2022]. Again a measurement basis is chosen to reduce the measurement
noise without accounting for the commensurate reduction in signal. I.e. the
experiment is designed so that $\mathrm{Pr}\left[X|\theta,t\right]\rightarrow
0$, but the loss of signal
$\frac{\partial\mathrm{Pr}\left[X|\theta,t\right]}{\partial\theta}\rightarrow
0$ is not fully accounted for, and is often assumed to be constant. To some
extent, the operation regime of Advanced LIGO [Aasi2015, Tse2022] and QND
proposals [Braginsky1980, Kimble2001] suffer from this error.

There are valid questions on how one should interpret this proof. For instance
it only addresses fundamental quantum measurement noise141414It also neglects
measurement back-action onto the signal. However back-action is minimised when
using a single particle sensor., other technical sources of experimental noise
should be reduced. One question is to what extent this is possible and whether
there is some practical regime where squeezing can be of advantage? As we note
in App. 3 invoking this explanation is tantamount to saying quantum mechanics
is not correct. Perhaps a more nuanced interpretation is to note that no
experiment can reach the precision bounds set by quantum mechanics. Reaching
these bounds requires perfect, instantaneous measurements with no decoherence.
But you can’t have it both ways. Claims that squeezing can surpass the SQL do
not present that message, indeed quite the opposite.

## Conclusion

At a fundamental level, quantum mechanics dictates that measurement
uncertainty is a statistical sampling phenomenon. One drawback of using
quantum correlations is that it provides less samples and this can lead to
worse uncertainty. In particular some statistical analyses of measurement
precision mistakenly associate the fundamental randomness of measurement
observations as a noise which should be reduced. Here we have shown that
approaches to improve the measurement uncertainty by reducing this noise
cannot succeed. To misquote Rolf Landauer – The signal is the noise!

By relating metrological performance to the number of separable states we have
proved that squeezed ensembles cannot outperform unentangled ensembles in
sensing. The proof was inspired by the proposition of a general quantum
mechanical uncertainty limit per wavefunction [McGuinness2021a]. Here
wavefunctions are considered the fundamental information unit in computers and
sensors, not particles. My position is that it this information per unit time
limit holds for all entangling procedures, not just squeezing, and applies to
general tasks in quantum information like computation. Finally, it should be
apparent that in the field of quantum physics, the peer-review process has
failed (see also [McGuinness2022, McGuinness2023, McGuinness2023a]) and this
has led to a loss of scientific progress.

## Appendix

### 1\. Photon interferometry and vacuum noise

In his seminal paper on quantum mechanical noise in an interferometer, Caves
noted a fundamental noise source that “can be attributed to vacuum (zero-
point) fluctuations in the electromagnetic field, which enter the
interferometer from the unused input port” [Caves1981]. One typical physical
description of this noise is that when an electromagnetic wave enters the
interferometer at the first beam-splitter, vacuum fluctuations entering the
unused port cause an unequal splitting of the electromagnetic field. Caves
noted an equivalent view that treats photons as particles and attributes this
imbalance to the “random scattering of the input photons at the beam-
splitter”. As a result more of the input light is directed to one arm of the
interferometer and less into the other, in turn affecting the output light
intensity at both ports. Yet another description is that photon shot-noise
randomises the arrival time of photons on the detector [Tse2019]. It is
commonly understood that this noise cannot be removed, except through
inserting squeezed vacuum into the unused input port to reduce these
fluctuations. Here I present a different physical description of the
interferometer, one that changes the emphasis of the noise and I believe,
helps better predict experimental results.

As in the main text, one can learn a lot by reducing the experiment down to
its individual components and providing a physical description of this basic
experiment. So what happens when a single photon enters the interferometer?
Using a particle description of the photon, which randomly scatters at the
beam-splitter, we encounter a problem. The photon only traverses one arm, thus
there is no interference. No matter the path difference, the photon exits the
unused port with 50% probability. This is not what is observed experimentally.
Furthermore, the idea of each photon traversing just one arm of the
interferometer leads one to infer that the output signal depends on
interference between separate photons. This does not capture the basics,
rather it is interference of a photon with itself that is relevant151515More
precisely it is interference between basis states of a single wavefunction,
which can include more than one photon for entangled states.. I also find the
suggestion that vacuum fluctuations somehow prevent one from creating an equal
superposition state hard to accept – the creation of equal superposition
states is a basic concept in quantum metrology and quantum information
processing. Importantly, this description also incorrectly predicts
experimental results. For example, the observed probability (up to
experimental imperfections) to detect a photon at the unused port is given by
$p=\sin^{2}(\phi/2)$, where $\phi$ is the relative phase difference. If vacuum
fluctuations always enter the unused port to randomise the observed photon
statistics, then how can one explain this observation? Take $\phi=0$. Without
inserting any squeezed vacuum into the unused port, one can predict the
experimental outcome with remarkable accuracy. Fringe visibilities in excess
of 90% have been demonstrated so far for single photon interference. This
should not be possible if vacuum fluctuations of order $\sqrt{N}$ are a real
and fundamental source of noise.

A different description to those presented so far is to assume the outgoing
photon wavefunction at the beam-splitter is a perfect superposition of basis
states (as far as technical tolerances allow), so that a relative phase
difference between returning basis states determines their complex
coefficients161616Here basis states are eigenstates of photon number in each
mode of the interferometer.. A photodetector, placed at the unused port,
performs a measurement of the photon wavefunction, and detects a photon with
probability given by the squared amplitude of the state in this mode. In this
description, there is no noise in the interferometer which we need to remove.
The noise is a statistical effect, due to probabilistic measurement and is the
result of the Born rule. When multiple independent photons are inserted into
the interferometer, the basis states of each individual wavefunction interfere
upon return, and the summation of many individual probabilistic measurements
yield the detected signal. We do not need to worry about balancing the light
intensity in each arm of the interferometer, except to manufacture an ideal
beam-splitter. Note, in this description, if the interferometer is operated so
that the unused port is dark, then although laser power noise and quantum
projection noise is reduced [Itano1993], so is the signal since
$\frac{\partial p}{\partial\phi}\rightarrow 0$.

To convincingly demonstrate that vacuum fluctuations are not a fundamental
noise source in photon interferometry, we can go one step further and remove
the beam-splitter, thereby preventing vacuum fluctuations from entering the
interferometer. To illustrate the idea, take photons produced from a pulsed
laser, and replace the beam-splitter with a flippable (double-sided) mirror,
starting flipped down so that incident photons enter the longitudinal arm of
the interferometer, after which the mirror is flipped up, so that the second
pulse enters the transverse arm. The mirror remains in this position so that
photons returning from the longitudinal arm are sent out the antisymmetric
(unused) port, after which it is flipped down allowing photons returning from
the transverse arm to output at the same port. Now we do want to compare the
relative phase between separated optical pulses. However, as optical
photodiodes are sensitive only to the light intensity and not the phase, this
mode of operation seems forbidden. It turns out that coherent ensembles of
spins can be used as phase sensitive detectors.

As a first approximation, let’s treat the photodetector quantum mechanically
and the optical pulses as classical control fields. In this description,
gravitational waves change the relative phase of control fields acting on spin
qubits. The first (longitudinal) optical pulse performs a $\pi/2$-rotation on
the ensemble, where the optical phase is stored coherently in the spin
direction of the ensemble. This phase can be ‘readout’ with the second
(transverse) optical pulse which performs a $\pi/2$-rotation around an axis
determined by the relative phase of the two pulses (see [McGuinness2021a] for
a demonstration with microwave photons). Finally the spin state along-$z$
should be projectively readout. For a single spin, the probability to remain
in the initial $\left|0\right>$ state is:
$p(\left|0\right>)\approx\sin^{2}(\phi/2)$, where $\phi$ now refers to the
phase difference between the two pulses. If the spin energy transition is
perfectly resonant to the laser frequency, time-delays between the pulses have
no effect on the final spin state.

An advantage of treating spins rather than photons, as the measuring device,
is that it is much easier to perform quantum non-demolition (QND) measurements
on spins. Not only that, sensitivity to power fluctuations can be removed by
operating near $\phi=\pi$ so that $p(\left|0\right>)\approx 1$ and the spins
are in a ‘dark fringe’. Consider fluctuations slow enough so that both the
longitudinal and transverse pulse have the same intensity. Now instead of
$\pi/2$-rotations, the spins are rotated through an angle $\Theta$. The
$\pi$-phase shift of the second pulse ensures the spins are brought back to
$\left|0\right>$. There is a fundamental physical principle behind this
observation, and it allows quadratic suppression of all amplitude
fluctuations. The photon wavefunction is mapped to the spin state, so that
conjugate observables of the photon wavefunction (phase and amplitude) are
mapped to conjugate spin observables ($\hat{S}_{z}$ and $\hat{S}_{x}$).
Measurements that obtain information on the photon phase cannot also reveal
information on, and are therefore insensitive to the photon amplitude.

To see this, assume we only want to detect gravitational waves along the
longitudinal axis, which corresponds to measuring the phase $\varphi$ of the
first optical pulse. The first pulse is mapped to the spin state resulting in:

$\left|\psi\right>=\cos(\Theta)\left|0\right>+e^{-{i\mkern
1.0mu}\varphi}\sin(\Theta)\left|1\right>.$

The pulse amplitude (proportional to $\Theta$) is mapped to $S_{z}$ and (the
cosine of) $\varphi$ mapped to $S_{x}$. As phase and amplitude are conjugate
observables, a $\hat{S}_{z}$ measurement gives no information on $\varphi$ –
one observes $p(\left|0\right>)=\cos^{2}(\Theta)$,
$p(\left|1\right>)=\sin^{2}(\Theta)$. Measurement of the laser phase
$\varphi$, requires measurement of a conjugate observable. In order to perform
such a measurement, we do not need to send an optical pulse into the
transverse arm of the interferometer, we just need to implement a
$\hat{S}_{x}$ (or $\hat{S}_{y}$) measurement. Although there are technical
challenges in performing such a perfect measurement, the postulates of quantum
mechanics, assume that that it is possible. Furthermore, we do not need to
analyse the back-action this measurement induces on the mirror and there is no
radiation pressure trade-off. We have now described a setup that allows
realisation of the QND type measurements originally proposed by Braginsky,
Vorontsov and Thorne [Braginsky1980] and more recently by Kimble et. al.
[Kimble2001]. In the language of uncertainty squeezing and QND measurements,
one would say that the meter remains in an eigenstate, and measurements of
this state have minimum uncertainty. However, when the measurement basis is
chosen to make the readout noise zero, the meter response is quadratic to
phase shifts and is insensitive to small phase shifts.

Finally, one should again consider the case, where just a single photon enters
the interferometer and is detected with just a single spin. The spin-photon
interaction should be tailored so that the combined system is in an entangled
superposition:
$\frac{1}{\sqrt{2}}\left(\left|0_{S}\right>\left|1_{P}\right>+e^{-{i\mkern
1.0mu}\varphi}\left|1_{S}\right>\left|0_{P}\right>\right)$, where the $S$, $P$
subscripts denote the spin and photon states respectively. In this scenario
there is the outstanding issue on defining the phase of a single photon. I
argue that $\varphi$ is a physical property of the experiment, and that it can
be observed by interfering this state with a suitable (single) photon state.

Although further details are required to make this discussion rigorous, that
does not make it useless. All physical descriptions involve some
approximations, the key is to find which ones are critical to obtaining
accurate predictions. One benefit of this perspective is to demonstrate that
vacuum fluctuations can be removed from an interferometer. We have reduced the
problem of gravitational wave detection to estimating spin direction, a
problem which is explicitly addressed in the main text. Of course, vacuum
fluctuations are replaced with something else – quantum projection noise when
the spin state is readout – but as we have shown, this noise cannot be removed
through squeezing. Furthermore, this shifts the focus, we should focus on
reducing technical noise – laser power fluctuations, beam-splitter, detector
efficiency. Most importantly, there is an achievable parameter space where the
predictions of this analysis diverge greatly from treatments of vacuum
fluctuations. For low input power into LIGO where number of photons is small
so radiation pressure is very small, we can define the minimum possible
sensitivity per unit time that one can achieve with $N$ input photons into the
interferometer. If we now implement squeezing and use the same number of input
photons into the interferometer and the same measurement time, we can ask
whether the interferometer sensitivity can surpass this limit? Our analysis
show that the answer to this question is no.

### 2\. A confusing history of early ideas to surpass the SQL

The idea of using special projective measurements (not states), with reduced
noise originated in discussions on improving the sensitivity of gravitational
wave detectors [Braginsky1974, Braginsky1977, Unruh1979, Braginsky1996]. These
low noise measurements, called quantum non-demolition (QND) measurements, are
performed by choosing a measurement operator so that the meter is in an
eigenstate of the observable and the variance of the measurement observable
goes to zero [Unruh1978, Braginsky1980]. However no entanglement is required
to observe such a variance since separable eigenstates also yield zero
variance results. One unfortunate relic of this early analysis which persists
today is that the SQL is often defined as the minimum obtainable variance on
one observable when the variance equals that of another conjugate observable.
Often this statement is incorrectly reduced to saying that the uncertainty in
measuring spin-direction is equivalent to the spin noise, giving the SQL:
$\Delta\accentset{\rightharpoonup}{S}=\sqrt{|S|}/2$. Thus when the spin noise
is $<\sqrt{|S|}/2$, the SQL for measuring spin direction has been surpassed.
Initially the idea was to take advantage of better measurements, not entangled
states, however around the time of Caves’ proposal for squeezed vacuum
fluctuations in the context of photon interferometry [Caves1981], emphasis on
states of the meter with reduced uncertainty started. These methods were then
extended to spin states in Ramsey interferometers, called spin-squeezed states
[Kitagawa1991, Wineland1992, Kitagawa1993].

We can summarize some reasons why QND measurements do not surpass the SQL.
First, the requirement of perfect prior knowledge about the state of the meter
in order to perform a QND measurement is self-defeating171717If we are being
pedantic, one can learn about _which_ eigenstate the meter is in, and perform
discrimination.. Put another way, if we want the noise to remain low, the
meter must remain in an eigenstate, but if the meter simply remains in an
eigenstate then nothing is happening?! Also there is an inherent lack of
recognition of the difference between ‘uncertainty’ and ‘noise’. In general,
QND schemes neglected to consider that the measurement signal and measurement
noise are not independent, measurements with reduced noise – where the meter
is near an eigenstate of the measurement operator – also have less signal. In
fact, Wootters showed in 1981 [Wootters1981], that even in a $2^{N}$
dimensional Hilbert space the optimal measurement yields a probability
function of the form:
$\mathrm{Pr}\left[1|\theta\right]=\mathrm{cos^{2}}\left[\frac{m}{2}(\theta-\theta_{0})\right]$,
where $m$ is an integer and $\theta_{0}$ is a constant; thus when the
probability to measure an observable goes to one or zero, the gradient
disappears (Unruh used related arguments in response initial proposals
[Unruh1978]). Moreso, Itano et. al. noted that since projective measurements
are not perfect, it is often preferable to choose a measurement that maximises
projection noise [Itano1993]. The message here is that simply reducing
measurement noise does not guarantee better precision, and in general
measurements which maximise noise are better. Finally, Giovannetti, Lloyd and
Maccone proved, for a restricted set of Hamiltonians, that entangled states
and not measurements are required to surpass the SQL [Giovannetti2006].

Another relic of early suggestions to use QND measurements to surpass the SQL
is their current181818It seems that my characterisation of QND measurements as
a discounted historical idea is not accurate. Carlton Caves informs me that
LIGO are still working on QND measurements as outlined here [Kimble2001]. use
in the preparation of squeezed spin states [Kuzmich2000, Appel2009,
Koschorreck2010, LouchetChauvet2010, SchleierSmith2010, Sewell2012,
Bohnet2014, Cox2016, Hosten2016, Hosten2016a, Braverman2019, Malia2020,
Greve2022, Malia2022]. The idea is to place an ensemble of spins initialised
in $\left|0\right>^{\otimes N}$ in an optical cavity, where a $\pi/2$-pulse
rotates $\accentset{\rightharpoonup}{S}$ into the $x-y$ plane. A QND
measurement is then performed on the ensemble to project a sub-ensemble of the
spins to either $\left|0\right>$ or $\left|1\right>$. This measurement is
generally a dispersive population measurement of the spins, where an optical
probe pulse is inserted into the cavity and the phase (frequency) shift of the
output light is recorded. The shift is proportional to the number of atoms
projected into $\left|0\right>,\left|1\right>$. Apparently, the ensemble
contains an entangled state with less noise and therefore improved measurement
precision.

I strongly disagree with this description. After the QND measurement, I would
describe a single state in the ensemble as being in a probability mixture of
$\left|0\right>,\left|1\right>$ and $\left|x\right>$191919Actually, not
exactly $\left|x\right>$, the electromagnetic field of the optical probe
slightly rotates this state around $z$.. To refute this description at least
one piece of information is required: a definition of the phase relationship
between entangled basis states. For example, one can write an entangled Bell
state as: $(\left|00\right>+\left|11\right>)/\sqrt{2}$, and although it is
generally overlooked, a well-defined phase between the states
$\left|00\right>$ and $\left|11\right>$ is required to generate and make use
of this entangled state (in this example the state can be written
$(\left|00\right>+e^{{i\mkern 1.0mu}\phi}\left|11\right>)/\sqrt{2}$ with
$\phi=0$). Otherwise we have a mixed state and the best we can say is that for
any measurement we will record either a ‘0’ or a ‘1’, each with 50%
probability. None of the papers that use QND preparation of a “squeezed” spin
state define or make use of any phase relationship between the entangled basis
states. It is true that knowledge of the phase of the initial $\pi/2$-pulse
and the intensity the probe light is required when measuring these states, but
these parameters do not determine $\phi$ (see Assumptions and FAQ’s sections
below).

### 3\. Proof assumptions

For completeness, we list some assumptions of the proof below. In particular
we assume:

  * •

The postulates of quantum mechanics.

  * •

The entire dependence on $\theta$ is encoded in $\hat{H}(\theta,t)$. The
starting state $\left|\psi_{0}\right>$ and the measurement operator are
independent of $\theta$.

  * •

Apart from the initial state vectors that evolve to
$\left|\psi_{SS}(\theta,t)\right>,\left|\psi_{1}(\theta,t)\right>$, everything
else in the external universe is the same when comparing unentangled and
entangled ensembles. The prior information on $\theta$ and the ability to
apply control fields is the same. In fact, my understanding is that the
squeezing community claim even unitary evolution is the same for these two
states, therefore the only difference of consequence is the starting state
(and the measurement).

  * •

Each copy of the same state is independent and measurements on copies of the
same state are identical. This assumption ensures that any enhancement over
the SQL comes solely from entanglement, not other correlations, additional
information or different control on copies.

Comment on Assumption 1: I discuss one assumption of the analysis presented
here and use a simple example to illustrate how it is critical to the proof.
The assumption is the (Born rule) postulate of quantum mechanics – the
probability of obtaining a measurement outcome $m$ when measuring a state
$\left|\psi\right>$ is equal to
$p(m)=\left<\psi\right|\hat{M}^{\dagger}_{m}\hat{M}_{m}\left|\psi\right>$,
where $\\{\hat{M}_{m}\\}$ is the collection of self-adjoint measurement
operators associated with outcomes $m$ (Postulate 3 in [Nielsen2000]). This
postulate does not account for the fact that measurements are never perfect in
experiments and that some quantum states are easier to measure.

If we compare readout of the following spin states, performed by scattering
photons and counting the total number of photons detected (a collective spin
measurement):

$\displaystyle\frac{1}{\sqrt{2}}(\left|0\right>$
$\displaystyle+\left|1\right>)$
$\displaystyle\frac{1}{\sqrt{2}}(\left|00\right>$
$\displaystyle+\left|11\right>)$

then the second state is easier to measure. For readout along $z$, the photon
difference for measurements of the second state is twice that when reading out
the first state202020One could argue that these measurement results have
greater variance, i.e. the variance of the operator $(\Delta S_{z})^{2}$ is
greater for the second state. Apart from the fact that we are comparing two
different operators here, at a fundamental level in terms of the amount of
information provided, I disagree. If the measurements are perfect, we can take
readout of the first state and multiply by an arbitrary large number to
increase the variance (or multiply by a small number to reduce the variance).
In terms of measurement statistics, this does not change the information
contained in the measurements.. Whenever state readout is not perfect, then
(under the fair sampling assumption) the second state provides more
information. A few points of emphasis, while this is a practical limitation in
every real experiment, the proposition that it represents a fundamental
precision limit in quantum squeezing is equivalent to saying quantum mechanics
is incorrect. Also, increasing the readout fidelity removes this loophole
since the information difference reduces as the readout improves. Importantly,
in spin squeezing literature, the second state (or its extension to $N$
particle Hilbert space) is not being used, the following is used instead:

$\frac{1}{\sqrt{2(1+\epsilon^{2})}}\left(\epsilon\left|00\right>+\left(\left|10\right>+\left|01\right>\right)+\epsilon\left|11\right>\right)$

where $\epsilon$ is real, $\epsilon<1$ for a squeezed state, $\epsilon\ll 1$
in the limit of strong squeezing, and $\left|10\right>$, $\left|01\right>$
cannot be distinguished by a collective spin measurement along $z$
[Kitagawa1993, Ma2011, Andre2002, Vuletic2023] (see FAQ’s section).

Assumptions for physical interpretations:

The following assumptions are not required to derive the mathematical
inequalities, but if one then uses those inequalities to claim that the
precision of a squeezed device cannot surpass the precision of an ideal
unentangled ensemble, that argument requires further assumptions. I.e. when
going from the maths to physical interpretations of the proof one makes
implicit use of these assumptions.

  * •

In general (quantum) measuring devices are not initialised in
$\left|\psi_{0}\right>$, therefore we should account for the time and
resources required to prepare this state. Using $\left|\psi_{0,1}\right>$,
$\left|\psi_{0,SS}\right>$ to differentiate between $\left|\psi_{0}\right>$
for a single spin and a squeezed state respectively, then we assume the time
and resources to generate $\left|\psi_{0,1}\right>$ is not greater than
$\left|\psi_{0,SS}\right>$. In every proposal and experiment that I am aware
of, at least one copy of $\left|\psi_{0,1}\right>$ is used to create
$\left|\psi_{0,SS}\right>$. For example, a $\pi/2$-rotation is performed on at
least one spin as the first step in creating entangled spin states. To create
squeezed photon states, single photons are inserted into a non-linear optical
element, or the interferometer together with an additional field. Thus the
assumption that, in general it is more technically demanding and time-
consuming to generate the ideal $\left|\psi_{0,1}\right>$ compared to the
ideal $\left|\psi_{0,SS}\right>$, seems correct. In fact, when generating
entangled NOON, GHZ and CAT type states, (I claim) it precisely because the
converse assumption does not hold that these physical devices do not
outperform a single particle. I would like to remove this assumption, but have
been unable to find a rigorous proof. One approach is to compare the distance
between the initialised basis state $\left|0\right>^{\otimes N}$ and
$\left|\psi_{0,1}\right>$ or $\left|\psi_{0,SS}\right>$. However,
$\mathrm{Dist}\left[\left|0\right>,\left|\psi_{0,1}\right>\right]\nless\mathrm{Dist}\left[\left|0\right>^{\otimes
N},\left|\psi_{0,SS}\right>\right]$ for any distance metric I have found.

  * •

A single spin produces no greater measurement back-action onto the signal
compared to a single squeezed state. Again it seems like this assumption can
be made rigorous by proving that a single spin produces minimal back-action
for the amount of information extracted, and showing this is not the case for
a single squeezed state. We can remove this assumption with respect to LIGO,
since a single photon has lower radiation pressure than a squeezed state. Our
treatment of the SQL assumes that measurement back-action is zero. I.e. we
treat the signal classically and assume it has a fixed value that does not
vary. This is standard in theoretical analyses of the Heisenberg limit,
especially metrology with spins. For LIGO, this is the low light power
operation regime. Again we emphasise that as squeezed states produce more
back-action than a single photon our results apply to the high power regime.

  * •

In going from (Fisher) information to measurement uncertainty, we make
assumptions on the estimator obtained from measurements of single spin
compared to measurements of a squeezed state. Note, we do not need to assume
that the minimal variance unbiased estimator exists [Kay1993], which may not
always be the case. For readout of $\left|\psi_{1}(\theta,t)\right>$ we just
need the existence of any estimator with lower mean squared error than the
best estimator obtained from readout of $\left|\psi_{SS}(\theta,t)\right>$. In
practice, this is easy to check, since state readout of a single spin only
gives two outcomes and finding the estimator is straightforward.

  * •

We assume that the information limit for ideal measurements of a single spin
can be reached. More precisely, if the bound is not strict, meaning there is a
gap between theory and practice, that squeezed states cannot fit in this gap.
In general, if we define $t$ as the total measurement time, the finite
$\pi/2$-rotation time required to create $\left|\psi_{0}\right>$, and the
finite readout time prevent $\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$
from being reached. However this is also the case for squeezed states. We also
assume no other technical noise or decoherence, or more precisely that
measurements of single particle states are less susceptible to this noise.
Importantly, this assumption allows us to define
$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$, and we note these assumptions
are generally used in derivations of the SQL and $\Delta\tilde{\theta}_{1}$.
Going beyond this assumption seems to require a modification of the postulates
of quantum mechanics. It is a position I strongly advocate.

### 4\. Alternative proofs of a noise independent uncertainty limit

The following gives some sketches for alternative approaches to obtaining the
same proof as in the main text. They are not rigorous and more work is
required to close technical details.

  1. (1)

We could make the observation that most experimental implementations of
squeezing such as quadrature detection in optical interferometers or spin
readout in Ramsey interferometers result in a binary outcome measurement. Thus
we can use the expression given in Eq. (5):

$\mathrm{I}\left[\theta,t\right]=\left|\frac{\partial\mathrm{Pr}\left[1|\theta,t\right]}{\partial\theta}\right|^{2}\frac{1}{\mathrm{Pr}\left[1|\theta,t\right]\left(1-\mathrm{Pr}\left[1|\theta,t\right]\right)}.$

It is left as an exercise to show that any modifications to
$\mathrm{Pr}\left[1|\theta,t\right]$ that increase
$\mathrm{I}\left[\theta,t\right]$ necessarily increase
$\left|\frac{\partial\mathrm{Pr}\left[1|\theta,t\right]}{\partial\theta}\right|$.
More involved is to show that this is the case even for probability
distributions taking $2^{N}$ discrete values, thus we do not need to restrict
ourselves to a two-dimensional probability space.

  2. (2)

We can follow Wootters [Wootters1981] and show that
$\mathrm{I}\left[\theta,t\right]$ is a distance metric on state vectors,
characterising the (angular) distance between the start and final state. As
this distance is parameterized by the state evolution it follows that if state
evolution is the same for two states, then the information is also the same.
Equivalently, the information provided by a path of states is given by the
gradient vector. We need to use the fact that state vectors are normalised, so
they have the same length.

  3. (3)

In Refs. [Wootters1981, Braunstein1994, Childs2000, Giovannetti2006,
Jones2010, Pang2017] amongst others, an uncertainty bound depending only on
the maximal difference in eigenvalues of
$\frac{\partial\hat{H}(\theta,t)}{\partial\theta}$ is provided. Minimising the
uncertainty requires placing the meter in an equal superposition of states
with extremal eigenvalues, $\mu_{\mathrm{max}}(t),\;\mu_{\mathrm{min}}(t)$:

$\left|\psi(t)\right>=\frac{1}{\sqrt{2}}\left(\left|\mu_{\mathrm{max}}(t)\right>+e^{{i\mkern
1.0mu}\Phi}\left|\mu_{\mathrm{min}}(t)\right>\right),$

where $\Phi$ is an arbitrary phase. The optimal state is isomorphic to a
single spin, since it is just a 2-dimensional space and the measurement
uncertainty per unit time is limited by the maximum response speed of the
meter to changes in $\theta$.

  4. (4)

We could observe that the uncertainty on $\theta$ in Eq. (1) is neatly divided
into two time-zones (this is in contrast to Eq. (5)). The first contribution
is during sensor interaction with the signal and the second occurs after
interaction and is just readout of the sensor. Now the benefit from squeezing
closely resembles the original proposals to surpass the SQL (see Appendix 2).
But Giovannetti, Lloyd and Maccone proved that entanglement cannot provide any
benefit at the measurement stage for a restricted class of Hamiltonians
[Giovannetti2006]. For a rigorous proof we also need to show that the two
terms in Eq. (1) are independent. For example states that interact faster also
happen to be states that cannot be measured well. This happens to be true when
we consider the meter as an ensemble of spins, putting the whole ensemble in a
NOON state reduces the ability to measure the direction of the meter as
compared to $N$ copies of the same state. However this is not the case for
squeezed states. The temporal separation achieves this to the extent required,
since after interaction, the first term is fixed and we do not yet need to
make any decision on the measurement.

  5. (5)

We could apply results from quantum state tomography, where the uncertainty in
estimating the spin direction of an unknown state just depends on the number
of copies of the state [ODonnell2016, Aaronson2018].

  6. (6)

We could observe the hint provided by looking at state discrimination, i.e.
when we know that $\theta$ can take only one of two discrete values. Childs,
Renes and Preskill [Childs2000] showed that driving the state to one of two
orthogonal states minimises the discrimination uncertainty and the best one
can do is to drive as quickly as possible. There is no benefit to reducing
readout noise here since it already goes to zero.

### 5\. Proof for average expected information and mixed states

The key to these extended proofs is that we are dealing with probability
distributions over non-negative real numbers, so we don’t have to worry about
any complex number magic. In comparing summations of non-negative real numbers
where each term in one sum is less than each corresponding term in another
sum, by convexity we have an inequality.

Mixed states: To prove the claim when the squeezed state is a mixed state
$\left|\rho_{0,SS}\right>$, we can write the initial state as probability
distribution over $L$ pure states, with probabilities $p_{l}$:
$\left|\rho_{0,SS}\right>=\sum_{l=1}^{L}p_{l}\left|\psi_{0,SS}\right>_{l}$. As
we have assumed that the initial state does not depend on $\theta$, we can use
the property that the Fisher information is convex over mixed states when
$p_{l}$ is independent of $\theta$ [Liu2020].

Average expected information: Eq. (3) just provides the information around a
given value of $\theta$. The expected information, averaged over the prior
probability distribution of $\theta$ is:

$\langle\mathrm{I}\left[\theta,t\right]\rangle=\int^{\Theta}\mathrm{d}\Theta
p(\theta=\Theta)\mathrm{I}\left[\Theta,t\right],\quad\langle\mathrm{I}\left[\theta,t\right]\rangle=\sum_{j=1}^{J}p(\theta=\Theta_{j})\mathrm{I}\left[\Theta_{j},t\right]$
(6)

where $p(\theta=\Theta)$ is the prior probability that $\theta$ takes on some
value $\Theta$. The LHS expression is the expected information when the prior
probability is continuous and the RHS when $\theta$ can take $J$ discrete
values.

Given that the information on any value of $\theta$ is not greater than a
single spin, we can show that this holds for the expected information averaged
over the probability distribution of $\theta$. First we note that the optimal
state is independent of the value of $\theta$ and therefore also independent
of the prior probability of $\theta$. It is not however the case that the
optimal measurement is independent of prior probability distribution of
$\theta$. To additionally prove that no measurement can produce more
information, we use the result of Giovannetti, Lloyd and Maccone who showed
that entanglement does not provide any advantage at the measurement stage
[Giovannetti2006]. Thus we only need to consider the amount of information
encoded in the state (before measurement).

### 6\. Proof - Arbitrary time-dependent Hamiltonian: $H(\theta,t)$

We motivate the proof by noting that if we could solve the unitary evolution
of a single spin and show that it remains in the optimal state for all
evolution times, then this would imply
$\left|\left<\psi_{1}(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}=0$.
However, without a general method of solving time-evolution, even when
restricting ourselves to a single spin, we cannot make that statement. An
equivalent approach is to use the information bound proved by Pang and Jordan
for general time-dependent Hamiltonians [Pang2017]:

$\mathrm{I}\left[\theta,t\right]\leq\left|\int^{t}_{0}\mathrm{d}t^{\prime}\left(\mu_{\mathrm{max}}(t^{\prime})-\mu_{\mathrm{min}}(t^{\prime})\right)/\hbar\,\right|^{2},$
(7)

where $\mu_{\mathrm{max}}(t),\;\mu_{\mathrm{min}}(t)$ are the maximum, minimum
eigenvalues of $\frac{\partial\hat{H}(\theta,t)}{\partial\theta}$. Eq. (7)
expresses a simple concept, the information per unit time is limited by the
maximum response speed of the meter to changes in $\theta$. This quantity is
characterised by the eigenvalues of
$\frac{\partial\hat{H}(\theta,t)}{\partial\theta}$ (c.f. how the eigenvalues
of $\hat{H}$ determine the evolution speed of any quantum state). Minimising
the uncertainty requires placing the meter in an equal superposition of states
with extremal eigenvalues, $\mu_{\mathrm{max}}(t),\;\mu_{\mathrm{min}}(t)$:

$\left|\psi^{\mathrm{Opt}}(t)\right>=\frac{1}{\sqrt{2}}\left(\left|\mu_{\mathrm{max}}(t)\right>+e^{{i\mkern
1.0mu}\Phi}\left|\mu_{\mathrm{min}}(t)\right>\right),$

where $\Phi$ is an arbitrary phase [Braunstein1996, Giovannetti2006,
Childs2000].

If we could solve the unitary evolution under $\hat{H}(\theta,t)$, show that
$\left|\psi_{1}(\theta,t)\right>=\left|\psi^{\mathrm{Opt}}(t)\right>$, and
find a projective measurement that saturates the bound, then we can saturate
Eq. (7) and show that
$\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]=4\left(\frac{\mathrm{d}\left<\psi_{1}(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)$.
Unfortunately in general this is not possible: for example when both the
eigenvalues and eigenstates of $\hat{H}$ depend on $\theta$. Therefore we need
to modify our claim to address not the gradient of $\left|\psi\right>$, but
rather the projection of
$\frac{\mathrm{d}\left|\psi\right>}{\mathrm{d}\theta}$ orthogonal to
$\left|\psi\right>$:
$\left(\frac{\mathrm{d}\left|\psi\right>}{\mathrm{d}\theta}\right)_{\perp}\equiv\frac{\mathrm{d}\left|\psi\right>}{\mathrm{d}\theta}-\left|\psi\right>\left<\psi\right|\left(\frac{\mathrm{d}\left|\psi\right>}{\mathrm{d}\theta}\right)$.

Claim: If
$\left(\frac{\mathrm{d}\left<\psi_{SS}(\theta,t)\right|}{\mathrm{d}\theta}\right)_{\perp}\left(\frac{\mathrm{d}\left|\psi_{SS}(\theta,t)\right>}{\mathrm{d}\theta}\right)_{\perp}=\left(\frac{\mathrm{d}\left<\psi_{1}(\theta,t)\right|}{\mathrm{d}\theta}\right)_{\perp}\left(\frac{\mathrm{d}\left|\psi_{1}(\theta,t)\right>}{\mathrm{d}\theta}\right)_{\perp}$
then
$\mathrm{I}_{\mathrm{SS}}\left[\theta,t\right]=\mathrm{I}_{\mathrm{1}}\left[\theta,t\right]$.

Proof: As noted by Braunstein, Caves and Milburn [Braunstein1996]

$\left(\frac{\mathrm{d}\left<\psi(\theta,t)\right|}{\mathrm{d}\theta}\right)_{\perp}\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)_{\perp}=\left(\frac{\mathrm{d}\left<\psi(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)-\left|\left<\psi(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}$

and Eq. (4) proves the claim. This statement is equivalent to looking at the
data-set $\\{X_{i}|\theta,t\\}$ resulting from projective measurements of
$\left|\psi_{SS}(\theta,t)\right>$. If this data-set shows no greater
dependence on $\theta$ than a single spin, then it cannot contain
fundamentally more information on $\theta$ than projective measurements of a
single spin.

### 7\. FAQs

I find it difficult to accept your statement that the whole field is wrong
without more evidence. Could you help point out some specific errors in
papers.

I have started doing that for many experiments, see here [McGuinness2021,
McGuinness2022, McGuinness2023, McGuinness2023a]. Let’s go through an explicit
example for spin-squeezing. All you need to do is ask some simple questions
about what is going on and see if the answers make sense. Some questions that
might help:

> In spin squeezing, what entangled state is being created?

It is surprisingly difficult to find a direct answer to this question. In
place of a mathematical definition of the squeezed state, a pictorial
representation on the Bloch sphere is often used. For a system consisting of
two particles, Mark Kasevich and Vladan Vuletić inform me that the state that
everyone is preparing is [Vuletic2023]:

$\frac{1}{\sqrt{2(1+\epsilon^{2})}}\left(\epsilon{\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\left|00\right>}+\left({\color[rgb]{0.5,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0.5,0,0}\left|10\right>+\left|01\right>}\right)+\epsilon{\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\left|11\right>}\right)$
(8)

where $\epsilon$ is real, $\epsilon<1$ for a squeezed state, $\epsilon\ll 1$
in the limit of strong squeezing, and $\left|10\right>$, $\left|01\right>$
cannot be distinguished by a collective spin measurement along $z$ (I have
added colours for emphasis). Equivalent definitions and the extension to $N$
particles are given in [14, 43, 97].

Let’s assume this is the input state to a Ramsey interferometer, where a
relative phase between basis states accumulates and is compared to the phase
of the final readout pulse in the interferometer. A few characteristics of
this state to note.

  * •

The green basis states accumulate a relative phase twice that of a single
particle. If the $N$ particle squeezed state has components:
${\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\left|0\right>^{\otimes
N}}$ and
${\color[rgb]{0,0.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0.5,0}\left|1\right>^{\otimes
N}}$, they accumulate a relative phase, $N$ times that of a single particle.

  * •

The relative phase accumulated by the red basis states is not observable by a
collective spin measurement along $z$. In fact, a perfect measurement of these
basis states always gives the same observable $S_{z}=0$, independent of the
phase of the final readout pulse.

  * •

For $\epsilon=1$, although one might say the state in Eq. (8) is not squeezed,
it is still entangled. On average, we would expect to observe Ramsey fringes
with a different contrast and period compared to the same number of
independent spins. For a two-spin ensemble, if the unentangled ensemble has a
period of $2\pi$ and the fringe contrast goes from $1$ to $-1$, then the
entangled ensemble should have a period of $\pi$ and fringes going from $1/2$
to $-1/2$.

  * •

When squeezing is increased, i.e. $\epsilon\rightarrow 0$, we would expect to
see contrast of the Ramsey fringes disappear when performing a collective spin
measurement. It is true that the measurement variance is reduced, since we
always obtain the same result $S_{z}=0$, but this gives no information on the
phase of the final readout pulse. Think about what is happening here,
squeezing just reduces the amplitude of the Ramsey fringes, without affecting
the period. How can this improve the measurement precision?

Again, points to emphasise. Measurements of this state give completely
different data, compared to measurement of the same number of unentangled
spins. In particular, the period is reduced and the contrast is reduced by the
same amount. Also, when squeezing is increased, although we observe that the
measurement variance reduces it is clear that this leads to worse phase
estimation.

If we accept that the squeezed state defined above is actually being created,
the next question one could ask is:

> What experimental evidence is presented, to verify the creation of this
> state?

Note, in the following papers, a collective spin measurement along $z$ is
performed on the spin squeezed state [Appel2009, Gross2010, Leroux2010,
Leroux2010a, LouchetChauvet2010, SchleierSmith2010, Bohnet2014, Muessel2014,
Strobel2014, Cox2016, Hosten2016, Linnemann2016, Braverman2019, Malia2020,
Bao2020, PedrozoPenafiel2020, Greve2022, Malia2022, Riedel2010, Hosten2016a,
Colombo2022]. No evidence is presented of a state with the above
characteristics (except that the Ramsey fringe contrast reduces and the
measurement variance reduces). Most strikingly, when a plot of the measurement
observable vs. readout phase is presented, the period is the same as for the
unentangled ensemble, see:

  1. (1)

Figure 2 of [Gross2010]

  2. (2)

Figure 2 of [Leroux2010]. Data is only presented for the variance as a
function of rotation angle.

  3. (3)

Figure 5 of [LouchetChauvet2010]

  4. (4)

Supplementary Figure A6 of [SchleierSmith2010]

  5. (5)

Figure 2 and Figure 3c of [Bohnet2014]

  6. (6)

Figure 1c and Figure 2 of [Muessel2014]. No data is presented for the
unsqueezed ensemble, but the expected dependence on the readout phase is the
same as presented for the squeezed ensemble (a period of $2\pi$).

  7. (7)

Figure 4A of [Strobel2014]. This data corresponds to measurements on a state,
which is prepared as a spin squeezed state and evolved for 25 ms. After 25 ms
squeezing is lost, although entanglement remains?! In Figure 2, data is
presented for squeezing as a function of tomography angle.

  8. (8)

Figure 3(d) of [Cox2016]. A plot of spin noise as a function of the phase
$\psi$ of the final readout pulse.

  9. (9)

Figure 2 of [Linnemann2016] where a $2\pi$ phase dependence is observed. See
also Figure 3(a) inset.

  10. (10)

Figure 2 of [Braverman2019], data is presented for the variance as a function
of rotation angle.

  11. (11)

Figure 2c of [PedrozoPenafiel2020], a plot of variance as a function of
tomography angle is presented for the spin squeezed state

  12. (12)

Figure 2a of [Riedel2010], the same angular dependence is observed for
squeezed and unsqueezed ensembles.

  13. (13)

Figure 4c of [Colombo2022] where the phase response for a squeezed and
unsqueezed ensemble are directly compared. See also Figure 4b for a plot of
spectroscopic enhancement as a function of rotation angle for the spin
squeezed state.

Note for measurements of noise or variance, twice the angular or phase
dependence is observed compared to population measurements (i.e. half the
period). The same is also observed for measurements of unentangled ensembles
when the correct rotation direction is chosen. Sometimes the sensor response
is pictorially represented, and no experimental data is presented, see:

  1. (1)

Figure 1 of [Appel2009]

  2. (2)

Figure 2 of [Leroux2010a]

In fact, these two points: no mathematical definition of the squeezed state, a
spin response the same as an unentangled state, are what forced me to take the
approach I did in the main text.

What should be abundantly clear, is that the squeezed state defined in Eq. (8)
(or the $N$ particle equivalent) is not created. So, the natural question to
ask is

> What state is being created?

Summarizing the experimental evidence that has been presented. Collective spin
measurements of “squeezed” ensembles have the same phase dependence as
unentangled ensembles and with lower contrast. In addition, the measurement
noise (spin variance) is observed to reduce for particular readout angles. An
important piece of information that the above papers do not mention, is that
the measurement noise (spin variance) also reduces for particular readout
angles of unentangled ensembles (see for example Figure 11 of [Itano1993]).
Despite many papers claiming the contrary, it is not correct that reduced
noise for particular projective measurements is evidence of squeezing or
entanglement. Exactly the same dependence is seen for measurements on
unentangled systems.

Finally, we can read the description of how the “squeezed” state is created.
In the following papers, a “squeezed” state is prepared by performing a QND
measurement on the ensemble, i.e. a projective measurement of the spin
population along $z$ [Appel2009, Leroux2010, Leroux2010a, LouchetChauvet2010,
SchleierSmith2010, Bohnet2014, Cox2016, Hosten2016, Braverman2019, Malia2020,
Bao2020, PedrozoPenafiel2020, Greve2022, Malia2022, Hosten2016a, Colombo2022].
In general, this measurement is weak, in that it only projects a small
proportion of spins to the $\left|0\right>,\left|1\right>$ eigenstates of $z$.
After performing the QND, the authors state that the ensemble is now in an
entangled “squeezed” state.

Based on the evidence presented in these papers, this is definitely not the
description I would use. I would describe some of the spins as being in
$\left|0\right>$ or $\left|1\right>$ conditional on the QND measurement
result. That is basic quantum mechanics, after a projective measurement, the
system is in an eigenstate of the measurement observable. The spins that are
not measured, are unaffected except to experience an a.c. Stark shift from the
QND measurement. So, this is my explanation of what happening in the above
papers. The spins are prepared in an unentangled state:
$\left|x\right>^{\otimes N}$. Some of these spins are projected to
$\left|0\right>$, $\left|1\right>$. The resulting ensemble can be described as
one where most of the spins are initialised to $\left|x\right>$ and some are
in $\left|0\right>$, $\left|1\right>$. This ensemble is then used to perform
Ramsey interferometry. The resulting contrast is worse than an ensemble
initialised to $\left|x\right>^{\otimes N}$ and the phase response is the
same. As far as I am aware, this is the only description that fits to all of
the evidence presented and is theoretically consistent.

Why do you restrict your analysis to (Fisher) information, why not analyse the
actual measurement uncertainty? Are you hiding something behind this
definition?

I analyse the information and not the measurement uncertainty, for a reason
that is both somewhat uninteresting and at the same time points to a bigger
issue in quantum metrology. The reason is that it is easy to violate bounds on
the measurement uncertainty, whereas the same is not true for the information.
This makes the uncertainty bound non-rigorous.

In general, the following inequality is used to relate information (from a
single measurement) to the estimation uncertainty [Braunstein1996,
Giovannetti2004, Giovannetti2006, DemkowiczDobrzanski2012, Pang2014, Yuan2015,
Pang2017]:

$\langle(\Delta\tilde{\theta})^{2}\rangle\geq\frac{1}{\mathrm{I}\left[\theta,t\right]},$
(9)

where $\langle(\Delta\tilde{\theta})^{2}\rangle$ is the expected mean squared
error of the estimator. Thus the uncertainty $\Delta\tilde{\theta}$ is the
positive square-root of the expected mean squared error of the estimator.

For $t=0$, no information is obtained on the signal $\theta$, thus one would
expect the uncertainty to be infinite. However if we have any amount of prior
information on $\theta$, then the uncertainty bound is violated, because at
$t=0$ the uncertainty is not infinite, instead it is given by our prior
uncertainty on $\theta$. Even if we take into account the prior information,
and modify Eq. (9) to express a relative uncertainty reduction (in comparison
to the prior uncertainty), it is possible to violate the bound. Take the
following discrete prior probability distribution of $\theta$:

$p(\theta=\pi/2)=1/2;\quad\quad p(\theta=\pi)=1/2,$

i.e. $\theta$ can take one of two values with equal probability. For this
prior, our initial uncertainty in $\theta$ is $\pi/4$. Even taking into
account this information, the uncertainty bound can be violated by an
arbitrary amount. To do so, we need drive a single spin sensor to one of two
orthogonal states (e.g. $\left|0\right>$, $\left|1\right>$) depending on the
value of $\theta$, so that if $\theta=\pi/2$, the spin ends up in
$\left|0\right>$ and if $\theta=\pi$, the spin ends up in $\left|1\right>$. If
the readout of the spin is perfect, then regardless of how long it takes to
perform unitary evolution, the final uncertainty on $\theta$ is zero. Similar
arguments have been observed in obtaining violations of the Heisenberg time-
energy uncertainty relation [Aharonov2002] and used as evidence that quantum
computers can exponentially outperform classical computers [Atia2017].

These examples show that if we really want to obtain a rigorous bound on the
measurement uncertainty, we need to be careful in considering the prior
information, because the final uncertainty depends on both the information
provided from the measurement and our prior uncertainty. Mathematically, the
reason why we can violate the Cramer-Rao uncertainty bound in the above
example is because the posterior probability distribution does not satisfy the
regularity condition

$\mathbb{E}\left[\frac{\partial\mathrm{ln}\left(\mathrm{Pr}\left[X|\theta,t\right]\right)}{\partial\theta}\right]=0\quad\mathrm{for\;all}\;\theta$

where the expectation is taken with respect to
$\mathrm{Pr}\left[X|\theta,t\right]$ (see section on Cramer-Rao lower bound in
[Kay1993]). This condition is assumed in derivation of the uncertainty bound
on the mean squared error of the estimator.

Performing the above analysis adds complexity and forces us to put
restrictions on the prior probability distribution to avoid pathological
priors. This also makes the analysis less general. Most importantly it is
distracting and does not address the issue at stake. We don’t care about the
prior probability distribution and how that impacts the final measurement
uncertainty, what we want to know is the following

> Do measurements of entangled squeezed states give more information on a
> signal than measurements of unentangled states?

The resulting uncertainty may violate the bound of Eq. (9), but unless
something strange occurs to violate our understanding of information laws, we
can say that: if the answer is no, then measurements of entangled squeezed
states will not provide a lower uncertainty than measurements unentangled
states (see also the discussion in App. 3).

What about the Heisenberg uncertainty relations for conjugate observables.
Don’t they disprove this work. Are you saying they are wrong?

Instead of operating in the Schrödinger picture we could analyse how unitary
evolution acts as a mapping of quantum states in the Heisenberg picture. Then,
for $\hat{H}(\theta,t)=\theta\hat{H}$, we have:
$\frac{\mathrm{d}\hat{U}(\theta,t)\left|\psi_{0}\right>}{\mathrm{d}\theta}=-{i\mkern
1.0mu}t\hat{H}\,\hat{U}(\theta,t)/\hbar\left|\psi_{0}\right>$. As $\hat{U}$
and $\hat{H}$ commute, expressing $\left|\psi_{0}\right>$ in terms of the
eigenstates of $\hat{H}$, we have:

$\begin{split}&\left(\frac{\mathrm{d}\left<\psi(\theta,t)\right|}{\mathrm{d}\theta}\right)\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)=\left<\psi_{0}\right|\left(-{i\mkern
1.0mu}t\hat{H}\,\hat{U}(\theta,t)/\hbar\right)^{\dagger}\left(-{i\mkern
1.0mu}t\hat{H}\,\hat{U}(\theta,t)/\hbar\right)\left|\psi_{0}\right>\\\
&=(t/\hbar)^{2}\sum_{k=1}^{K}\left<\psi_{E_{k}}\right|\left(\mathrm{Exp}\left[{i\mkern
1.0mu}\theta
tE_{k}/\hbar\right]E_{k}\alpha_{i}^{\ast}\right)\left(\alpha_{i}E_{k}\mathrm{Exp}\left[-{i\mkern
1.0mu}\theta tE_{k}/\hbar\right]\right)\left|\psi_{E_{k}}\right>\\\
&=(t/\hbar)^{2}\sum_{k=1}^{K}E_{k}^{2}|\alpha_{i}|^{2},\\\ \end{split}$

and

$\left|\left<\psi(\theta,t)\right|\left(\frac{\mathrm{d}\left|\psi(\theta,t)\right>}{\mathrm{d}\theta}\right)\right|^{2}=\left|-{i\mkern
1.0mu}t/\hbar\left<\psi_{0}\right|\hat{U}^{\dagger}\hat{H}\,\hat{U}\left|\psi_{0}\right>\right|^{2}=(t/\hbar)^{2}\left(\sum_{k=1}^{K}E_{k}|\alpha_{i}|^{2}\right)^{2}.$

Thus, for this Hamiltonian, maximising information on $\theta$ means
maximising the variance of $\hat{H}$212121In general we want to maximise the
variance of the generator of translation with respect to $\theta$.:
$(\Delta\hat{H})^{2}\equiv\langle\hat{H}^{2}\rangle-\langle\hat{H}\rangle^{2}$
with respect to $\left|\psi_{0}\right>$. For unitary evolution this also means
maximising the variance of $\hat{H}$ on the output state. We have:

$\mathrm{I}\left[\theta\right]=4\left(\Delta\hat{H}\right)^{2}.$

Due to the correspondence between different pictures, increasing
$\left(\Delta\hat{H}\right)^{2}$ in this context only comes about through
increasing the state response to $\theta$. In this analysis, it is incorrect
to equate $\left(\Delta\hat{H}\right)^{2}$ as the uncertainty in estimating
the energy eigenvalues or the Hamiltonian $\hat{H}$. It is the variance of the
measurement results, assuming perfect measurements. Importantly, we see that
states with large variance in measurement outcomes (i.e. noise) give the most
information on $\theta$. This is in direct contradiction with the analysis
provided by the squeezing community.

There are some technical aspects to this analysis which means that working in
the Schrödinger picture is preferable. In the main text, we assume that two
state vectors have the same dependence on $\theta$. However in the Heisenberg
picture, the dependence is included in the operator not the state vector, so
we cannot make that assumption. Also when comparing a single spin with a
squeezed state, we are comparing the variance of two different operators with
different dimension. We don’t have that issue in the Schrödinger picture
because we are comparing inner products on normalised vectors. Even when the
vector spaces are different dimensions, this is still ok. On a related point,
defining the measurement variance is a tricky endeavour. Processing of raw
experimental data is always required to obtain meaningful estimates, we don’t
just apply a $\hat{H}$ operator to our system and receive the outcome $E_{k}$.
There is nothing to stop us from multiplying the data by a large number and
artificially increasing the variance. This operation does not improve our
measurement precision and it can be hard to identify this artificial
enhancement and separate it from actual improvements.

Take a phase estimation experiment. In one experiment, the state to be
measured is $\left(\left|1\right>+e^{{i\mkern
1.0mu}\varphi}\left|0\right>\right)/\sqrt{2}$. Compare to another experiment,
with the state $\left(\left|11\right>+e^{{i\mkern
1.0mu}\varphi}\left|00\right>\right)/\sqrt{2}$. The measurement outcomes of
the second experiment produce twice as many photons as the first. But in the
limit of perfect fidelity readout, these extra photons do not provide more
information on $\varphi$ – they are correlated! Squeezing people might claim
that because this measurement observable has twice the variance of the single
spin state, then by measuring a conjugate observable we would obtain a better
uncertainty. But this is an incorrect argument in many respects. For one, it
is a simple rescaling error. The error becomes even harder to identify when we
start talking about measurements on ensembles where the contribution of each
state gets blurred out. It is much easier to pick up on this error when
comparing individual normalised states in the Schrödinger picture (see the
comment on Assumption 1 in App. 3).