April 2023 .

Correlation function for the punctual state of the fermion string in the space
of dimension $D=10$.

Vladimir S. Dotsenko
LPTHE, Sorbonne Université, CNRS, UMR 7589
4 place Jussieu,75252 Paris Cedex 05, France.

$\@@underline{\hbox{Abstract.}}$

Correlation function is defined and calculated for the punctual states of the
fermion supersymmetric string $(N=1)$, in its critical dimension $D=10$.

##  1 Introduction.

Usually, for the strings, one studies the scattering amplitudes of physical
onshell states. On the contrary, in this paper we want to adresse the question
of the correlation functions for strings, in Euclidian space, of critical
dimensions $D=26$ for the bosonic string and $D=10$ for the fermionic string.
And this is for a particular case of the two-point correlation function of
punctual states.

For the bosonic string, in dimension $D=26$, this function has already been
considered in the literature long ago [1]. It is known to be pathological,
divergent.

We want to study a similar two point function, but for the fermionic string in
its critical dimension $D=10$.

But to start with, we shall reconsider in the next Section, with our
technique, the bosonic string two point function for the punctual states. This
is with the fully covariant formulation, with the functional integral for the
$X^{\mu}(t,x)$ string position fields, where all $\mu$ components are
integrated over, plus the ghost fields
$b(t,x),\bar{b}(t,x),c(t,x),\bar{c}(t,x)$ which are also integrated over,
which are propagating in a way that they are compensating the two nonphysical
modes of the $X^{\mu}$ fields.

In the Section 3 we shell develop a similar technique for the fermion string
two-point function, with $X^{\mu}$ and $\psi^{\mu}$ fields plus the
corresponding ghosts. We shall define and calculate the two-point correlation
function for the punctual states of the fermion string. The correlation
function we obtain is not divergent, is well defined.

The Section 4 is devoted to discussions and conclusions.

##  2 Bosonic string two-point function

The field $X^{\mu}(t,x)$ and the ghosts
$b(t,x),\bar{b}(t,x),c(t,x),\bar{c}(t,x)$, are defined on the cylinder – on
the strip $0<x<1$, $0<t<T$, Fig.1, all the fields being periodic functions on
$x$:

$\displaystyle X^{\mu}(t,1)=X^{\mu}(t,0),\quad b(t,1)=b(t,0)$ $\displaystyle
c(t,1)=c(t,0),\quad\bar{b}(t,1)=\bar{b}(t,0),\quad\bar{c}(t,1)=\bar{c}(t,0).$
(2.1)

On $t$, the field $X^{\mu}(t,x)$ has fixed boundary conditions:

$X^{\mu}(0,x)=0^{\mu},\quad X^{\mu}(T,x)=R^{\mu}$ (2.2)

which corresponds to the string being punched at points $0^{\mu}$ and
$R^{\mu}$ in the $X^{\mu}$ space, Fig.2.

$T$$t$$0$$1$$x$ Figure 1: The strip $0<x<1,0<t<T$ in the parameter space of
the variables $x,t$. $\vec{0}$$\vec{R}$$X^{\mu}-$ space Figure 2: The punctual
states correlator $<p(\vec{R})p(\vec{0})>$ of the bosonic string, eq. 2.3.

We are looking for the correlation function which we define as

$\displaystyle<p(R^{\mu})\cdot
p(0^{\mu})>=\int^{\infty}_{0}dTe^{{-A}[X_{cl}]}G_{X}(R;T)\cdot G_{b,c}(R;T)$
$\displaystyle=\int_{0}^{\infty}dT\int DX^{\mu}(t,x)\int Db(t,x)\int
D\bar{b}(t,x)\int Dc(t,x)\int D\bar{c}(t,x)$
$\displaystyle\exp\\{-A[X^{\mu}]-B[b,\bar{b},c,\bar{c}]\\}$ (2.3)

with the actions:

$A[X^{\mu}]=\int^{T}_{0}dt\int^{1}_{0}dx((\partial_{t}X)^{2}+(\partial_{x}X)^{2})$
(2.4)
$B[b,\bar{b},c,\bar{c}]=\int^{T}_{0}dt\int^{1}_{0}dx(b\bar{\partial}c+\bar{b}\partial\bar{c})$
(2.5)

1.

We calculate first the $X^{\mu}$ part of the correlator (2.3), with the
propogator, when $T$ is fixed,

$G_{X}(\vec{R};T)=\int DX^{\mu}(t,x)\exp\\{-A[X^{\mu}]\\}$ (2.6)

We take

$X^{\mu}(x,t)=X_{cl}(x,t)+X^{\mu}_{qu}(x,t)$ (2.7)
$X^{\mu}_{cl}(x,t)=(X^{\mu}_{(2)}-X^{\mu}_{(1)})\frac{t}{T}+X^{\mu}_{(1)}$
(2.8)

with

$X^{\mu}_{(1)}=0^{\mu},\quad X^{\mu}_{(2)}=R^{\mu}$ (2.9)
$X^{\mu}_{qu}(0,x)=X^{\mu}_{qu}(T,x)=0^{\mu}$ (2.10)

We have, evidently, for $A[X^{\mu}]$ in (2.4), the classical part

$A[X^{\mu}_{cl}]=T\cdot\frac{1}{T^{2}}(X^{\mu}_{(2)}-X^{\mu}_{(1)})^{2}=\frac{1}{T}(R^{\mu})^{2}$
(2.11) $\exp(-A[X^{\mu}_{cl}])=\exp(-\frac{1}{T}(R^{\mu})^{2})$ (2.12)

For the quantum part, with $X^{\mu}_{qu}(t,x)$ being periodic function of $x$,
it could be decomposed as follows:

$\displaystyle
X^{\mu}_{qu}(t,x)=\sum^{\infty}_{n=1}a^{\mu}_{n}(t)\sqrt{2}\sin(2\pi
nx)+\sum^{\infty}_{n=1}b^{\mu}_{n}(t)\sqrt{2}\cos(2\pi nx)+b^{\mu}_{0}(t)$
(2.13)

One obtains:

$\displaystyle\partial_{x}X^{\mu}_{qu}(t,x)=\sum^{\infty}_{n=1}a^{\mu}_{n}(t)\sqrt{2}\cos(2\pi
nx)\cdot 2\pi n$
$\displaystyle-\sum^{\infty}_{n=1}b^{\mu}_{n}(t)\sqrt{2}\sin(2\pi nx)\cdot
2\pi n$ (2.14)

$\displaystyle\partial_{t}X^{\mu}_{qu}(t,x)=\sum^{\infty}_{n=1}\partial_{t}a^{\mu}_{n}(t)\sqrt{2}\sin(2\pi
nx)$
$\displaystyle+\sum^{\infty}_{n=1}\partial_{t}b^{\mu}_{n}(t)\sqrt{2}\cos(2\pi
nx)+\partial_{t}b^{\mu}_{0}(t)$ (2.15)

Integrating over $x$, one obtains:

$\int^{1}_{0}dx(\partial_{x}X^{\mu}_{qu}(t,x))^{2}=\sum^{\infty}_{n=1}4\pi^{2}n^{2}[(a^{\mu}_{n}(t))^{2}+(b^{\mu}_{n}(t))^{2}]$
(2.16)

$\int^{1}_{0}dx(\partial_{t}X^{\mu}_{qu}(t,x))^{2}=\sum^{\infty}_{n=1}[(\partial_{t}a^{\mu}_{n}(t))^{2}+(\partial_{t}b_{n}^{\mu}(t))^{2}]+(\partial_{t}b_{0}^{\mu}(t))^{2}$
(2.17)

$\displaystyle
A[X^{\mu}_{qu}]=\int^{T}_{0}dt\int^{1}_{0}dx[(\partial_{x}X^{\mu}_{qu})^{2}+(\partial_{t}X^{\mu}_{qu})^{2}]$
$\displaystyle=\int^{T}_{0}dt\sum^{\infty}_{n=1}[4\pi^{2}n^{2}(a^{\mu}_{n}(t))^{2}+(\partial_{t}a^{\mu}_{n}(t))^{2}]$
$\displaystyle+\int^{T}_{0}dt\sum^{\infty}_{n=1}[4\pi
n^{2}(b^{\mu}_{n}(t))^{2}+(\partial_{t}b^{\mu}_{n}(t))^{2}]+\int^{T}_{0}dt(\partial_{t}b^{\mu}_{0}(t))^{2}$
(2.18)

With the boundary conditions zero, at $t=0$ and at $t=T$, the coefficients
$a^{\mu}_{n}(t)$, $b^{\mu}_{n}(t)$, $b^{\mu}_{0}(t)$, in $X^{\mu}_{qu}(t,x)$,
eq.(2.13), can be decomposed on sinus:

$\displaystyle
a^{\mu}_{n}(t)=\sum^{\infty}_{m=1}a^{\mu}_{nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ $\displaystyle
b^{\mu}_{n}(t)=\sum^{\infty}_{m=1}b^{\mu}_{nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ $\displaystyle
b^{\mu}_{0}(t)=\sum^{\infty}_{m=1}b^{\mu}_{0m}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ (2.19)
$\displaystyle\partial_{t}a_{n}(t)=\sum^{\infty}_{m=1}a^{\mu}_{nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\cdot\frac{\pi m}{T}$
$\displaystyle\partial_{t}b_{n}(t)=\sum^{\infty}_{m=1}b^{\mu}_{nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\cdot\frac{\pi m}{T}$
$\displaystyle\partial_{t}b_{0}(t)=\sum^{\infty}_{m=1}b^{\mu}_{0m}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\cdot\frac{\pi m}{T}$ (2.20)

For the integrals on $t$, one obtains:

$\displaystyle\int^{T}_{0}dt(a^{\mu}_{n}(t))^{2}=\sum^{\infty}_{m=1}(a^{\mu}_{nm})^{2}$
$\displaystyle\int^{T}_{0}dt(b^{\mu}_{n}(t))^{2}=\sum^{\infty}_{m=1}(b^{\mu}_{nm})^{2}$
(2.21)
$\displaystyle\int^{T}_{0}dt(\partial_{t}a^{\mu}_{n}(t))^{2}=\sum^{\infty}_{m=1}(a^{\mu}_{nm})^{2}\frac{\pi^{2}m^{2}}{T^{2}}$
$\displaystyle\int^{T}_{0}dt(\partial_{t}b^{\mu}_{n}(t))^{2}=\sum^{\infty}_{m=1}(b^{\mu}_{nm})^{2}\frac{\pi^{2}m^{2}}{T^{2}}$
$\displaystyle\int^{T}_{0}dt(\partial_{t}b^{\mu}_{0}(t))^{2}=\sum^{\infty}_{m=1}(b^{\mu}_{0m})^{2}\frac{\pi^{2}m^{2}}{T^{2}}$
(2.22)

The part of $a^{\mu}$ in $A[X^{\mu}_{qu}]$, eq.(2.18):

$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}[4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}](a^{\mu}_{nm})^{2}$
(2.23)

The part of $b^{\mu}$ in $A[X^{\mu}_{qu}]$, eq.(2.18):

$\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}[4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}](b^{\mu}_{nm})^{2}$
(2.24)

The part of $b_{0}^{\mu}$ in $A[X^{\mu}_{qu}]$, eq.(2.18):

$\sum^{\infty}_{m=1}\frac{\pi^{2}m^{2}}{T^{2}}(b_{0m}^{\mu})^{2}$ (2.25)

Altogether, by the eq.(2.18), and the sum of the equations (2.23), (2.24),
(2.25) we obtain:

$\displaystyle
A[X_{qu}]=\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}[4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}](a^{\mu}_{nm})^{2}$
$\displaystyle+(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})^{2}(b^{\mu}_{nm})^{2}]+\sum^{\infty}_{m=1}\frac{\pi^{2}m^{2}}{T^{2}}(b^{\mu}_{0m})^{2}$
(2.26)

The propagator of $X^{\mu}_{qu}$, for one $\mu$ component:

$\displaystyle G_{X}(R;T)=\int DX_{qu}\exp(-A[X_{qu}])$
$\displaystyle=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(\int^{+\infty}_{-\infty}da_{nm}\int^{+\infty}_{-\infty}db_{nm})\times\prod^{\infty}_{m=1}\int^{+\infty}_{-\infty}db_{0m}$
$\displaystyle\times\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\exp\\{-(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})a^{2}_{nm}\\}$
$\displaystyle\times\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\exp\\{-(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})b^{2}_{nm}\\}$
$\displaystyle\times\prod^{\infty}_{m=1}\exp\\{-\frac{\pi^{2}m^{2}}{T^{2}}b^{2}_{0m}\\}$
$\displaystyle\propto\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}}\times\prod^{\infty}_{m=1}\frac{T}{\pi
m}$ (2.27)
$G_{X}(R;T)\propto\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}}\times\prod^{\infty}_{m=1}\frac{T}{\pi
m}\propto\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4n^{2}+\frac{m^{2}}{T^{2}}}\times\prod^{\infty}_{m=1}\frac{T}{m}$
(2.28)

One obtains from eq.(2.28):

$G_{X}(R;T)\propto\frac{1}{\sqrt{T}}\frac{\exp\\{\frac{\pi}{6}T\\}}{\prod^{\infty}_{n=1}(1-e^{-4\pi
nT})}$ (2.29)

The formulas which have been used:

$\prod^{\infty}_{m=1}\frac{T}{m}\propto\frac{1}{\sqrt{T}}$ (2.30)
$\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4n^{2}+\frac{m^{2}}{T^{2}}}\propto
e^{\frac{\pi T}{6}}\prod^{\infty}_{n=1}\frac{1}{1-e^{-4\pi nT}}$ (2.31)

In calculation of the products in (2.30), (2.31) the Riemann $\zeta(z)$
function have been used. The details are given in the Appendix A.

2.

The ghosts part of the correlation function (2.3), with the propagator:

$\displaystyle C_{b,c}(R;T)=\int Db(t,x)\int D\bar{b}(t,x)\int Dc(t;x)\int
D\bar{c}(t;x)$ $\displaystyle\exp\\{-B[b,\bar{b},c,\bar{c}]\\}$ (2.32)

The action $B(b,\bar{b},c,\bar{c})$, eq.(2.5), can be expressed in real
components:

$\displaystyle
B(b,\bar{b},c,\bar{c})=\int^{T}_{0}dt\int^{1}_{0}dx(b\bar{\partial}c+\bar{b}\partial\bar{c})$
$\displaystyle=\int^{T}_{0}dt\int^{1}_{0}dx[b_{1}(\partial_{1}c_{1}-\partial_{2}c_{2})-b_{2}(\partial_{2}c_{1}+\partial_{1}c_{2})]$
(2.33)

We have used the expressions:

$\displaystyle
b=b_{1}+ib_{2},\,\,\,\bar{b}=b_{1}-ib_{2},\,\,\,c=c_{1}+ic_{2},\,\,\,\bar{c}=c_{1}-ic_{2}$
$\displaystyle\bar{\partial}=\frac{1}{2}(\partial_{1}+i\partial_{2}),\,\,\,\partial=\frac{1}{2}(\partial_{1}-i\partial_{2})$
(2.34)

The decompositions of the fields $b_{1},b_{2},c_{1},c_{2}$, being periodic in
$x$ with the period $0<x<1$, will be taken of the form:

$\displaystyle b_{1}(x,t)=\sum^{\infty}_{n=1}[b_{1,n}(t)\sqrt{2}\cos(2\pi
nx)+\tilde{b}_{1,n}(t)\sqrt{2}\sin(2\pi nx)+b_{1,0}(t)$ $\displaystyle
c_{1}(x,t)=\sum^{\infty}_{n=1}[c_{1,n}(t)\sqrt{2}\sin(2\pi
nx)+\tilde{c}_{1,n}(t)\sqrt{2}\cos(2\pi nx)+\tilde{c}_{1,0}(t)$ $\displaystyle
b_{2}(x,t)=\sum^{\infty}_{n=1}[b_{2,n}(t)\sqrt{2}\cos(2\pi
nx)+\tilde{b}_{2,n}(t)\sqrt{2}\sin(2\pi nx)+b_{2,0}(t)$ $\displaystyle
c_{2}(x,t)=\sum^{\infty}_{n=1}[c_{2,n}(t)\sqrt{2}\sin(2\pi
nx)+\tilde{c}_{2,n}(t)\sqrt{2}\cos(2\pi nx)+\tilde{c}_{2,0}(t)$ (2.35)

The derivatives of the fields $c_{1}(x,t),\,\,c_{2}(x,t)$, which appear in the
eq.(2.33), will be given by:

$\displaystyle\partial_{1}c_{1}(x,t)=\partial_{x}c_{1}(x,t)$
$\displaystyle=\sum^{\infty}_{n=1}[c_{2,n}(t)\sqrt{2}\cos(2\pi
nx)-\tilde{c}_{1,n}(t)\sqrt{2}\sin(2\pi nx)]2\pi n$
$\displaystyle\partial_{1}c_{2}(x,t)=\partial_{x}c_{2}(x,t)$
$\displaystyle=\sum^{\infty}_{n=1}[c_{2,n}(t)\sqrt{2}\cos(2\pi
nx)-\tilde{c}_{2,n}(t)\sqrt{2}\sin(2\pi nx)]2\pi n$
$\displaystyle\partial_{2}c_{1}(x,t)=\partial_{t}c_{1}(x,t)$
$\displaystyle=\sum^{\infty}_{n=1}[\partial_{t}c_{1,n}(t)\sqrt{2}\sin(2\pi
nx)+\partial_{t}\tilde{c}_{1,n}(t)\sqrt{2}\cos(2\pi
nx)]+\partial_{t}\tilde{c}_{1,0}(t)$
$\displaystyle\partial_{2}c_{2}(x,t)=\partial_{t}c_{2}(x,t)$
$\displaystyle=\sum^{\infty}_{n=1}[\partial_{t}c_{2,n}(t)\sqrt{2}\sin(2\pi
nx)+\partial_{t}\tilde{c}_{2,n}(t)\sqrt{2}\cos(2\pi
nx)]+\partial_{t}\tilde{c}_{2,0}(t)$ (2.36)

In the Action $B(b,\bar{b},c,\bar{c})$, eq.(2.33), when integrated on $x$,
appear four terms:

$\displaystyle\int^{1}_{0}dx\,\,b_{1}\partial_{1}c_{1}=\sum^{\infty}_{n=1}[b_{1,n}(t)c_{1,n}(t)-\tilde{b}_{1,n}(t)\tilde{c}_{1,n}(t)]2\pi
n$
$\displaystyle-\int^{1}_{0}dx\,\,b_{1}\partial_{2}c_{2}=-\sum^{\infty}_{n=1}[b_{1,n}\partial_{t}\tilde{c}_{2,n}(t)+\tilde{b}_{1,n}\partial_{t}c_{2,n}(t)]-b_{1,0}(t)\partial_{t}\tilde{c}_{2,0}(t)$
$\displaystyle-\int^{1}_{0}dx\,\,b_{2}\partial_{2}c_{1}=-\sum^{\infty}_{n=1}[b_{2,n}\partial_{t}\tilde{c}_{1,n}(t)+\tilde{b}_{2,n}(t)\partial_{t}c_{1,n}(t)]-b_{2,0}(t)\partial_{t}\tilde{c}_{1,0}(t)$
$\displaystyle-\int^{1}_{0}dx\,\,b_{2}\partial_{1}c_{2}=\sum^{\infty}_{n=1}[-b_{2,n}(t)c_{2,n}(t)+\tilde{b}_{2,n}(t)\tilde{c}_{2,n}(t)]2\pi
n$ (2.37)

Next we decompose the fields in the above expressions in $t$ modes and we
integrate them on $t$.

We start with the fields of the $x$–zero modes, $\tilde{c}_{1,0}(t)$,
$\tilde{c}_{2,0}(t)$, $b_{1,0}(t)$, $b_{2,0}(t)$, with the terms, in (2.37),

$-b_{1,0}(x)\partial_{t}\tilde{c}_{2,0}(t),\quad-
b_{2,0}(t)\partial_{t}\tilde{c}_{1,0}(t)$ (2.38)

We take the following reduced decompositions on $t$ modes:

$\displaystyle\tilde{c}_{1,0}(t)=\sum^{\infty}_{m=1}\tilde{c}_{1,0m}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$
$\displaystyle\tilde{c}_{2,0}(t)=\sum^{\infty}_{m=1}\tilde{c}_{2,0m}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ $\displaystyle
b_{1,0}(t)=\sum^{\infty}_{m=1}b_{1,0m}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$ $\displaystyle
b_{2,0}(t)=\sum^{\infty}_{m=1}b_{2,0m}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$ (2.39)
$\displaystyle\partial_{t}\tilde{c}_{1,0}(t)=\sum^{\infty}_{m=1}\tilde{c}_{1,0m}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\times\frac{\pi m}{T}$
$\displaystyle\partial_{t}\tilde{c}_{2,0}(t)=\sum^{\infty}_{m=1}\tilde{c}_{2,0m}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\times\frac{\pi m}{T}$ (2.40)

We put (2.39), (2.40) into (2.38) and we integrate them on $t$. We obtain:

$\displaystyle-\int_{0}^{T}dtb_{1,0}(x)\partial_{t}\tilde{c}_{2,0}(t)=\sum^{\infty}_{m=1}b_{1,0m}c_{2,0m}\times\frac{\pi
m}{T}$
$\displaystyle-\int_{0}^{T}dtb_{2,0}(x)\partial_{t}\tilde{c}_{1,0}(t)=\sum^{\infty}_{m=1}b_{2,0m}\tilde{c}_{2,0m}\times\frac{\pi
m}{T}$ (2.41)

The $x$ zero modes part of the functional integral for the propagator is given
by:

$\displaystyle\int Db_{1,0}(t)\int D\tilde{c}_{2,0}(t)\int Db_{2,0}(t)\int
D\tilde{c}_{1,0}(t)$
$\displaystyle\times\prod^{\infty}_{m=1}\exp\\{b_{1,0m}\tilde{c}_{1,0m}\frac{\pi
m}{T}\\}\prod^{\infty}_{m=1}\exp\\{b_{2,0m}\tilde{c}_{1,om}\frac{\pi m}{T}\\}$
$\displaystyle=\prod^{\infty}_{m=1}\int db_{1.0m}\prod_{m=1}^{\infty}\int
d\tilde{c}_{2.0m}\prod^{\infty}_{m=1}\int db_{2.0m}\prod^{\infty}_{m=1}\int
d\tilde{c}_{1,m}$
$\displaystyle\times\prod^{\infty}_{m=1}\exp\\{b_{1,0m}\tilde{c}_{2,om}\frac{\pi
m}{T}\\}\prod^{\infty}_{m=1}\exp\\{b_{2,0m}\tilde{c}_{1,om}\frac{\pi m}{T}\\}$
(2.42)

One obtains, for the zero modes part of the propagator,

$\prod^{\infty}_{m=1}(\frac{\pi m}{T})^{2}$ (2.43)

Next, for non-zero $x$ modes, remains the following expressions in eq.(2.37):

$\displaystyle\sum^{\infty}_{n=1}[b_{1,n}(t)c_{1,n}(t)+\tilde{b}_{1,n}(t)\tilde{c}_{1.n}(t)]2\pi
n$
$\displaystyle-\sum^{\infty}_{n=1}[b_{1,n}(t)\partial_{t}\tilde{c}_{2,n}(t)+\tilde{b}_{1,n}(t)\partial_{t}c_{2.n}(t)]$
$\displaystyle-\sum^{\infty}_{n=1}[b_{2,n}(t)\partial_{t}\tilde{c}_{1,n}(t)+\tilde{b}_{2,n}(t)\partial_{t}c_{1.n}(t)]$
$\displaystyle-\sum^{\infty}_{n=1}[b_{2,n}(t)c_{2,n}(t)+\tilde{b}_{2,n}(t)\tilde{c}_{2.n}(t)]2\pi
n$ (2.44)

For non-zero $x$ modes we take the following reduced expansions on $t$ modes,
similar to those for the zero $x$ modes in eq.(2.39):

$\displaystyle
c_{1,n}(t)=\sum^{\infty}_{m=1}c_{1,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$ $\displaystyle
c_{2,n}(t)=\sum^{\infty}_{m=1}c_{2,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$
$\displaystyle\tilde{c}_{1,n}(t)=\sum^{\infty}_{m=1}\tilde{c}_{1,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$
$\displaystyle\tilde{c}_{2,n}(t)=\sum^{\infty}_{m=1}\tilde{c}_{2,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ $\displaystyle
b_{1,n}(t)=\sum^{\infty}_{m=1}b_{1,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$ $\displaystyle
b_{2,n}(t)=\sum^{\infty}_{m=1}b_{2,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$
$\displaystyle\tilde{b}_{1,n}(t)=\sum^{\infty}_{m=1}\tilde{b}_{1,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$
$\displaystyle\tilde{b}_{2,n}(t)=\sum^{\infty}_{m=1}\tilde{b}_{2,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$ (2.45)

We integrate on $t$ the expression (2.44), with the decomposition on $t$ modes
in (2.45). We get:

$\displaystyle\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}[(b_{1,nm}c_{1,nm}+\tilde{b}_{1,nm}\tilde{c}_{1,nm})2\pi
n$ $\displaystyle-(b_{1,nm}\tilde{c}_{2,nm}+\tilde{b}_{1,nm}c_{2,nm})\frac{\pi
m}{T}$
$\displaystyle-(b_{2,nm}\tilde{c}_{1,nm}+\tilde{b}_{2,nm}c_{1,nm})\frac{\pi
m}{T}$ $\displaystyle-(b_{2,nm}c_{2,nm}+\tilde{b}_{2,nm}\tilde{c}_{2,nm})2\pi
n]$ (2.46)

Above is the action for ghosts, for non-zero $x$ modes.

The propagator, the non-zero $x$ modes part of ghosts, is obtained by the
functional integral of the exponential of (2.46):

$\displaystyle\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}[\int db_{1,nm}\int
d\tilde{b}_{1,nm}\int db_{2,nm}\int d\tilde{b}_{2,nm}$ $\displaystyle\int
dc_{1,nm}\int d\tilde{c}_{1,nm}\int dc_{2,nm}\int d\tilde{c}_{2,nm}]$
$\displaystyle\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\exp\\{-(b_{1,nm}c_{1,nm}+\tilde{b}_{1,nm}\tilde{c}_{1,nm})2\pi
n$ $\displaystyle+(b_{1,nm}\tilde{c}_{2,nm}+\tilde{b}_{1,nm}c_{2,nm})\frac{\pi
m}{T}$
$\displaystyle+(b_{2,nm}\tilde{c}_{1,nm}+\tilde{b}_{2,nm}c_{1,nm})\frac{\pi
m}{T}$ $\displaystyle+(b_{2,nm}c_{2,nm}+\tilde{b}_{2,nm}\tilde{c}_{2,nm})2\pi
n\\}$ (2.47)

One obtains, after integration:

$\displaystyle\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(16\pi^{4}n^{4}+8\pi^{4}n^{2}\frac{m^{2}}{T^{2}}+\frac{\pi^{4}m^{4}}{T^{4}})$
$\displaystyle=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})^{2}$
(2.48)

For the zero $x$ modes we have obtained the expression in the eq.(2.43).
Together, we find the following expression for the ghosts propagator:

$G_{b,c}(R;T)=(\prod^{\infty}_{m=1}\frac{\pi
m}{T})^{2}\times(\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}))^{2}$
(2.49)

Earlier we have obtained, eq.(2.28), when the product over the $\mu$
components is taken, the following expression for the propagator of the
$X^{\mu}_{qu}(t,x)$ field:

$G_{X}(R;T)=\prod^{26}_{\mu=1}\int
DX^{\mu}_{qu}e^{-A[X^{\mu}_{qu}]}=(\prod^{\infty}_{m=1}\frac{T}{\pi
m})^{26}(\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})})^{26}$
(2.50)

We can conclude that the ghosts compensate, exactly, the contribution to the
propagator of two components of the $X^{\mu}_{qu}$ fields, the two non-
physical fluctuations of this field. In particular, this justifies the reduced
decompositions, on $t$ \- modes, (2.39), (2.45) of the ghosts. The correlation
function $<p(R)p(0)>$ is defined by the equation (2.3). We obtain:

$\displaystyle<p(R)p(0)>=\int^{\infty}_{0}dTe^{-\frac{1}{T}R^{2}}(\prod^{\infty}_{m=1}\frac{T}{\pi
m})^{24}\times((\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T})^{2}})^{24}$
$\displaystyle\propto\int^{\infty}_{0}dTe^{-\frac{R^{2}}{T}}(\frac{1}{\sqrt{T}})^{24}(\frac{\exp\\{\frac{\pi
T}{6}\\}}{(1-e^{-4\pi nT})})^{24}$
$\displaystyle=\int^{\infty}_{0}dTe^{-\frac{1}{T}R^{2}}\frac{1}{T^{12}}\times\frac{\exp\\{4\pi
T\\}}{\prod^{\infty}_{n=1}(1-e^{-4\pi nT})^{24}}$ (2.51)

We have reobtained the well known result [1].

The expression (2.51) is highly divergent at $T\rightarrow\infty$. It is
pathological, in case of the bosonic string.

In the next Section we shall define the correlation function $<p(R)p(0)>$ for
the fermionic string, in its critical dimension $D=10$. In that case we shall
find the function which is well defined, with the $T$ integral convergent.

##  3 Fermion string two-point function.

Correlation function of the fermionic string punched at points
$\vec{X}_{1}=\vec{0}$ and $\vec{X}_{2}=\vec{R}$ of the $D$ dimensional space
$(D=10)$ is given by the general expression:

$<p(R)p(0)>=\int^{\infty}_{0}dT\,\,G(R;T)$ (3.1) $\displaystyle G(R;T)=\int
DX^{\mu}(t,x)\int D\psi^{\mu}(t,x)\int D\bar{\psi}^{\mu}(t,x)$
$\displaystyle\int Db(t,x)\int D\bar{b}(t,x)\int Dc(t,x)\int D\bar{c}(t,x)$
$\displaystyle\int D\beta(t,x)\int D\gamma(t,x)$
$\displaystyle\times\exp\\{-A_{X}[X^{\mu}]-A_{\psi,\psi}[\psi^{\mu},\bar{\psi}^{\mu}]$
$\displaystyle-A_{b,c}[b,\bar{b},c,\bar{c}]-A_{\beta,\gamma}[\beta,\gamma]\\}$
(3.2)

We have seen in Sec.2 that, for our function, the role of the ghosts,
$b,\bar{b},c,\bar{c}$ is just to compensate for two component of $X^{\mu}$.

It will be similar for our present function in (3.1), (3.2). We shall suppose
that the role of the ghosts $\beta,\gamma$ in (3.1), (3.2) will be similar,
just to compensate for two components of the fermion fields $\psi^{\mu}$ and
$\bar{\psi}^{\mu}$. Then our expression for the propagator (3.2) can be taken
in the reduced form:

$\displaystyle G(R;T)=\int DX^{\mu}(t,x)\int D\bar{\psi}^{\mu}(t,x)$
$\displaystyle\times\exp\\{-A_{X}[X^{\mu}]-A_{\psi,\bar{\psi}}[\psi^{\mu},\bar{\psi}^{\mu}]\\}$
(3.3)

with $\mu=1,...,8$ both for $X^{\mu}$ and for $\psi^{\mu}$, $\bar{\psi}^{\mu}$
fields in (3.3), with the actions:

$A_{X}[X^{\mu}]=\int^{T}_{0}dt[(\partial_{t}X^{\mu})^{2}+(\partial_{x}X^{\mu})^{2}]$
(3.4)
$A_{\psi,\bar{\psi}}[\psi^{\mu},\bar{\psi}^{\mu}]=\int^{T}_{0}dt\int^{1}_{0}dx(\psi^{\mu}\bar{\partial}\psi^{\mu}+\bar{\psi}^{\mu}\partial\bar{\psi}^{\mu})$
(3.5)

The $\mu$ components of $X^{\mu}$ and $\psi^{\mu}$, $\bar{\psi}^{\mu}$ are
independent for our correlation functions. For one $\mu$ component:

$A_{X}[X]=\int^{T}_{0}dt\int^{1}_{0}dx[(\partial_{t}X)^{2}+(\partial_{x}X)^{2}]$
(3.6)
$A_{\psi.\bar{\psi}}[\psi,\bar{\psi}]=\int^{T}_{0}dt\int^{1}_{0}dx(\psi\bar{\partial}\psi+\bar{\psi}\partial\bar{\psi})$
(3.7)

For the $X$ part of the correlation function the calculations have already
been done in the Sec.2, with the results:

$\displaystyle
G_{X}(R;T)=e^{-\frac{R^{2}}{T}}\times\prod^{\infty}_{m=1}\frac{T}{\pi
m}\times\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})}$
$\displaystyle\propto
e^{-\frac{1}{T}R^{2}}\times\frac{1}{\sqrt{T}}\frac{\exp\\{\frac{\pi
T}{6}\\}}{\prod^{\infty}_{n=1}(1-e^{-4\pi nT})}$ (3.8)

– equations (2.28), (2.29), Sec.2.

We shall now calculate the $\psi$, $\bar{\psi}$ part of the propagator
$G(R;T)$, equations (3.3) – (3.5):

$\displaystyle G_{\psi,\bar{\psi}}=\int
D\psi(t,x)D\bar{\psi}(t,x)\exp\\{-A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]\\}$
$\displaystyle=\int D\psi(t,x)\int
D\bar{\psi}(t,x)\exp\\{-\int^{T}_{0}dt\int^{1}_{0}dx(\psi\bar{\partial}\psi+\bar{\psi}\partial\bar{\psi})\\}$
(3.9)

We shall give the real form for the action
$A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]$, eq.(3.7).

$\displaystyle\psi=\psi_{1}+i\psi_{2},\quad\bar{\psi}=\psi_{1}-i\psi_{2}$
$\displaystyle\bar{\partial}=\frac{1}{2}(\partial_{1}+i\partial_{2}),\quad\partial=\frac{1}{2}(\partial_{1}-i\partial_{2}),$
$\displaystyle\partial_{1}=\partial_{x},\quad\partial_{2}=\partial_{t}$ (3.10)

$\psi_{1}$, $\psi_{2}$ above are real. We find

$A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]=\int^{T}_{0}dt\int^{1}_{0}dx(\psi_{1}\partial_{1}\psi_{1}-\psi_{1}\partial_{2}\psi_{2}-\psi_{2}\partial_{2}\psi_{1}-\psi_{2}\partial_{1}\psi_{2})$
(3.11)

We shall denote as $I_{\psi,\bar{\psi}}$ the $\psi,\bar{\psi}$ part of the
propagator:

$\displaystyle I_{\psi,\bar{\psi}}=\int D\psi_{1}\int
D\psi_{2}\exp\\{-A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]\\}$ $\displaystyle=\int
D\psi_{1}\int
D\psi_{2}\exp\\{-\int^{T}_{0}dt\int^{1}_{0}dx(\psi_{1}\partial_{1}\psi_{1}-\psi_{1}\partial_{2}\psi_{2}-\psi_{2}\partial_{2}\psi_{1}-\psi_{2}\partial_{1}\psi_{2})\\}$
(3.12)

The components $\psi_{1}(t,x)$, $\psi_{2}(t,x)$ are periodic in $x$,
$x\rightarrow x+1$. They can be decomposed as follows:

$\displaystyle\psi_{1}(t,x)=\sum^{\infty}_{n=1}[\psi_{1,n}(t)\sqrt{2}\sin(2\pi
nx)+\tilde{\psi}_{1,n}(t)\sqrt{2}\cos(2\pi nx)]+\tilde{\psi}_{1,0}(t)$
$\displaystyle\psi_{2}(t,x)=\sum^{\infty}_{n=1}[\psi_{2,n}(t)\sqrt{2}\sin(2\pi
nx)+\tilde{\psi}_{2,n}(t)\sqrt{2}\cos(2\pi nx)]+\tilde{\psi}_{2,0}(t)$ (3.13)

Their derivatives:

$\displaystyle\partial_{1}\psi_{1}(t,x)=\sum^{\infty}_{n=1}[\psi_{1,n}(t)\sqrt{2}\cos(2\pi
nx)-\tilde{\psi}_{1,n}(t)\sqrt{2}\sin(2\pi nx)]2\pi n$
$\displaystyle\partial_{1}\psi_{2}(t,x)=\sum^{\infty}_{n=1}[\psi_{2,n}(t)\sqrt{2}\cos(2\pi
nx)-\tilde{\psi}_{2,n}(t)\sqrt{2}\sin(2\pi nx)]2\pi n$
$\displaystyle\partial_{2}\psi_{1}(t,x)=\sum^{\infty}_{n=1}[\partial_{t}\psi_{1,n}(t)\sqrt{2}\sin(2\pi
nx)+\partial_{t}\tilde{\psi}_{1,n}(t)\sqrt{2}\cos(2\pi
nx)]+\partial_{t}\tilde{\psi}_{1,0}(t)$
$\displaystyle\partial_{2}\psi_{2}(t,x)=\sum^{\infty}_{n=1}[\partial_{t}\psi_{2,n}(t)\sqrt{2}\sin(2\pi
nx)+\partial_{t}\tilde{\psi}_{2,n}(t)\sqrt{2}\cos(2\pi
nx)]+\partial_{t}\tilde{\psi}_{2,0}(t)$ (3.14)

We calculate next $A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]$, eq.(3.11), by
integrating for the moment on $x$ alone. We find:

$\displaystyle
A_{\psi,\bar{\psi}}[\psi,\bar{\psi}]=\int^{T}_{0}dt\\{\sum^{\infty}_{n=1}[-2\psi_{1,n}(t)\tilde{\psi}_{1,n}(t)\times
2\pi n$
$\displaystyle-\psi_{1,n}(t)\partial_{t}\psi_{2,n}(t)-\tilde{\psi}_{1,n}(t)\partial_{t}\psi_{2,n}(t)$
$\displaystyle-\psi_{2,n}(t)\partial_{t}\psi_{1,n}(t)-\tilde{\psi}_{2,n}(t)\partial_{t}\psi_{1,n}(t)$
$\displaystyle+2\psi_{2,n}(t)\tilde{\psi}_{2,n}(t)\times 2\pi n]$
$\displaystyle-\tilde{\psi}_{1,0}(t)\partial_{t}\tilde{\psi}_{2,0}(t)-\tilde{\psi}_{2,0}(t)\partial_{t}\tilde{\psi}_{1,0}(t)\\}$
(3.15)

We decompose now $\psi_{1,n}(t)$, $\psi_{2,n}(t)$, $\tilde{\psi}_{1,n}(t)$,
$\tilde{\psi}_{2,n}(t)$ on $t$ modes. First we do it for the non-zero $x$
modes. We take the following reduced decompositions:

$\displaystyle\psi_{1,n}(t)=\sum^{\infty}_{m=1}\psi_{1,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$
$\displaystyle\psi_{2,n}(t)=\sum^{\infty}_{m=1}\psi_{2,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$
$\displaystyle\tilde{\psi}_{1,n}(t)=\sum^{\infty}_{m=1}\tilde{\psi}_{1,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})$
$\displaystyle\tilde{\psi}_{2,n}(t)=\sum^{\infty}_{m=1}\tilde{\psi}_{2,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})$ (3.16)

Their derivatives:

$\displaystyle\partial_{t}\psi_{1,n}(t)=\sum^{\infty}_{m=1}\psi_{1,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\times(\frac{\pi m}{T})$
$\displaystyle\partial_{t}\psi_{2,n}(t)=\sum^{\infty}_{m=1}\psi_{2,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})\times(-\frac{\pi m}{T})$
$\displaystyle\partial_{t}\tilde{\psi}_{1,n}(t)=\sum^{\infty}_{m=1}\tilde{\psi}_{1,nm}\sqrt{\frac{2}{T}}\cos(\frac{\pi
mt}{T})\times(\frac{\pi m}{T})$
$\displaystyle\partial_{t}\tilde{\psi}_{2,n}(t)=\sum^{\infty}_{m=1}\tilde{\psi}_{2,nm}\sqrt{\frac{2}{T}}\sin(\frac{\pi
mt}{T})\times(-\frac{\pi m}{T})$ (3.17)

Next we integrate on $t$ the action $A_{\psi\bar{\psi}}(\psi,\bar{\psi})$ in
eq.(3.15), its non zero $x$ modes part.

We find:

$\displaystyle(A_{\psi\bar{\psi}}(\psi,\bar{\psi}))_{n.z.}=\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}\\{-2\psi_{1,nm}\tilde{\psi}_{1,nm}\times
2\pi n$
$\displaystyle+(\psi_{1,nm}\psi_{2,nm}+\tilde{\psi}_{1,nm}\tilde{\psi}_{2,nm})\times\frac{\pi
m}{T}$
$\displaystyle-(\psi_{2,nm}\psi_{1,nm}+\tilde{\psi}_{2,nm}\tilde{\psi}_{1,nm})\times\frac{\pi
m}{T}$ $\displaystyle+2\psi_{2,nm}\tilde{\psi}_{2,nm}\times 2\pi n\\}$
$\displaystyle=\sum^{\infty}_{n=1}\sum^{\infty}_{m=1}\\{(-2\psi_{1,nm}\tilde{\psi}_{1,nm}+2\tilde{\psi}_{2,nm}\tilde{\psi}_{2,nm})2\pi
n$
$\displaystyle+(2\psi_{1,nm}\psi_{2,nm}+2\tilde{\psi}_{1,nm}\tilde{\psi}_{2,nm})\times\frac{\pi
m}{T}\\}$ (3.18)

Next,

$\displaystyle(I_{\psi,\bar{\psi}})_{n.z.}=\int D\psi\int
D\bar{\psi}\exp\\{-A[\psi,\bar{\psi}]\\}$
$\displaystyle=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\int d\psi_{1,nm}\int
d\psi_{2,nm}\int d\tilde{\psi}_{1,nm}\int d\tilde{\psi}_{2,nm}$
$\displaystyle\times\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\exp\\{(2\psi_{1,nm}\tilde{\psi}_{1,nm}-2\psi_{2,nm}\tilde{\psi}_{2,nm})2\pi
n$
$\displaystyle-(2\psi_{1,nm}{\psi}_{2,nm}+2\tilde{\psi}_{1,nm}\tilde{\psi}_{2,nm})\frac{\pi
m}{T}\\}$
$\displaystyle=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(16\pi^{2}n^{2}+4\frac{\pi
m^{2}}{T^{2}})=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}4(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}}),$
(3.19)

We have obtained $I_{\psi,\bar{\psi}}$, non-zero $x$ modes,

$(I_{\psi,\bar{\psi}})_{n.z.}=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}4(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})$
(3.20)

We remind that: $I_{X}$, non-zero $x$ modes,

$(I_{X})_{n.z.}=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})}$
(3.21)

– eq.(3.8).

We find that

$(I_{X})_{n.z.}\times(I_{\psi,\bar{\psi}})_{n.z.}\propto 1$ (3.22)

We have to treate now the zero $x$ modes of the fermions, $\psi$,
$\bar{\psi}$.

We shall consider the correlation function $<p(R)p(0)>$ of the fermionic
string in which the $x$ zero modes of the fermions $\tilde{\psi}_{1,0}(t)$,
$\tilde{\psi}_{2,0}(t)$, in eq.(3.15), have fixed values at $t=0$ and at
$t=T$, which means at $\vec{0}$, and at $\vec{R}$ in the $\vec{X}$ space. This
corresponds, in particular, to the limiting case of two small loops, at points
$\vec{0}$ and $\vec{R}$, the loops with the fermions having fixed values at
these loops, the loops which tend to points, at $\vec{0}$ and at $\vec{R}$.

The correlation function of this type corresponds to the energy-energy
correlation function $<\varepsilon(\vec{R})\varepsilon(\vec{0})>$ of the 3D
Ising model, in its presentation by the fermionic string, on the lattice [2] –
[6].

In our present calculations, at the critical dimension $D=10$, we are very far
from the non-critial $3D$ fermionic string. But still we consider, at $D=10$,
the object which is similar to the $<\varepsilon(R)\varepsilon(0)>$
correlation function. This is for a vague motivation.

The action for the zero $x$ modes of $\tilde{\psi}_{1}$, $\tilde{\psi}_{2}$ is
given by eq.(3.15):

$(A[\psi,\bar{\psi}])_{0}=-\int_{0}^{T}dt[\tilde{\psi}_{1,0}(t)\partial_{t}\tilde{\psi}_{2.0}(t)+\tilde{\psi}_{2,0}(t)\partial_{t}\tilde{\psi}_{1,0}(t)]$
(3.23)

We take $\tilde{\psi}_{1,0}(t)$, $\tilde{\psi}_{2,0}(t)$ in the form:

$\displaystyle\tilde{\psi}_{1,0}(t)=\Psi_{1,0}+\tilde{\tilde{\psi}}_{1,0}(t)$
$\displaystyle\tilde{\psi}_{2,0}(t)=\Psi_{2,0}+\tilde{\tilde{\psi}}_{2,0}(t)$
(3.24)

where $\Psi_{1,0}$, $\Psi_{2,0}$ are constantes and
$\tilde{\tilde{\psi}}_{1,0}(t)$, $\tilde{\tilde{\psi}}_{2,0}(t)$ are having
zero boundary values at $t=0$ and $t=T$:

$\displaystyle\tilde{\tilde{\psi}}_{1,0}(0)=\tilde{\tilde{\psi}}_{1,0}(T)=0$
$\displaystyle\tilde{\tilde{\psi}}_{2,0}(0)=\tilde{\tilde{\psi}}_{2,0}(T)=0$
(3.25)

$\tilde{\psi}_{1,0}(t)$ and $\tilde{\psi}_{2,0}(t)$ of this form, eq.(3.24),
have fixed boundary values at $t=0$ and $t=T$:

$\displaystyle\tilde{\psi}_{1,0}(0)=\tilde{\psi}_{1,0}(T)=\Psi_{1,0}$
$\displaystyle\tilde{\psi}_{2,0}(0)=\tilde{\psi}_{2,0}(T)=\Psi_{2,0}$ (3.26)

We put equations (3.24) into eq.(3.23) and we calculate the action (3.23) of
$\tilde{\psi}_{1,0}(t)$, $\tilde{\psi}_{2,0}(t)$ in eq.(3.24).

$\displaystyle(A(\psi,\bar{\psi}))_{0}=-\int^{T}_{0}dt[(\Psi_{1,0}+\tilde{\tilde{\psi}}_{1,0}(t))\partial_{t}\tilde{\tilde{\psi}}_{2,0}$
$\displaystyle+(\Psi_{2,0}+\tilde{\tilde{\psi}}_{2,0}(t))\partial_{t}\tilde{\tilde{\psi}}_{1,0}(t)]$
$\displaystyle=-\int_{0}^{T}dt[\tilde{\tilde{\psi}}_{1,0}(t)\partial_{2}\tilde{\tilde{\psi}}_{2,0}(t)+\tilde{\tilde{\psi}}_{2,0}(t)\partial_{t}\tilde{\tilde{\psi}}_{1,0}(t)]$
$\displaystyle=-2\int_{0}^{T}dt\tilde{\tilde{\psi}}_{1,0}(t)\partial_{t}\tilde{\tilde{\psi}}_{2,0}(t)$
(3.27)

Next, $\tilde{\tilde{\psi}}_{1,0}(t)$, $\tilde{\tilde{\psi}}_{2,0}(t)$ having
zero boundary values at $t=0$ and at $t=T$, eq.(3.25), they expand in sinus,
$\\{\sin(\frac{\pi mt}{T})$, $m=1,2,...\infty\\}$. Then the derivatives
$\partial_{t}\tilde{\tilde{\psi}}_{1,0}(t)$,
$\partial_{t}\tilde{\tilde{\psi}}_{2,0}(t)$, they expand in
cosinus,$\\{\cos(\frac{\pi mt}{T})$, $m=1,2,...\infty\\}$. In this case the
integral in eq.(3.27) will always be zero. So that the action of zero $x$
modes, eq.(3.23), will always be zero for $\tilde{\psi}_{1,0}(t)$,
$\tilde{\psi}_{2,0}(t)$ having fixed boundary conditions, eq.(3.26). In this
case for the propagator $(I_{\psi},\bar{\psi})_{0}$ we obtain:

$(I_{\psi,\bar{\psi}})_{0}=\exp\\{-(A[\psi,\bar{\psi}])_{0}\\}=1$ (3.28)

Now, on the whole, we have found

$I_{X}\equiv G_{X}(R;T)=e^{-\frac{R^{2}}{T}}\prod_{m=1}^{\infty}\frac{T}{\pi
m}\times\prod^{\infty}_{n=1}\prod_{m=1}^{\infty}\frac{1}{(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})}$
(3.29)

– eq.(3.8),

$I_{\psi,\bar{\psi}}=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}4(4\pi^{2}n^{2}+\frac{\pi^{2}m^{2}}{T^{2}})$
(3.30)

– eq.(3.20). Putting them together, we obtain

$G(R;T)=I_{X}\times I_{\psi,\bar{\psi}}\propto
e^{-\frac{R^{2}}{T}}\prod^{\infty}_{m=1}\frac{T}{\pi m}\propto
e^{-\frac{R^{2}}{T}}\frac{1}{\sqrt{T}}$ (3.31)

This is the result for one $\mu$ component, $\mu=1,2,...D-2=1,2,...8$. For
$D-2$ $\mu$ components,

$G(R;T)\propto
e^{-\frac{R^{2}}{T}}\times\frac{1}{(T)^{\frac{D-2}{2}}}=e^{-\frac{R^{2}}{t}}\times\frac{1}{(T)^{4}}$
(3.32)

For the correlation function $<p(R)p(0)>$, eq.(3.1),we obtain

$<p(R)p(0)>=\int^{\infty}_{0}dT\,G(R;T)=\int^{\infty}_{0}dT\,e^{-\frac{R^{2}}{T}}\times\frac{1}{T^{4}}=\frac{2}{R^{6}}$
(3.33)

The result is simple but well defined, the $T$ integral is convergent.

##  4 Conclusions.

We have defined and calculated the correlation function of punctual states of
the fermionic string theory, in its critical dimension $D=10$.

Compared to the bosonic string, the correlation function is no longer
divergent, because of compensation between the bosonic and fermionic
excitations. The correlation function is found to be simple and well defined.

Divergence of the $T$ integral for the bosonic string correlator signify that
the surfaces of the propagation of the bosonic string (bosonic surfaces)
degenerate into the long thin tubes. The bosonic surfaces are unstable in this
respect.

This does not happen for the fermion string correlator. The surfaces of its
propagation remain extended surfaces.

Back to the $3D$ Ising.

The self avoiding surfaces (the domain walls) of the partition sum for the
$3D$ Ising model, they degenerate (their dominant configurations) into the
branched polymers, the long thin tubes again, branched this time, instead of
extended surfaces [9] – [11]. The surfaces are self avoiding, but still they
are bosonic, having the same unstability.

We would expect that the fermion string surfaces, of the $3D$ Ising model [2]
– [8], will be stable, will remain extended surfaces, like their partners in
the critical dimension $D=10$.

Acknowledgments.

I am grateful to Marco Picco for the discussions and supports, and to
Valentina Dotsenko for much help in the preparation of the manuscript.

.

##  Appendix A Infinite products.

1)

$\prod^{\infty}_{m=1}\frac{T}{m}=\exp\\{\sum^{\infty}_{m=1}\log\frac{T}{m}\\}$
(A.1) $\sum^{\infty}_{m=1}\log\frac{T}{m}=\sum^{\infty}_{m=1}\log
T+\sum^{\infty}_{m=1}\frac{1}{m}$ (A.2) $\sum^{\infty}_{m=1}\log
T=\zeta(0)\log T=-\frac{1}{2}\log T=\log\frac{1}{\sqrt{T}}$ (A.3)
$\displaystyle\sum^{\infty}_{m=1}\log\frac{1}{m}=\lim_{z\rightarrow
0}\sum^{\infty}_{m=1}\frac{1}{z}[(\frac{1}{m})^{z}-1]$
$\displaystyle=\lim_{z\rightarrow
0}\frac{1}{z}[\sum^{\infty}_{m=1}(\frac{1}{m^{z}}-1)]$
$\displaystyle=\lim_{z\rightarrow
0}\frac{1}{z}[\zeta(z)-\zeta(0)]=\zeta^{\prime}(0)=-\frac{1}{2}\log
2\pi=\log\frac{1}{\sqrt{2\pi}}$ (A.4)

Eq.(A.2):

$\sum^{\infty}_{m=1}\log\frac{T}{m}=\log\frac{1}{\sqrt{T}}+\log\frac{1}{\sqrt{2\pi}}=\log\frac{1}{\sqrt{2\pi
T}}$ (A.5)

Eq.(A.1):

$\prod^{\infty}_{m=1}\frac{t}{m}=\exp\\{\log\frac{1}{\sqrt{2\pi
T}}\\}=\frac{1}{\sqrt{2\pi T}}$ (A.6)
$\prod^{\infty}_{m=1}\frac{T^{2}}{m^{2}}=\frac{1}{2\pi T}$ (A.7)

Equation (A.6) could also be be factorized:

$\prod^{\infty}_{m=1}\frac{1}{m}=\frac{1}{\sqrt{2\pi}}$ (A.8)
$\prod^{\infty}_{m=1}T=\frac{1}{\sqrt{T}}$ (A.9)

2)

$\displaystyle\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4n^{2}+\frac{m^{2}}{T^{2}}}$
$\displaystyle=\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}(\frac{T^{2}}{m^{2}}\times\frac{1}{1+\frac{4n^{2}T^{2}}{m^{2}}})$
$\displaystyle=\prod^{\infty}_{n=1}(\prod^{\infty}_{m-1}\frac{T^{2}}{m^{2}}\times\prod^{\infty}_{m=1}\frac{1}{1+\frac{4n^{2}T^{2}}{m^{2}}})$
(A.10) $\prod^{\infty}_{m=1}\frac{T^{2}}{m^{2}}=\frac{1}{2\pi T}$ (A.11)
$\prod^{\infty}_{m=1}\frac{1}{1+\frac{4n^{2}T^{2}}{m^{2}}}=(\prod^{\infty}_{m=1}(1+\frac{4n^{2}T^{2}}{m^{2}}))^{-1}$
(A.12)

we shall use the formula:

$\prod^{\infty}_{m=1}(1+\frac{a^{2}}{m^{2}})=\frac{\sinh\pi a}{\pi
a}=\frac{1}{2\pi a}(e^{\pi a}-e^{-\pi a})=\frac{e^{\pi a}}{2\pi a}(1-e^{-2\pi
a})$ (A.13)

We find:

$\prod^{\infty}_{m=1}(1+\frac{4n^{2}T^{2}}{m^{2}})^{-1}=[\frac{1}{2\pi\cdot
2nT}e^{2\pi nT}(1-e^{-4\pi nT})]^{-1}=4\pi nT\frac{e^{-2\pi nT}}{1-e^{-4\pi
nT}}$ (A.14)

We put (A.11) and (A.14) into (A.10):

$\displaystyle\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4n^{2}+\frac{m^{2}}{T^{2}}}=\prod^{\infty}_{n=1}(\frac{1}{2\pi
T}\times 4\pi nT\frac{e^{-2\pi nT}}{1-e^{-4\pi nT}})$
$\displaystyle=\prod^{\infty}_{n=1}2n\times\exp\\{-2\pi
T\sum^{\infty}_{n=1}n\\}\times\frac{1}{\prod^{\infty}_{n=1}(1-e^{-4\pi nT})}$
(A.15)

We shall use once again the $\zeta$ function value:

$\sum^{\infty}_{n=1}n=\zeta(-1)=-\frac{1}{12}$ (A.16)

Finally:

$\prod^{\infty}_{n=1}\prod^{\infty}_{m=1}\frac{1}{4n^{2}+\frac{m^{2}}{T^{2}}}=\prod^{\infty}_{n=1}2n\times\frac{e^{\frac{\pi}{6}T}}{\prod^{\infty}_{n=1}(1-e^{-4\pi
nT})}\propto\frac{e^{\frac{\pi}{6}T}}{\prod^{\infty}_{n=1}(1-e^{-4\pi nT)}}$
(A.17)

We have used:

$\prod^{\infty}_{n=1}2n=\prod^{\infty}_{n=1}2\times\prod^{\infty}_{n=1}n=\frac{1}{\sqrt{2}}\times\sqrt{2\pi}=\sqrt{\pi}\propto
1$ (A.18)

## References

  * [1] J.Polchinski, Preprint, unpublished.
  * [2] A.M.Polyakov, Phys.Lett.82B (1979), 247.
  * [3] Vl.S.Dotsenko and A.M.Polyakov, Advanced Stadies in Pure Mathematics. v.16 (1988), 171.
  * [4] S.Samuel, J.Math. Phys. 21 (1980) 2806,2815,2820.
  * [5] E.Fradkin, M.Srednicki and L.Susskind, Phys.Rev. D21 (1980) 2885.
  * [6] A.Casher, D.Foerster and P. Windey, Nucl.Phys. B251, (1985) 29.
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# Sequential product on standard effect algebra ${\cal E}(H)$††thanks: This
project is supported by Natural Science Found of China (10771191 and
10471124).

Shen Jun1,2, Wu Junde1 E-mail<EMAIL_ADDRESS>
###### Abstract

A quantum effect is an operator $A$ on a complex Hilbert space $H$ that
satisfies $0\leq A\leq I$, ${\cal E}(H)$ is the set of all quantum effects on
$H$. In 2001, Professor Gudder and Nagy studied the sequential product $A\circ
B=A^{\frac{1}{2}}BA^{\frac{1}{2}}$ of $A,B\in{\cal E}(H)$. In 2005, Professor
Gudder asked: Is $A\circ B=A^{\frac{1}{2}}BA^{\frac{1}{2}}$ the only
sequential product on ${\cal E}(H)$? Recently, Liu and Wu presented an example
to show that the answer is negative. In this paper, firstly, we characterize
some algebraic properties of the abstract sequential product on ${\cal E}(H)$;
secondly, we present a general method for constructing sequential products on
${\cal E}(H)$; finally, we study some properties of the sequential products
constructed by the method.

1Department of Mathematics, Zhejiang University, Hangzhou 310027, P. R. China

2Department of Mathematics, Anhui Normal University, Wuhu 241003, P. R. China

Key Words. Quantum effect, standard effect algebra, sequential product.

1\. Introduction

Sequential effect algebra is an important model for studying the quantum
measurement theory ([1-7]). A sequential effect algebra is an effect algebra
which has a sequential product operation. Firstly, we recall some elementary
notations and results.

An effect algebra is a system $(E,0,1,\oplus)$, where 0 and 1 are distinct
elements of $E$ and $\oplus$ is a partial binary operation on $E$ satisfying
that [8]:

(EA1) If $a\oplus b$ is defined, then $b\oplus a$ is defined and $b\oplus
a=a\oplus b$.

(EA2) If $a\oplus(b\oplus c)$ is defined, then $(a\oplus b)\oplus c$ is
defined and

$(a\oplus b)\oplus c=a\oplus(b\oplus c).$

(EA3) For each $a\in E$, there exists a unique element $b\in E$ such that
$a\oplus b=1$.

(EA4) If $a\oplus 1$ is defined, then $a=0$.

In an effect algebra $(E,0,1,\oplus)$, if $a\oplus b$ is defined, we write
$a\bot b$. For each $a\in(E,0,1,\oplus)$, it follows from (EA3) that there
exists a unique element $b\in E$ such that $a\oplus b=1$, we denote $b$ by
$a^{\prime}$. Let $a,b\in(E,0,1,\oplus)$, if there exists a $c\in E$ such that
$a\bot c$ and $a\oplus c=b$, then we say that $a\leq b$. It follows from [8]
that $\leq$ is a partial order of $(E,0,1,\oplus)$ and satisfies that for each
$a\in E$, $0\leq a\leq 1$, $a\bot b$ if and only if $a\leq b^{\prime}$.

Let $(E,0,1,\oplus,\circ)$ be an effect algebra and $a\in E$. If $a\wedge
a^{\prime}=0$, then $a$ is said to be a sharp element of $E$. We denote
$E_{S}$ the set of all sharp elements of $E$ ([9-10]).

A sequential effect algebra is an effect algebra $(E,0,1,\oplus)$ with another
binary operation $\circ$ defined on it satisfying [2]:

(SEA1) The map $b\mapsto a\circ b$ is additive for each $a\in E$, that is, if
$b\bot c$, then $a\circ b\bot a\circ c$ and $a\circ(b\oplus c)=a\circ b\oplus
a\circ c$.

(SEA2) $1\circ a=a$ for each $a\in E$.

(SEA3) If $a\circ b=0$, then $a\circ b=b\circ a$.

(SEA4) If $a\circ b=b\circ a$, then $a\circ b^{\prime}=b^{\prime}\circ a$ and
$a\circ(b\circ c)=(a\circ b)\circ c$ for each $c\in E$.

(SEA5) If $c\circ a=a\circ c$ and $c\circ b=b\circ c$, then $c\circ(a\circ
b)=(a\circ b)\circ c$ and $c\circ(a\oplus b)=(a\oplus b)\circ c$ whenever
$a\bot b$.

If $(E,0,1,\oplus,\circ)$ is a sequential effect algebra, then the operation
$\circ$ is said to be a sequential product on $(E,0,1,\oplus)$. If
$a,b\in(E,0,1,\oplus,\circ)$ and $a\circ b=b\circ a$, then $a$ and $b$ is said
to be sequentially independent and is denoted by $a|b$ ([1-2]).

Let $H$ be a complex Hilbert space, ${\cal B}(H)$ be the set of all bounded
linear operators on $H$, ${\cal P}(H)$ be the set of all projections on $H$,
${\cal E}(H)$ be the set of all self-adjoint operators on $H$ satisfying that
$0\leq A\leq I$. For $A,B\in{\cal E}(H)$, we say that $A\oplus B$ is defined
if $A+B\in{\cal E}(H)$, in this case, we define $A\oplus B=A+B$. It is easy to
see that $({\cal E}(H),0,I,\oplus)$ is an effect algebra, we call it standard
effect algebra ([8]). Each element $A$ in ${\cal E}(H)$ is said to be a
quantum effect, the set ${\cal E}(H)_{S}$ of all sharp elements of $({\cal
E}(H),0,I,\oplus)$ is just ${\cal P}(H)$ ([2, 9]).

Let $A\in{\cal B}(H)$, we denote $Ker(A)=\\{x\in H\mid Ax=0\\}$,
$Ran(A)=\\{Ax\mid x\in H\\}$, $P_{Ker(A)}$ denotes the projection onto
$Ker(A)$. Let $x\in H$ be a unit vector, $P_{x}$ denotes the projection onto
the one-dimensional subspace spanned by $x$.

In 2001 and 2002, Professor Gudder, Nagy and Greechie showed that for any two
quantum effects $A$ and $B$, if we define $A\circ
B=A^{\frac{1}{2}}BA^{\frac{1}{2}}$, then the operation $\circ$ is a sequential
product on the standard effect algebra $({\cal E}(H),0,I,\oplus)$, moreover,
they studied some properties of this special sequential product on $({\cal
E}(H),0,I,\oplus)$ ([1,2]).

In 2005, Professor Gudder asked ([4]): Is $A\circ
B=A^{\frac{1}{2}}BA^{\frac{1}{2}}$ the only sequential product on standard
effect algebra $({\cal E}(H),0,I,\oplus)$?

In 2009, Liu and Wu constructed a new sequential product on $({\cal
E}(H),0,I,\oplus)$, thus answered Gudder’s problem negatively ([7]). This new
sequential product on $({\cal E}(H),0,I,\oplus)$ motivated us to study the
following topics: (1) Characterize the algebraic properties of abstract
sequential product on $({\cal E}(H),0,I,\oplus)$. (2) Present a general method
for constructing sequential product on $({\cal E}(H),0,I,\oplus)$. (3)
Characterize some elementary properties of the sequential product constructed
by the method. Our results generalize many conclusions in [1,3,7,14].

2\. Abstract sequential product on $({\cal E}(H),0,I,\oplus)$

In this section, we study some elementary properties of the abstract
sequential product on the standard effect algebra $({\cal E}(H),0,I,\oplus)$.

Lemma 2.1 ([2]). Let $(E,0,1,\oplus,\circ)$ be a sequential effect algebra,
$a\in E$. Then the following conditions are all equivalent:

(1) $a\in E_{S}$;

(2) $a\circ a^{\prime}=0$;

(3) $a\circ a=a$.

Lemma 2.2 ([2]). Let $(E,0,1,\oplus,\circ)$ be a sequential effect algebra,
$a\in E$, $b\in E_{S}$. Then the following conditions are all equivalent:

(1) $a\leq b$;

(2) $a\circ b=b\circ a=a$.

Lemma 2.3 ([2, 8]). Let $(E,0,1,\oplus,\circ)$ be a sequential effect algebra,
$a,b,c\in E$.

(1) If $a\perp b$, $a\perp c$ and $a\oplus b=a\oplus c$, then $b=c$.

(2) $a\circ b\leq a$.

(3) If $a\leq b$, then $c\circ a\leq c\circ b$.

Lemma 2.4 ([7]). Let $\circ$ be a sequential product on the standard effect
algebra $({\cal E}(H),0,I,\oplus)$. Then for any $A,B\in{\cal E}(H)$ and real
number $t$, $0\leq t\leq 1$, we have $(tA)\circ B=A\circ(tB)=t(A\circ B)$.

Lemma 2.5 ([1]). Let $A,B,C\in{\cal B}(H)$ and $A,B,C$ be self-adjoint
operators. If for every unit vector $x\in H$, $\langle Cx,x\rangle=\langle
Ax,x\rangle\langle Bx,x\rangle$, then $A=tI$ or $B=tI$ for some real number
$t$.

Lemma 2.6 ([11]). Let $A\in{\cal B}(H)$ have the following operator matrix
form

$A=\left(\begin{array}[]{cc}A_{11}&A_{12}\\\ A_{21}&A_{22}\\\
\end{array}\right)$

with respect to the space decomposition $H=H_{1}\oplus H_{2}$. Then $A\geq 0$
iff

(1) $A_{ii}\in{\cal B}(H_{i})$ and $A_{ii}\geq 0$, $i=1,2$;

(2) $A_{21}=A_{12}^{*}$;

(3) there exists a linear operator $D$ from $H_{2}$ into $H_{1}$ such that
$||D||\leq 1$ and $A_{12}=A_{11}^{\frac{1}{2}}DA_{22}^{\frac{1}{2}}$.

Theorem 2.1. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $B\in{\cal E}(H)$, $E\in{\cal P}(H)$. Then $E\circ B=EBE$.

Proof. For $A\in{\cal E}(H)$, let $\Phi_{A}:{\cal E}(H)\longrightarrow{\cal
E}(H)$ be defined by $\Phi_{A}(C)=A\circ C$ for each $C\in{\cal E}(H)$. It
follows from Lemma 2.4 and (SEA1) that $\Phi_{A}$ is affine on the convex set
${\cal E}(H)$. Note that ${\cal E}(H)$ generates algebraically the vector
space ${\cal B}(H)$, so $\Phi_{A}$ has a unique linear extension to ${\cal
B}(H)$, which we also denote by $\Phi_{A}$. Then $\Phi_{A}$ is a positive
linear operator on ${\cal B}(H)$ and $\Phi_{A}(I)=A$. Thus $\Phi_{A}$ is
continuous.

Note that $E\in{\cal P}(H)={\cal E}(H)_{S}$, it follows from Lemma 2.1 that
$E\circ(I-E)=0$ and so $\Phi_{E}(I-E)=0$. By composing $\Phi_{E}$ with all
states on ${\cal B}(H)$ and using Schwarz’s inequality, we conclude that
$\Phi_{E}(B)=\Phi_{E}(EBE)$. Since $EBE\in{\cal E}(H)$, $E\in{\cal E}(H)_{S}$
and $EBE\leq E$, by Lemma 2.2 we have $E\circ(EBE)=EBE$. Thus $E\circ
B=\Phi_{E}(B)=\Phi_{E}(EBE)=E\circ(EBE)=EBE$.

Theorem 2.2. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $A,B\in{\cal E}(H)$ and $AB=BA$. Then $A\circ B=B\circ
A=AB$.

Proof. We use the notations as in the proof of Theorem 2.1.

Suppose $E\in{\cal P}(H)$ and $E\in\\{A\\}^{\prime}$, i.e., $EA=AE$. Note that
$EAE,(I-E)A(I-E)\in{\cal E}(H)$, $EAE\leq E$ and $(I-E)A(I-E)\leq I-E$, by
Lemma 2.2, it follows that $EAE|E$ and $(I-E)A(I-E)|(I-E)$. Since
$A=EAE+(I-E)A(I-E)$, by (SEA4) and (SEA5) we have $A|E$. By Theorem 2.1 we
conclude that $A\circ E=E\circ A=EAE=AE$. Thus, $\Phi_{A}(E)=AE$. Since
$\Phi_{A}$ is a continuous linear operator and $\\{A\\}^{\prime}$ is a von
Neumann algebra, we conclude that $\Phi_{A}(B)=AB$. That is, $A\circ B=AB$.
Similarly, we have $B\circ A=BA$. Thus $A\circ B=B\circ A=AB$.

Theorem 2.3. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $A,B\in{\cal E}(H)$. Then the following conditions are all
equivalent:

(1) $AB=BA=B$;

(2) $A\circ B\geq B$;

(3) $A\circ B=B$;

(4) $B\circ A=B$;

(5) $B\leq P_{Ker(I-A)}$;

(6) $B\leq A^{n}$ for each positive integer $n$.

Proof. (1)$\Rightarrow$(3) and (1)$\Rightarrow$(4): By Theorem 2.2.

(3)$\Rightarrow$(2) is obvious.

(4)$\Rightarrow$(3): By Theorem 2.2, $B\circ A=B=B\circ I$. Thus, it follows
from Lemma 2.3 that $B\circ(I-A)=0$. By (SEA3), $B|(I-A)$. By (SEA4), $B|A$.
So $A\circ B=B\circ A=B$.

(2)$\Rightarrow$(6): By using Theorem 2.2 and Lemma 2.3 repeatedly, we have:

$B\leq A\circ B\leq A\circ I=A$;

$A\circ B\leq A\circ(A\circ B)\leq A\circ A=A^{2}$;

$A\circ(A\circ B)\leq A\circ\Big{(}A\circ(A\circ B)\Big{)}\leq A\circ
A^{2}=A^{3}$;

$\vdots$

$A\circ\cdots\circ(A\circ B)\leq A\circ\Big{(}A\circ\cdots\circ(A\circ
B)\Big{)}\leq A\circ A^{n-1}=A^{n}$.

The above showed that $B\leq A^{n}$ for each positive integer $n$.

(6)$\Rightarrow$(5): Let $\chi_{\\{1\\}}$ be the characteristic function of
$\\{1\\}$. Note that $0\leq A\leq I$, it is easy to know that $\\{A^{n}\\}$
converges to $\chi_{\\{1\\}}(A)=P_{Ker(I-A)}$ in the strong operator topology.
Thus $B\leq P_{Ker(I-A)}$.

(5)$\Rightarrow$(1): Since $0\leq B\leq P_{Ker(I-A)}$, we have
$Ker(P_{Ker(I-A)})\subseteq Ker(B)$. So $Ran(B)\subseteq
Ran(P_{Ker(I-A)})=Ker(I-A)$. Thus $(I-A)B=0$. That is, $AB=B$. Taking adjoint,
we get $AB=BA=B$.

Theorem 2.4. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $A,B\in{\cal E}(H)$. Then the following conditions are all
equivalent:

(1) $C\circ(A\circ B)=(C\circ A)\circ B$ for every $C\in{\cal E}(H)$;

(2) $\langle(A\circ B)x,x\rangle=\langle Ax,x\rangle\langle Bx,x\rangle$ for
every $x\in H$ with $\|x\|=1$;

(3) $A=tI$ or $B=tI$ for some real number $0\leq t\leq 1$.

Proof. By Lemma 2.5, we conclude that (2)$\Rightarrow$(3). By Theorem 2.2 and
Lemma 2.4, (3)$\Rightarrow$(1) is trivial.

(1)$\Rightarrow$(2): If (1) hold, then $P_{x}\circ(A\circ B)=(P_{x}\circ
A)\circ B$ for every $x\in H$ with $\|x\|=1$. By Theorem 2.1,
$P_{x}\circ(A\circ B)=P_{x}(A\circ B)P_{x}=\langle(A\circ B)x,x\rangle P_{x}$.
By Theorem 2.1 and Lemma 2.4, $(P_{x}\circ A)\circ B=(P_{x}AP_{x})\circ
B=(\langle Ax,x\rangle P_{x})\circ B=\langle Ax,x\rangle(P_{x}\circ B)=\langle
Ax,x\rangle P_{x}BP_{x}=\langle Ax,x\rangle\langle Bx,x\rangle P_{x}$. Thus
(2) hold.

Theorem 2.5. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $B\in{\cal E}(H)$, $E\in{\cal P}(H)$. Then the following
conditions are all equivalent:

(1) $E\circ B\leq B$;

(2) $EB=BE$;

(3) $E\circ B=B\circ E$.

Proof. (2)$\Rightarrow$(3): By Theorem 2.2.

(3)$\Rightarrow$(1): By Lemma 2.3.

(1)$\Rightarrow$(2): Since $E\in{\cal P}(H)$, by Theorem 2.1, $E\circ B=EBE$.
Thus, $B-EBE\geq 0$. Note that

$B-EBE=\left(\begin{array}[]{cc}0&EB(I-E)\\\ (I-E)BE&(I-E)B(I-E)\\\
\end{array}\right)$

with respect to the space decomposition $H=Ran(E)\oplus Ker(E)$, so by Lemma
2.6 we have $EB(I-E)=(I-E)BE=0$. Thus $B=EBE+(I-E)B(I-E)$. So $EB=BE$.

Theorem 2.6. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $A,B,C\in{\cal E}(H)$. If $A$ is invertible, then the
following conditions are all equivalent:

(1) $B\leq C$;

(2) $A\circ B\leq A\circ C$.

Proof. (1)$\Rightarrow$(2): By Lemma 2.3.

(2)$\Rightarrow$(1): It is easy to see that $\|A^{-1}\|^{-1}A^{-1}\in{\cal
E}(H)$.

By Lemma 2.3, $(\|A^{-1}\|^{-1}A^{-1})\circ(A\circ
B)\leq(\|A^{-1}\|^{-1}A^{-1})\circ(A\circ C)$.

By Theorem 2.2, $(\|A^{-1}\|^{-1}A^{-1})|A$ and $(\|A^{-1}\|^{-1}A^{-1})\circ
A=\|A^{-1}\|^{-1}I$.

By (SEA4) and Theorem 2.2, we have

$(\|A^{-1}\|^{-1}A^{-1})\circ(A\circ B)=\Big{(}(\|A^{-1}\|^{-1}A^{-1})\circ
A\Big{)}\circ B=(\|A^{-1}\|^{-1}I)\circ B=\|A^{-1}\|^{-1}B$,

$(\|A^{-1}\|^{-1}A^{-1})\circ(A\circ C)=\Big{(}(\|A^{-1}\|^{-1}A^{-1})\circ
A\Big{)}\circ C=(\|A^{-1}\|^{-1}I)\circ C=\|A^{-1}\|^{-1}C$.

So $B\leq C$.

Corollary 2.1. Let $\circ$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, $A,B,C\in{\cal E}(H)$. If $A$ is invertible, then the
following conditions are all equivalent:

(1) $B=C$;

(2) $A\circ B=A\circ C$.

3\. General method for constructing sequential products on ${\cal E}(H)$

In the sequel, unless specified, suppose $H$ be a finite dimensional complex
Hilbert space, ${\mathbf{C}}$ be the set of complex numbers, ${\mathbf{R}}$ be
the set of real numbers, for each $A\in{\cal E}(H)$, $sp(A)$ be the spectra of
$A$ and $B(sp(A))$ be the set of all bounded complex Borel functions on
$sp(A)$.

Let $A,B\in{\cal B}(H)$, if there exists a complex constant $\xi$ such that
$|\xi|=1$ and $A=\xi B$, then we denote $A\approx B$.

In [7], Liu and Wu showed that if we define $A\circ
B=A^{\frac{1}{2}}f_{i}(A)Bf_{-i}(A)A^{\frac{1}{2}}$ for $A,B\in{\cal E}(H)$,
where $f_{z}(t)=\exp z(\ln t)$ if $t\in(0,1]$ and $f_{z}(0)=0$, then $\circ$
is a sequential product on $({\cal E}(H),0,I,\oplus)$, this result answered
Gudder’s problem negatively.

Now, we present a general method for constructing sequential products on
${\cal E}(H)$.

For each $A\in{\cal E}(H)$, take a $f_{A}\in B(sp(A))$.

Define $A\diamond B=f_{A}(A)B\overline{f_{A}}(A)$ for $A,B\in{\cal E}(H)$.

we say the set $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition if the following two conditions hold:

(i) For every $A\in{\cal E}(H)$ and $t\in sp(A)$, $|f_{A}(t)|=\sqrt{t}$;

(ii) For any $A,B\in{\cal E}(H)$, if $AB=BA$, then $f_{A}(A)f_{B}(B)\approx
f_{AB}(AB)$.

If $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product condition,
then it is easy to see that

(1) $f_{A}(A)\overline{f_{A}}(A)=\overline{f_{A}}(A)f_{A}(A)=A$,
$(f_{A}(A))^{*}=\overline{f_{A}}(A)$.

(2) If $0\in sp(A)$, then $f_{A}(0)=0$.

(3) If $A=\sum\limits^{n}_{k=1}\lambda_{k}E_{k}$, where
$\\{E_{k}\\}^{n}_{k=1}$ are pairwise orthogonal projections, then
$f_{A}(A)=\sum\limits^{n}_{k=1}f_{A}(\lambda_{k})E_{k}$.

(4) For each $E\in{\cal P}(H)$, $f_{E}(E)=f_{E}(0)(I-E)+f_{E}(1)E=f_{E}(1)E$.

(5) for any $A,B\in{\cal E}(H)$, $A\diamond B\in{\cal E}(H)$.

Lemma 3.1 ([12]). Let $H$ be a complex Hilbert space, $A,B\in{\cal B}(H)$,
$A,B,AB$ be three normal operators, and at least one of $A,B$ be a compact
operator. Then $BA$ is also a normal operator.

Lemma 3.2 ([13]). If $M,N,T\in{\cal B}(H)$, $M,N$ are normal operators and
$MT=TN$, then $M^{*}T=TN^{*}$.

Lemma 3.3. Suppose $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfy sequential product
condition and $A,B\in{\cal E}(H)$. If $A\diamond B=B\diamond A$ or $A\diamond
B=\overline{f_{B}}(B)Af_{B}(B)$, then $AB=BA$.

Proof. If $A\diamond B=B\diamond A$, that is,
$f_{A}(A)B\overline{f_{A}}(A)=f_{B}(B)A\overline{f_{B}}(B)$, then
$f_{A}(A)\overline{f_{B}}(B)f_{B}(B)\overline{f_{A}}(A)=f_{B}(B)\overline{f_{A}}(A)f_{A}(A)\overline{f_{B}}(B)$,
so $f_{A}(A)\overline{f_{B}}(B)$ is normal. By Lemma 3.1, we have
$\overline{f_{B}}(B)f_{A}(A)$ is also normal. Note that
$\Big{(}f_{A}(A)\overline{f_{B}}(B)\Big{)}f_{A}(A)=f_{A}(A)\Big{(}\overline{f_{B}}(B)f_{A}(A)\Big{)}$,
by using Lemma 3.2, we have
$\Big{(}f_{A}(A)\overline{f_{B}}(B)\Big{)}^{*}f_{A}(A)=f_{A}(A)\Big{(}\overline{f_{B}}(B)f_{A}(A)\Big{)}^{*}$.
That is, $f_{B}(B)A=Af_{B}(B)$. Taking adjoint, we have
$\overline{f_{B}}(B)A=A\overline{f_{B}}(B)$. Thus,
$AB=A\overline{f_{B}}(B)f_{B}(B)=\overline{f_{B}}(B)Af_{B}(B)=\overline{f_{B}}(B)f_{B}(B)A=BA$.

If $A\diamond B=\overline{f_{B}}(B)Af_{B}(B)$, that is,
$f_{A}(A)B\overline{f_{A}}(A)=\overline{f_{B}}(B)Af_{B}(B)$, the proof is
similar, we omit it.

Lemma 3.4. Suppose $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfy sequential product
condition and $A,B\in{\cal E}(H)$. If $AB=BA$, then $A\diamond B=B\diamond
A=AB$.

Proof. Since $AB=BA$, by sequential product condition (i) we have $A\diamond
B=f_{A}(A)B\overline{f_{A}}(A)=|f_{A}|^{2}(A)B=AB$. Similarly, $B\diamond
A=f_{B}(B)A\overline{f_{B}}(B)=|f_{B}|^{2}(B)A=AB$. Thus $A\diamond
B=B\diamond A=AB$.

Lemma 3.5. Suppose $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfy sequential product
condition and $A,B\in{\cal E}(H)$. If $AB=BA$, then for every $C\in{\cal
E}(H)$, $A\diamond(B\diamond C)=(A\diamond B)\diamond C$.

Proof. By Lemma 3.4, $A\diamond B=AB$. By sequential product condition (ii),
there exists a complex constant $\xi$ such that $|\xi|=1$ and
$f_{A}(A)f_{B}(B)=\xi f_{AB}(AB)$. Taking adjoint, we have
$\overline{f_{B}}(B)\overline{f_{A}}(A)=\overline{\xi}\,\overline{f_{AB}}(AB)$.
Thus,
$f_{A}(A)f_{B}(B)C\overline{f_{B}}(B)\overline{f_{A}}(A)=f_{AB}(AB)C\overline{f_{AB}}(AB)=f_{A\diamond
B}(A\diamond B)C\overline{f_{A\diamond B}}(A\diamond B)$. That is,
$A\diamond(B\diamond C)=(A\diamond B)\diamond C$.

Lemma 3.6 ([1]). If $y,z\in H$ and $|\langle y,x\rangle|=|\langle z,x\rangle|$
for every $x\in H$, then there exists a $c\in{\mathbf{C}}$, $|c|=1$, such that
$y=cz$.

Lemma 3.7 ([14]). Let $f:H\longrightarrow{\mathbf{C}}$ be a mapping,
$T\in{\cal B}(H)$. If the operator $S:H\longrightarrow H$ defined by
$S(x)=f(x)T(x)$ is linear, then $f(x)=f(y)$ for every $x,y\not\in Ker(T)$.

Lemma 3.8. Let $f:H\longrightarrow{\mathbf{C}}$ be a mapping, $T\in{\cal
B}(H)$. If the operator $S:H\longrightarrow H$ defined by $S(x)=f(x)T(x)$ is
linear, then there exists a constant $\xi\in{\mathbf{C}}$ such that $S(x)=\xi
T(x)$ for every $x\in H$.

Proof. By Lemma 3.7, there exists a constant $\xi\in{\mathbf{C}}$ such that
$S(x)=\xi T(x)$ for every $x\not\in Ker(T)$. Of course, $S(x)=0=\xi T(x)$ for
every $x\in Ker(T)$. So $S(x)=\xi T(x)$ for every $x\in H$.

Our main result in the section is the following.

Theorem 3.1. For each $A\in{\cal E}(H)$, take a $f_{A}\in B(sp(A))$. Define
$A\diamond B=f_{A}(A)B\overline{f_{A}}(A)$ for $A,B\in{\cal E}(H)$. Then
$\diamond$ is a sequential product on $({\cal E}(H),0,I,\oplus)$ iff the set
$\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product condition.

Proof. (1) Firstly we suppose $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfy
sequential product condition, we show that $({\cal E}(H),0,I,\oplus,\diamond)$
is a sequential effect algebra.

(SEA1) is obvious.

By Lemma 3.4, $I\diamond B=B$ for each $B\in{\cal E}(H)$, so (SEA2) hold.

We verify (SEA3) as follows:

If $A\diamond B=0$, then $f_{A}(A)B\overline{f_{A}}(A)=0$, so
$f_{A}(A)B^{\frac{1}{2}}=0$, thus, we have
$AB=\overline{f_{A}}(A)f_{A}(A)B^{\frac{1}{2}}B^{\frac{1}{2}}=0$. Taking
adjoint, we have $AB=BA$. So $A\diamond B=B\diamond A$.

We verify (SEA4) as follows:

If $A\diamond B=B\diamond A$, then by Lemma 3.3, $AB=BA$. So $A(I-B)=(I-B)A$.
By Lemma 3.4, we have $A\diamond(I-B)=(I-B)\diamond A$. By Lemma 3.5,
$A\diamond(B\diamond C)=(A\diamond B)\diamond C$ for every $C\in{\cal E}(H)$.

We verify (SEA5) as follows:

If $C\diamond A=A\diamond C$ and $C\diamond B=B\diamond C$, then by Lemma 3.3,
$AC=CA$, $BC=CB$. So (SEA5) follows easily by Lemma 3.4.

Thus, we proved that $({\cal E}(H),0,I,\oplus,\diamond)$ is a sequential
effect algebra.

(2) Now we suppose $\diamond$ be a sequential product on $({\cal
E}(H),0,I,\oplus)$, we show that the set $\\{f_{A}\\}_{A\in{\cal E}(H)}$
satisfies sequential product condition.

Since $({\cal E}(H),0,I,\oplus,\diamond)$ is a sequential effect algebra, by
Theorem 2.2, for each $A\in{\cal E}(H)$, $A\diamond I=A$, thus
$|f_{A}|^{2}(A)=A$. If $A=\sum\limits^{n}_{k=1}\lambda_{k}E_{k}$, where
$\\{E_{k}\\}^{n}_{k=1}$ are pairwise orthogonal projections,
$\sum\limits^{n}_{k=1}E_{k}=I$, then $sp(A)=\\{\lambda_{k}\\}$,
$|f_{A}|^{2}(A)=\sum\limits^{n}_{k=1}|f_{A}(\lambda_{k})|^{2}E_{k}$. Thus
$|f_{A}(\lambda_{k})|=\sqrt{\lambda_{k}}$ and $\\{f_{A}\\}_{A\in{\cal E}(H)}$
satisfies sequential product condition (i).

To prove $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition (ii), let $A,B\in{\cal E}(H)$ and $AB=BA$. By Theorem 2.2, we have
$A\diamond B=B\diamond A=AB$. Thus by (SEA4), $A\diamond(B\diamond
C)=(A\diamond B)\diamond C$ for every $C\in{\cal E}(H)$.

Let $x\in H$, $\|x\|=1$, $C=P_{x}$. Then for every $y\in H$, we have

$\langle
f_{A}(A)f_{B}(B)P_{x}\overline{f_{B}}(B)\overline{f_{A}}(A)y,y\rangle$
$=\langle\Big{(}A\diamond(B\diamond P_{x})\Big{)}y,y\rangle$
$=\langle\Big{(}(A\diamond B)\diamond P_{x}\Big{)}y,y\rangle$
$=\langle\Big{(}(AB)\diamond P_{x}\Big{)}y,y\rangle$ $=\langle
f_{AB}(AB)P_{x}\overline{f_{AB}}(AB)y,y\rangle\ .$

Since

$\langle
f_{A}(A)f_{B}(B)P_{x}\overline{f_{B}}(B)\overline{f_{A}}(A)y,y\rangle=|\langle\overline{f_{B}}(B)\overline{f_{A}}(A)y,x\rangle|^{2}\
,$ $\langle
f_{AB}(AB)P_{x}\overline{f_{AB}}(AB)y,y\rangle=|\langle\overline{f_{AB}}(AB)y,x\rangle|^{2}\
,$

we have
$|\langle\overline{f_{B}}(B)\overline{f_{A}}(A)y,x\rangle|=|\langle\overline{f_{AB}}(AB)y,x\rangle|$
for every $x,y\in H$.

By Lemma 3.6, there exists a complex function g on $H$ such that $|g(x)|\equiv
1$ and $\overline{f_{B}}(B)\overline{f_{A}}(A)x=g(x)\overline{f_{AB}}(AB)x$
for every $x\in H$. By Lemma 3.8, there exists a constant $\xi\in{\mathbf{C}}$
such that $|\xi|=1$ and
$\overline{f_{B}}(B)\overline{f_{A}}(A)x=\xi\overline{f_{AB}}(AB)x$ for every
$x\in H$. So we conclude that
$\overline{f_{B}}(B)\overline{f_{A}}(A)=\xi\overline{f_{AB}}(AB)$. Taking
adjoint, we have $f_{A}(A)f_{B}(B)=\overline{\xi}f_{AB}(AB)$. Thus
$f_{A}(A)f_{B}(B)\approx f_{AB}(AB)$. This showed that the set
$\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product condition.

Theorem 3.1 present a general method for constructing sequential products on
${\cal E}(H)$. Now, we give two examples.

Example 3.1. Let $g$ be a bounded complex Borel function on $[0,1]$ such that

$|g(t)|=\sqrt{t}$ for each $t\in[0,1]$ , $g(t_{1}t_{2})=g(t_{1})g(t_{2})$ for
any $t_{1},t_{2}\in[0,1]$ .

For each $A\in{\cal E}(H)$, let $f_{A}=g|_{sp(A)}$. Then it is easy to know
that $\\{f_{A}\\}$ satisfies sequential product condition. So by Theorem 3.1,
$A\diamond B=f_{A}(A)B\overline{f_{A}}(A)=g(A)B\overline{g}(A)$ defines a
sequential product on the standard effect algebra $({\cal E}(H),0,I,\oplus)$.

It is clear that Example 3.1 generalized Liu and Wu’s result in [7].

Example 3.2. Let $H$ be a 2-dimensional complex Hilbert space,
$\Gamma=\\{\gamma\mid\gamma$ be a decomposition of $I$ into two rank-one
orthogonal projections$\\}$. For each $\gamma\in\Gamma$, we can represent
$\gamma$ by a pair of rank-one orthogonal projections $(E_{1},E_{2})$, if
$A\in{\cal E}(H)$, $A\not\in span\\{I\\}$ and
$A=\sum\limits^{2}_{k=1}\lambda_{k}E_{k}$, then we say that $A$ can be
diagonalized by $\gamma$.

For each $\gamma\in\Gamma$, we take a $\xi(\gamma)\in\mathbf{R}$. If
$A\in{\cal E}(H)$, $A\not\in span\\{I\\}$ and $A$ can be diagonalized by
$\gamma$, let $f_{A}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for $t\in sp(A)$.

If $A\in{\cal E}(H)$ and $A=\lambda I$, let $f_{A}(t)=\sqrt{t}$ for $t\in
sp(A)$.

Then the set $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition (see the proof below). So by Theorem 3.1, $A\diamond
B=f_{A}(A)B\overline{f_{A}}(A)$ defines a sequential product on the standard
effect algebra $({\cal E}(H),0,I,\oplus)$.

Proof. Obviously $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition (i).

Now we show that $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition (ii). Let $A,B\in{\cal E}(H)$, $AB=BA$.

(1) If $A=\sum\limits^{2}_{k=1}\lambda_{k}E_{k}$,
$B=\sum\limits^{2}_{k=1}\mu_{k}E_{k}$, $\lambda_{1}\neq\lambda_{2}$,
$\mu_{1}\neq\mu_{2}$, let $\gamma=(E_{1},E_{2})$, we have
$f_{A}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for $t\in sp(A)$,
$f_{B}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for $t\in sp(B)$. So
$f_{A}(A)=A^{\frac{1}{2}+\xi(\gamma)i}=\sum\limits^{2}_{k=1}\lambda_{k}^{\frac{1}{2}+\xi(\gamma)i}E_{k}$,
$f_{B}(B)=B^{\frac{1}{2}+\xi(\gamma)i}=\sum\limits^{2}_{k=1}\mu_{k}^{\frac{1}{2}+\xi(\gamma)i}E_{k}$.

(1a) If $\lambda_{1}\mu_{1}=\lambda_{2}\mu_{2}$, then
$AB=\lambda_{1}\mu_{1}I$, so $f_{AB}(t)=t^{\frac{1}{2}}$ for $t\in sp(AB)$,
thus we have $f_{AB}(AB)=(AB)^{\frac{1}{2}}=\sqrt{\lambda_{1}\mu_{1}}I$,
$f_{A}(A)f_{B}(B)=\sum\limits^{2}_{k=1}(\lambda_{k}\mu_{k})^{\frac{1}{2}+\xi(\gamma)i}E_{k}=(\lambda_{1}\mu_{1})^{\frac{1}{2}+\xi(\gamma)i}I=(\lambda_{1}\mu_{1})^{\xi(\gamma)i}f_{AB}(AB)\approx
f_{AB}(AB)$.

(1b) If $\lambda_{1}\mu_{1}\neq\lambda_{2}\mu_{2}$, then
$AB=\sum\limits^{2}_{k=1}\lambda_{k}\mu_{k}E_{k}$, so
$f_{AB}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for $t\in sp(AB)$,
$f_{AB}(AB)=(AB)^{\frac{1}{2}+\xi(\gamma)i}=\sum\limits^{2}_{k=1}(\lambda_{k}\mu_{k})^{\frac{1}{2}+\xi(\gamma)i}E_{k}$,
thus we have
$f_{A}(A)f_{B}(B)=\sum\limits^{2}_{k=1}(\lambda_{k}\mu_{k})^{\frac{1}{2}+\xi(\gamma)i}E_{k}=f_{AB}(AB)$.

(2) If $A=\lambda I$, $B=\sum\limits^{2}_{k=1}\mu_{k}E_{k}$,
$\mu_{1}\neq\mu_{2}$, let $\gamma=(E_{1},E_{2})$. Then we have
$f_{A}(t)=t^{\frac{1}{2}}$ for $t\in sp(A)$,
$f_{B}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for $t\in sp(B)$. So
$f_{A}(A)=A^{\frac{1}{2}}=\sqrt{\lambda}I$,
$f_{B}(B)=B^{\frac{1}{2}+\xi(\gamma)i}=\sum\limits^{2}_{k=1}\mu_{k}^{\frac{1}{2}+\xi(\gamma)i}E_{k}$,
$AB=\sum\limits^{2}_{k=1}\lambda\mu_{k}E_{k}$.

(2a) If $\lambda=0$, then $AB=0$, $f_{AB}(t)=t^{\frac{1}{2}}$ for $t\in
sp(AB)$, so $f_{AB}(AB)=(AB)^{\frac{1}{2}}=0$. Thus
$f_{A}(A)f_{B}(B)=0=f_{AB}(AB)$.

(2b) If $\lambda\neq 0$, then $f_{AB}(t)=t^{\frac{1}{2}+\xi(\gamma)i}$ for
$t\in sp(AB)$. So
$f_{AB}(AB)=(AB)^{\frac{1}{2}+\xi(\gamma)i}=\lambda^{\frac{1}{2}+\xi(\gamma)i}\sum\limits^{2}_{k=1}(\mu_{k})^{\frac{1}{2}+\xi(\gamma)i}E_{k}$.
Thus
$f_{A}(A)f_{B}(B)=\sqrt{\lambda}\sum\limits^{2}_{k=1}(\mu_{k})^{\frac{1}{2}+\xi(\gamma)i}E_{k}\approx
f_{AB}(AB)$.

(3) If $A=\lambda I$, $B=\mu I$, then $f_{A}(t)=t^{\frac{1}{2}}$ for $t\in
sp(A)$, $f_{B}(t)=t^{\frac{1}{2}}$ for $t\in sp(B)$. So
$f_{A}(A)=A^{\frac{1}{2}}=\sqrt{\lambda}I$,
$f_{B}(B)=B^{\frac{1}{2}}=\sqrt{\mu}I$. $AB=\lambda\mu I$,
$f_{AB}(t)=t^{\frac{1}{2}}$ for $t\in sp(AB)$,
$f_{AB}(AB)=(AB)^{\frac{1}{2}}=\sqrt{\lambda\mu}I$. Thus
$f_{A}(A)f_{B}(B)=f_{AB}(AB)$.

It follows from (1)-(3) that the set $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies
sequential product condition (ii).

4\. Properties of the sequential product $\diamond$ on $({\cal
E}(H),0,I,\oplus)$

Now, we study some elementary properties of the sequential product $\diamond$
defined in Section 3.

In this section, unless specified, we follow the notations in Section 3. We
always suppose $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition. So by Theorem 3.1 $\diamond$ is a sequential product on the
standard effect algebra $(\varepsilon(H),0,I,\oplus)$.

Lemma 4.1. If $C\in{\cal E}(H)$, $0\leq t\leq 1$, then $f_{tC}(tC)\approx
f_{tI}(t)f_{C}(C)$.

Proof. Since $\\{f_{A}\\}_{A\in{\cal E}(H)}$ satisfies sequential product
condition, $f_{tC}(tC)\approx f_{tI}(tI)f_{C}(C)=f_{tI}(t)f_{C}(C)$.

Lemma 4.2. Let $A\in{\cal E}(H)$, $x\in H$, $\|x\|=1$, $\|f_{A}(A)x\|\neq 0$,
$y=\frac{f_{A}(A)x}{\|f_{A}(A)x\|}$. Then $A\diamond
P_{x}=\|f_{A}(A)x\|^{2}P_{y}$.

Proof. For each $z\in H$, $(A\diamond
P_{x})z=f_{A}(A)P_{x}\overline{f_{A}}(A)z=\langle\overline{f_{A}}(A)z,x\rangle
f_{A}(A)x=\langle z,f_{A}(A)x\rangle f_{A}(A)x=\|f_{A}(A)x\|^{2}P_{y}z$. So
$A\diamond P_{x}=\|f_{A}(A)x\|^{2}P_{y}$.

Lemma 4.3. Let $M\subseteq{\cal B}(H)$ be a von Neumann algebra, $P$ be a
minimal projection in $M$, $A\in M$, $x\in Ran(P)$, $\|x\|=1$. Then
$PAP=\omega_{x}(A)P$, where $\omega_{x}(A)=\langle Ax,x\rangle$.

Proof. Since $P$ is a minimal projection in $M$, by [15, Proposition 6.4.3],
$PAP=\lambda P$ for some complex number $\lambda$. Thus $\langle
PAPx,x\rangle=\langle\lambda Px,x\rangle$. So $\lambda=\omega_{x}(A)$.

Theorem 4.1. Let $A,B\in{\cal E}(H)$. Then the following conditions are all
equivalent:

(1) $AB=BA$;

(2) $A\diamond B=B\diamond A$;

(3) $A\diamond(B\diamond C)=(A\diamond B)\diamond C$ for every $C\in{\cal
E}(H)$.

Proof. (1)$\Rightarrow$(2): By Theorem 2.2.

(2)$\Rightarrow$(1): By Lemma 3.3.

(1)$\Rightarrow$(3): By Lemma 3.5.

(3)$\Rightarrow$(1): Let $x\in H$, $\|x\|=1$, $C=P_{x}$. Then for each $y\in
H$,

$\langle
f_{A}(A)f_{B}(B)P_{x}\overline{f_{B}}(B)\overline{f_{A}}(A)y,y\rangle$
$=\langle\Big{(}A\diamond(B\diamond P_{x})\Big{)}y,y\rangle$
$=\langle\Big{(}(A\diamond B)\diamond P_{x}\Big{)}y,y\rangle$ $=\langle
f_{A\diamond B}(A\diamond B)P_{x}\overline{f_{A\diamond B}}(A\diamond
B)y,y\rangle\ .$

Since

$\langle
f_{A}(A)f_{B}(B)P_{x}\overline{f_{B}}(B)\overline{f_{A}}(A)y,y\rangle=|\langle\overline{f_{B}}(B)\overline{f_{A}}(A)y,x\rangle|^{2}\
,$ $\langle f_{A\diamond B}(A\diamond B)P_{x}\overline{f_{A\diamond
B}}(A\diamond B)y,y\rangle=|\langle\overline{f_{A\diamond B}}(A\diamond
B)y,x\rangle|^{2}\ ,$

we have
$|\langle\overline{f_{B}}(B)\overline{f_{A}}(A)y,x\rangle|=|\langle\overline{f_{A\diamond
B}}(A\diamond B)y,x\rangle|$ for every $x,y\in H$.

By Lemma 3.6, there exists a complex function $g$ on $H$ such that $|g(x)|=1$
and $\overline{f_{B}}(B)\overline{f_{A}}(A)x=g(x)\overline{f_{A\diamond
B}}(A\diamond B)x$ for every $x\in H$.

By Lemma 3.8, there exists a constant $\xi$ such that $|\xi|=1$ and
$\overline{f_{B}}(B)\overline{f_{A}}(A)x=\xi\overline{f_{A\diamond
B}}(A\diamond B)x$ for every $x\in H$.

So we conclude that
$\overline{f_{B}}(B)\overline{f_{A}}(A)=\xi\overline{f_{A\diamond
B}}(A\diamond B)$.

Taking adjoint, we have $f_{A}(A)f_{B}(B)=\overline{\xi}f_{A\diamond
B}(A\diamond B)$. Thus
$\overline{f_{B}}(B)Af_{B}(B)=\overline{f_{B}}(B)\overline{f_{A}}(A)f_{A}(A)f_{B}(B)=\xi\overline{f_{A\diamond
B}}(A\diamond B)\overline{\xi}f_{A\diamond B}(A\diamond B)=A\diamond B$. That
is, $A\diamond B=\overline{f_{B}}(B)Af_{B}(B)$, so by Lemma 3.3, we have
$AB=BA$.

Theorem 4.2. Let $A,B\in{\cal E}(H)$. If $A\diamond B\in{\cal P}(H)$, then
$AB=BA$.

Proof. If $A\diamond B=0$, then by (SEA3) we have $A\diamond B=B\diamond A$,
so by Theorem 4.1 we have $AB=BA$.

If $A\diamond B\neq 0$. Firstly, we let $x\in Ran(A\diamond B)$ and $\|x\|=1$.
Then $f_{A}(A)B\overline{f_{A}}(A)x=x$. So $\langle
B\overline{f_{A}}(A)x,\overline{f_{A}}(A)x\rangle=1$. By Schwarz inequality,
we conclude that $B\overline{f_{A}}(A)x=\overline{f_{A}}(A)x$. Thus
$Ax=f_{A}(A)\overline{f_{A}}(A)x=f_{A}(A)B\overline{f_{A}}(A)x=x$. So $1\in
sp(A)$ and $B\overline{f_{A}}(A)x=\overline{f_{A}}(A)x=\overline{f_{A}}(1)x$.

Next, we let $x\in Ker(A\diamond B)$ and $\|x\|=1$. Then
$f_{A}(A)B\overline{f_{A}}(A)x=0$. So $\langle
B\overline{f_{A}}(A)x,\overline{f_{A}}(A)x\rangle=0$. We conclude that
$B\overline{f_{A}}(A)x=0$.

Thus, we always have $B\overline{f_{A}}(A)=\overline{f_{A}}(1)(A\diamond B)$.
That is, $f_{A}(1)B\overline{f_{A}}(A)=A\diamond B$.

Taking adjoint, we have
$f_{A}(1)B\overline{f_{A}}(A)=\overline{f_{A}}(1)f_{A}(A)B$.

By Lemma 3.2, we have
$\overline{f_{A}}(1)Bf_{A}(A)=f_{A}(1)\overline{f_{A}}(A)B$. So
$f_{A}(1)\overline{f_{A}}(A)B$ is self-adjoint. By [15, Proposition 3.2.8], we
have

$sp\Big{(}f_{A}(1)\overline{f_{A}}(A)B\Big{)}\backslash\\{0\\}=sp\Big{(}f_{A}(1)B\overline{f_{A}}(A)\Big{)}\backslash\\{0\\}=sp(A\diamond
B)\backslash\\{0\\}\subseteq\mathbf{R^{+}}$.

Thus we conclude that $f_{A}(1)\overline{f_{A}}(A)B\geq 0$.

Since
$\Big{(}f_{A}(1)\overline{f_{A}}(A)B\Big{)}^{2}=\Big{(}\overline{f_{A}}(1)Bf_{A}(A)\Big{)}\Big{(}f_{A}(1)\overline{f_{A}}(A)B\Big{)}=BAB=\Big{(}f_{A}(1)B\overline{f_{A}}(A)\Big{)}\Big{(}\overline{f_{A}}(1)f_{A}(A)B\Big{)}=(A\diamond
B)^{2}$, by the uniqueness of positive square root, we have
$f_{A}(1)\overline{f_{A}}(A)B=A\diamond B$. That is,
$f_{A}(1)\overline{f_{A}}(A)B=\overline{f_{A}}(1)Bf_{A}(A)=f_{A}(1)B\overline{f_{A}}(A)=\overline{f_{A}}(1)f_{A}(A)B=A\diamond
B$. Thus,
$BA=f_{A}(1)B\overline{f_{A}}(A)\overline{f_{A}}(1)f_{A}(A)=f_{A}(1)\overline{f_{A}}(A)B\overline{f_{A}}(1)f_{A}(A)=f_{A}(1)\overline{f_{A}}(A)\overline{f_{A}}(1)f_{A}(A)B=AB$.

Theorem 4.3. Let $A,B\in{\cal E}(H)$. Then the following conditions are all
equivalent:

(1) $A\diamond(C\diamond B)=(A\diamond C)\diamond B$ for every $C\in{\cal
E}(H)$;

(2) $C\diamond(A\diamond B)=(C\diamond A)\diamond B$ for every $C\in{\cal
E}(H)$;

(3) $\langle(A\diamond B)x,x\rangle=\langle Ax,x\rangle\langle Bx,x\rangle$
for every $x\in H$ with $\|x\|=1$;

(4) $A=tI$ or $B=tI$ for some $0\leq t\leq 1$.

Proof. By Theorem 2.4, we conclude that
(2)$\Longleftrightarrow$(3)$\Longleftrightarrow$(4).

(4)$\Rightarrow$(1) follows from Lemma 2.4 and Theorem 2.2 easily.

(1)$\Rightarrow$(4): If (1) hold, then $A\diamond(P_{x}\diamond B)=(A\diamond
P_{x})\diamond B$ for each $x\in H$ with $\|x\|=1$. Without lose of
generality, we suppose $\|f_{A}(A)x\|\neq 0$. Let
$y=\frac{f_{A}(A)x}{\|f_{A}(A)x\|}$.

By Lemma 4.2 and Theorem 2.1,

$A\diamond(P_{x}\diamond B)=f_{A}(A)(P_{x}BP_{x})\overline{f_{A}}(A)$
$=f_{A}(A)(\langle Bx,x\rangle P_{x})\overline{f_{A}}(A)$ $=\langle
Bx,x\rangle(A\diamond P_{x})$ $=\|f_{A}(A)x\|^{2}\langle Bx,x\rangle P_{y}\ .$

By Lemma 4.1 and Lemma 4.2,

$(A\diamond P_{x})\diamond B=(\|f_{A}(A)x\|^{2}P_{y})\diamond B$
$=f_{\|f_{A}(A)x\|^{2}P_{y}}(\|f_{A}(A)x\|^{2}P_{y})B\overline{f_{\|f_{A}(A)x\|^{2}P_{y}}}(\|f_{A}(A)x\|^{2}P_{y})$
$=f_{\|f_{A}(A)x\|^{2}I}(\|f_{A}(A)x\|^{2})f_{P_{y}}(P_{y})B\overline{f_{\|f_{A}(A)x\|^{2}I}}(\|f_{A}(A)x\|^{2})\overline{f_{P_{y}}}(P_{y})$
$=\|f_{A}(A)x\|^{2}P_{y}BP_{y}$ $=\|f_{A}(A)x\|^{2}\langle By,y\rangle P_{y}\
.$

Thus $\langle Bx,x\rangle=\langle By,y\rangle$. So we have
$\langle\overline{f_{A}}(A)Bf_{A}(A)x,x\rangle=\langle Ax,x\rangle\langle
Bx,x\rangle$. By Lemma 2.5, we conclude that (4) hold.

Theorem 4.4. Let $A\in{\cal E}(H)$, $E\in{\cal P}(H)$. Then the following
conditions are all equivalent:

(1) $A\diamond E\leq E$;

(2) $E\overline{f_{A}}(A)(I-E)=0$.

Proof. Since $E\in{\cal P}(H)$ and $\|\overline{f_{A}}(A)\|\leq 1$, we have

$A\diamond E\leq E\Longleftrightarrow\langle
f_{A}(A)E\overline{f_{A}}(A)x,x\rangle\leq\langle Ex,x\rangle\hbox{ for every
$x\in H$}$ $\Longleftrightarrow\|E\overline{f_{A}}(A)x\|\leq\|Ex\|\hbox{ for
every $x\in H$}$ $\Longleftrightarrow\overline{f_{A}}(A)\mid_{Ker(E)}\subseteq
Ker(E)$ $\Longleftrightarrow E\overline{f_{A}}(A)(I-E)=0\ .$

Corollary 4.1 [14]. Let $A\in{\cal E}(H)$, $E\in{\cal P}(H)$. Then the
following conditions are all equivalent:

(1) $A^{\frac{1}{2}}EA^{\frac{1}{2}}\leq E$;

(2) $AE=EA$.

Proof. (2)$\Rightarrow$(1) is trivial.

(1)$\Rightarrow$(2): Let $f_{B}(t)=\sqrt{t}$ for each $B\in{\cal E}(H)$ and
$t\in sp(B)$, then $\\{f_{B}\\}_{B\in{\cal E}(H)}$ satisfies sequential
product condition. For this sequential product, $A\diamond
E=A^{\frac{1}{2}}EA^{\frac{1}{2}}$. So by Theorem 4.4 we have
$EA^{\frac{1}{2}}(I-E)=0$. That is, $EA^{\frac{1}{2}}=EA^{\frac{1}{2}}E$.
Taking adjoint, we have $EA^{\frac{1}{2}}=A^{\frac{1}{2}}E$. Thus $AE=EA$.

Corollary 4.2. Let $M\subseteq{\cal B}(H)$ be a von Neumann algebra, ${\cal
E}(M)=\\{A\in M|0\leq A\leq I\\}$, $P$ or $I-P$ be a minimal projection in
$M$. Then for every $A\in{\cal E}(M)$, the following conditions are all
equivalent:

(1) $A\diamond P\leq P$;

(2) $AP=PA$.

Proof. (2)$\Rightarrow$(1): By Theorem 2.2, $A\diamond P=AP=PAP\leq P$.

(1)$\Rightarrow$(2): If $P$ is a minimal projection in $M$, then by Theorem
4.4 we have $P\overline{f_{A}}(A)(I-P)=0$, that is,
$P\overline{f_{A}}(A)=P\overline{f_{A}}(A)P$.

Let $x\in Ran(P)$ with $\|x\|=1$. Then by Lemma 4.3 we have
$P\overline{f_{A}}(A)P=\omega_{x}(\overline{f_{A}}(A))P$. So
$P\overline{f_{A}}(A)=\omega_{x}(\overline{f_{A}}(A))P$. Taking adjoint, we
have $f_{A}(A)P=\omega_{x}(f_{A}(A))P$. By Lemma 3.2, we have
$Pf_{A}(A)=\overline{\omega_{x}(\overline{f_{A}}(A))}P=\omega_{x}(f_{A}(A))P$.
Thus $Pf_{A}(A)=f_{A}(A)P$. Taking adjoint, we have
$P\overline{f_{A}}(A)=\overline{f_{A}}(A)P$. So,
$PA=Pf_{A}(A)\overline{f_{A}}(A)=f_{A}(A)P\overline{f_{A}}(A)=f_{A}(A)\overline{f_{A}}(A)P=AP$.

If $I-P$ is a minimal projection in $M$. By Theorem 4.4 we have
$P\overline{f_{A}}(A)(I-P)=0$. Taking adjoint, we have $(I-P)f_{A}(A)P=0$.
That is, $(I-P)f_{A}(A)=(I-P)f_{A}(A)(I-P)$. Similar to the proof above, we
conclude that $(I-P)A=A(I-P)$. So $AP=PA$.

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