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Definition of Quantum
Phase Space Distribution Functions

One possible motivation for introducing quantum distribution functions is their utility for the comparison of classical and quantum mechanics, as mentioned above. In addition to this there is another, equally important motivation for studying these functions: they can be used to compute expectation values in a comparatively simple way, where the computational simplification is mainly due to the fact that the corresponding formulae depend on scalars only, as opposed to conventional expressions of quantum expectation values which typically involve operators.

Consider a classical observable $A(x,p)$ that depends on the position and momentum variables $x$ and $p$. The expectation value of $A$ can be computed as

\begin{displaymath}
\left< A \right>_t \; = \; \int\!\! {\mbox{d}}x\! \int\!\! {\mbox{d}}p\; A(x,p) \, F(x,p,t),
\end{displaymath} (A.1)

where the classical phase space probability density $F(x,p,t)$ describes the state of the system at time $t$, and can be obtained, for example, by solving the LIOUVILLE equation. (All integrals in this appendix are from $-\infty$ to $\infty$, and the index ${}_t$ explicitly indicates the time dependence of the expectation value.) The objective of the following considerations is to find a quantum mechanical expression which is analogous to equation (A.1).

Rather than discussing a general quantum observable ${\hat{A}}({\hat{x}},{\hat{p}})$, with the position and the momentum operators ${\hat{x}}$ and ${\hat{p}}$, I begin by considering a particular operator instead, namely $e^{i(\xi{\hat{x}}+\eta{\hat{p}})}$, with constants $\xi,\eta\in\mathbb{R}$. Exponentials of this type are used below in the FOURIER expansion (A.7) to construct any other operator. In order to obtain an expression analogous to equation (A.1) the operator $e^{i(\xi{\hat{x}}+\eta{\hat{p}})}$ somehow has to be substituted by a corresponding scalar expression. For example one could set
\begin{subequations}
\begin{equation}
\left< e^{i(\xi{\hat{x}}+\eta{\hat{p}})} ...
...! {\mbox{d}}p\; e^{i(\xi x+\eta p)} F_2(x,p,t)
\end{equation}\end{subequations}
as well, with another distribution function $F_2(x,p,t)$. But due to the fact that ${\hat{x}}$ and ${\hat{p}}$ do not commute one has

\begin{displaymath}
e^{i(\xi{\hat{x}}+\eta{\hat{p}})} \; = \; e^{i\xi{\hat{x}}}e...
...\hbar\xi\eta}
\; \neq \; e^{i\xi{\hat{x}}}e^{i\eta{\hat{p}}},
\end{displaymath} (A.1)

as is easily confirmed using the BAKER-CAMPBELL-HAUSDORFF formula.A.2Therefore, in general the distribution functions $F_1$ and $F_2$ are not identical, the reason being that in the integrands of equations (A.2a) and (A.2b) the same scalar function $e^{i(\xi x+\eta p)}$ has been associated with the two different operators $e^{i(\xi{\hat{x}}+\eta{\hat{p}})}$ and
$e^{i\xi{\hat{x}}}e^{i\eta{\hat{p}}}$, respectively.

In order to avoid this ambiguity one first chooses a complex-valued kernel function $f(\xi,\eta)$; then the scalar $e^{i(\xi x+\eta p)}$ is defined to be associated with the operator $f(\xi,\eta)e^{i(\xi{\hat{x}}+\eta{\hat{p}})}$ exclusively,

\begin{displaymath}
f(\xi,\eta)e^{i(\xi{\hat{x}}+\eta{\hat{p}})} \;\; \widehat{=} \;\;
e^{i(\xi x+\eta p)},
\end{displaymath} (A.3)

thereby establishing a one-to-one correspondence between scalars and operators. In this way the above-mentioned ambiguity is shifted towards the definition of the kernel $f(\xi,\eta)$. Depending on the rule of association specified by this function, one can define different distribution functions $F^f(x,p,t)$ via
\begin{displaymath}
\left< f(\xi,\eta)e^{i(\xi{\hat{x}}+\eta{\hat{p}})} \right>_...
...{d}}x\! \int\!\! {\mbox{d}}p\; e^{i(\xi x+\eta p)} F^f(x,p,t).
\end{displaymath} (A.4)

The quantization rule (A.5) not only defines how to associate exponential operators with scalars, but is much more general, as it can be applied to each term of the FOURIER expansion of any operator ${\hat{A}}$,

\begin{displaymath}
{\hat{A}}({\hat{x}},{\hat{p}}) \; = \; \frac{1}{2\pi} \int\!...
...\eta\;
\tilde{A}(\xi,\eta) e^{i(\xi{\hat{x}}+\eta{\hat{p}})}.
\end{displaymath} (A.5)

Therefore the scalar function associated with ${\hat{A}}$ obviously is
\begin{displaymath}
A^f(x,p) \; := \; \frac{1}{2\pi} \int\!\! {\mbox{d}}\xi\! \i...
...
\frac{\tilde{A}(\xi,\eta)}{f(\xi,\eta)} e^{i(\xi x+\eta p)},
\end{displaymath} (A.6)

which is defined unambiguously.A.3Using the representation (A.8) of the classical observable associated with the operator ${\hat{A}}$, it is then straightforward to compute its quantum mechanical expectation value:
\begin{subequations}
\begin{eqnarray}
% \Erwart{\A(\x,\p)}_t \; =
\big< {\hat{...
...int\!\! {\mbox{d}}p\; A^f(x,p) \, F^f(x,p,t),
\end{eqnarray}\end{subequations}
where equation (A.9b) is the desired expression analogous to equation (A.1).

An explicit expression for the distribution function $F^f$ can be obtained from the implicit definition (A.6) by FOURIER transformation. For a system in the state $\left\vert \psi(t) \right>$ at time $t$, the expectation values are given by $\left< \,\cdot\, \right>_t=\left< \psi(t) \left\vert \,\cdot\, \right\vert \psi(t) \right>$, and one gets

\begin{displaymath}
\int\!\! {\mbox{d}}\xi\! \int\!\! {\mbox{d}}\eta\;
\left< ...
...(\xi,\eta) \, e^{-i(\xi x+\eta p)} \; = \;
4\pi^2 F^f(x,p,t),
\end{displaymath}

which by insertion of the identity operator $\, \mathbbm{1}=\int {\mbox{d}}x''\, \left\vert x'' \right>\left< x'' \right\vert \,$ can also be written as

\begin{displaymath}
F^f(x,p,t) = \frac{1}{4\pi^2}
\int\!\! {\mbox{d}}\xi\! \int...
... \right\vert x'' \right>
f(\xi,\eta) \, e^{-i(\xi x+\eta p)}.
\end{displaymath}



(A.6)


The exponential $e^{i\eta{\hat{p}}}$ acts as a translation operator in position space (cf. equation (5.23));
\begin{displaymath}
e^{i\eta{\hat{p}}}\left\vert x'' \right> \; = \; \hat{T}(-\...
...ft\vert x'' \right>
\; = \; \left\vert x''-\hbar\eta \right>,
\end{displaymath} (A.7)

such that
\begin{displaymath}
e^{i(\xi{\hat{x}}+\eta{\hat{p}})}\left\vert x'' \right>
\; ...
...{2}i\hbar\xi\eta}e^{i\xi x''}\left\vert x''-\hbar\eta \right>.
\end{displaymath} (A.8)

Inserting this into equation (A.10) and substituting $x':=x''-\hbar\eta/2$, I finally obtain a convenient explicit formula for the distribution function $F^f(x,p,t)$:
\begin{displaymath}
\hspace*{-0.4cm}
\fbox{$ \displaystyle
\begin{array}{rcl...
...}.
\rule[-0.5cm]{0.0cm}{1.0cm}\hspace*{0.1cm}
\end{array} $}
\end{displaymath} (A.9)




Footnotes

... formula.A.2
A simplified BAKER-CAMPBELL-HAUSDORFF (BCH) formula


\begin{displaymath}
e^{\hat{A}}e^{\hat{B}}\; = \; e^{{\hat{A}}+{\hat{B}}}e^{\frac{1}{2}[{\hat{A}},{\hat{B}}]}
\end{displaymath} (A.2)



holds in the special case when the operators ${\hat{A}}$ and ${\hat{B}}$ both commute with their commutator $[{\hat{A}},{\hat{B}}]$ (which is true in the present case, because $[{\hat{x}},{\hat{p}}]=i\hbar$ is a c-number). See [Per93] for a proof, and [Wil67,Ote91] for more on BCH formulae.
... unambiguously.A.3
The question needs to be addressed if the theory, and in particular the evaluation of integrals like that in equation (A.8), could be spoiled by kernel functions $f(\xi,\eta)$ that have zeroes or are singular for some values of $\xi,\eta$. I do not discuss this issue here, but refer the reader to [SPM99] where it is shown that the formalism can be applied smoothly even in such notorious cases.

Note that the kernel function can also be chosen, more generally, as a functional of the quantum state $\left\vert \psi \right>$ of the system itself: $f(\xi,\eta,\psi)$. COHEN gives an example for such a kernel function $f^{\rm C}$ that, despite being quite complicated, leads to the very simple COHEN distribution function $F^{\rm C}(x,p,t) = \left\vert\left< q \left\vert \psi(t) \right> \right.\right\vert^2 \left\vert\left< p \left\vert \psi(t) \right> \right.\right\vert^2$ that nicely combines the position and momentum representations of $\left\vert \psi \right>$ in an intuitive way [Coh66]. However, since choosing a $\psi$-dependent $f$ has a number of unfavourable consequences -- for example, equation (A.8) indicates that in this case the function $A^{\rm C}$ associated with the operator ${\hat{A}}$ becomes $\psi$-dependent, too -- I do not further discuss kernels that are functionals of $\left\vert \psi \right>$.


next up previous contents
Next: Special Distribution Functions Up: Quantum Phase Space Distribution Previous: Quantum Phase Space Distribution   Contents
Martin Engel 2004-01-01