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{\noindent\small Mathematical Physics and Quantum Field Theory,\newline 
 Electronic Journal of Differential Equations, Conf. 04, 2000, pp. 155--158.\newline 
http://ejde.math.swt.edu  or http://ejde.math.unt.edu 
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2000 Southwest Texas State University  and 
University of North Texas.} \vspace{1cm} 

\title[Decay times in quantum mechanics]
{Decay times in quantum mechanics} 
\author[Richard Lavine] {Richard Lavine} 

\address{Richard Lavine \hfill\break\indent
Department of Mathematics, University of Rochester \hfill\break\indent
Rochester, NY 14627, USA  } 
\email{lavine@math.rochester.edu} 

\thanks{Published July 12, 2000.}
\subjclass{81U05, 81P15} 
\keywords{Schr\"odinger operator, measurement} 

\begin{abstract}
We consider the problem of determining the probability
distribution for the time of decay for a one-dimensional 
Schr\"odinger operator.
\end{abstract}
\maketitle

An important feature of quantum physics is that times of occurrences
like decay of a radioactive nucleus are random; typically the decay time
is observed to have an exponential probability distribution.  But
quantum mechanics does not directly prescribe a probability distribution
for such times.  Recall how probability distributions arise:  Any
``observable'' $\mathcal{A}$ is represented by a self-adjoint operator
$A$ on a Hilbert space $\mathcal{H}$; if at time $t$ the system is in a 
state represented by the vector $\varphi$ in $\mathcal{H}$, and 
$f (\mathcal{A})$ is measured
(for any reasonable function $f$) then the expected value is
$$ E(f (\mathcal{A})) = \langle \varphi | f(A)\varphi \rangle \;.$$
By the spectral theory of self-adjoint operators this determines a
probability distribution function $F$ such that 
$$ E(f(\mathcal{A})) = \int f (\lambda) d F (\lambda)$$
so that the probability of finding the value of $\mathcal{A}$ in
$(\lambda_1, \lambda_2]$ is $F(\lambda_2) - F(\lambda_1)$.

An ``observable'' is something that can, in principle, be measured at
any given instant.  But the most straightforward way to measure the time
of decay would be to place detectors that respond to the products of the
decay whenever that happens; this is not a measurement made at a
specific time.  Still, the probability distributions for such
experiments must be determined in principle by quantum mechanics, since
the detector could be regarded as a part of the quantum system, and its
result read off (``measured'') at some later time.  But the quantum
mechanical description of such an apparatus seems hopelessly
complicated, and its details should be irrelevant.

For a simple model, the Schr\"odinger operator $ H = - d^2 /
dr^2 + V(r)$ on the half-line $[0, \infty)$, with $V \geq 0$, we will
propose a self-adjoint operator, on the Hilbert space $L^2 ([0,
\infty))$ of the model, whose probability distribution should be that of
the decay time.

A potential $V(r)$ with a barrier that tends to trap a particle inside
$[0,R]$ is often used to model decay.  It has been shown \cite{L1} that in
this case there is a state $\varphi$ located inside $[0,R]$ such that $|
\langle \varphi | e^{-iHt} \varphi \rangle |^2$, the probability of the
particle being found in the initial state at time $t$, is approximately
given by an exponentially decaying function.  (This gives the
probability of a result if a certain measurement is made at time $t$,
not the probability distribution for when decay will happen.)

First note that certain operators have been considered as measuring
trapping time:  
\begin{equation}
 T_{[0,R]} = \int^\infty_{- \infty} e^{iHt} \chi_{[0,R]} e^{- iHt} dt
\end{equation}
represents the time spent in $[0,R]$, where $\chi_{[0,R]}$ is
(multiplication by) the characteristic function of $[0,R]$.  This
operator commutes with $H$.

The time delay $T_D$ also commutes with $H$; it satisfies \cite{L2}, \cite{N}
\begin{equation}
 HT_D = \int^\infty_{-\infty} e^{iHt} (V + \frac r2 V') e^{- iHt} dt
\;.\end{equation}

The time spent in state $\varphi$ is
\begin{equation} T_\varphi = \int^\infty_{-\infty} e^{iHt} | \varphi \rangle  
\; \langle\varphi | e^{- iHt} dt \;,
\end{equation}
which also commutes with $H$.

None of these operators has probability distribution close to
exponential; in fact they are all bounded operators, at least when
restricted to a spectral subspace for any closed energy interval
contained in $(0, \infty)$.  The probability of finding a time longer
than $\| T \|$ is zero while the exponential distribution allows
arbitrarily large times.

So it appears that particles can not be trapped for more than a given
finite time!  This does not necessarily contradict the arbitrarily long
times observed until decay happens, because the particle is not
unequivocally inside $[0,R]$ until the time of decay.

To find an operator to represent a physical quantity, we can try to
quantize a classical function of $p$ and $q$, or lay down enough
requirements to determine the operator.  Following the latter approach,
we might require
\begin{equation}
e^{iHt} T e^{- iHt} = T - t 
\end{equation}
if $T$ gives the time until decay happens.  This implies
\begin{equation}
i [ H,T] = - I
\end{equation}
which is notoriously impossible for $H$ bounded below.  But it
is possible to find operators $A$ such that $e^{iHt} A e^{- iHt}$
decreases at a constant rate; in fact if $H_0 = - d^2 / dr^2\; (V =0)$
and
\begin{equation}
A_0 = \frac i4 \left( r \frac d{dr} + \frac d{dr} r \right) \;,
\end{equation}
we have
\begin{equation}
i [ H_0, A_0 ] = - H_0 \;.
\end{equation}
The most general operator with this commutation relation is $A_0 + f
(H_0)$, so the time until decay, multiplied by energy must be of this
form. A metastable state is typically highly concentrated in energy, so that
(7) is close to (5), up to a scalar multiple.

Taking the approach via classical mechanics, the time a free particle,
moving on the half-line, takes to reach a curve $r = f (p^2)$ (on its
way out) is $2|p|^{-1} f (p^2) - (2p)^{-1} r $.  Multiplication by the
energy $p^2$ gives $\frac 12 [ |p| f  (p^2) - pr]$.  In particular, for
the time to reach $r =R$ we get $ \frac 12 ( |p| R - pr)$; quantization
gives the operator $ \frac 12 H^{1/2}_0 R + A_0$. 

If $V$ is short range, e.g. $|V (r) | \leq (1 + r)^{- \alpha}$ with
$\alpha >1$, then the wave operators $\Omega_\pm$ exist (and are unitary
since $V \geq 0)$ and the operators
$$ A_\pm = \Omega_+ A_0 \Omega^*_\pm$$
satisfy
\begin{equation}
i [ H, A_\pm ] = - H \;.
\end{equation}

The probability distribution for $A_0$ in a state $\varphi$ can be found
by obtaining a spectral representation for $A_0$, i.e., a unitary $W$
diagonalizing $A_0$ so that $WAW^*$ is a multiplication operator.  In
the standard spectral representation of $H_0$ as multiplication by
$\lambda$ on $L^2 ([0, \infty))$, $A_0$ is given by
\begin{equation}
 - \frac i8 \left[ \lambda \frac d{d\lambda} + \frac d{d\lambda} \lambda
\right]\;.
\end{equation}
Then the operator $W$ given by
\begin{equation}
\begin{split}
W \varphi (\tau) & = (2 \pi)^{-1/2} \int^\infty_0 \lambda^{- 4 i \tau -
\frac 12 } \varphi (\lambda) d \lambda \\
& = (2 \pi)^{- 1/2} \int^\infty_{-\infty} e^{- is\tau} e^{s/8}
\varphi ( e^{s/4}) ds \; 
\end{split}
\end{equation}
is unitary from $L^2 ( [0,\infty))$ to $L^2 ({\mathbb R})$, since it is the
composition of the Fourier transform and a unitary change of variables.
It gives a spectral representation of $A_0$, since
\begin{equation}
(WA_0 \varphi) (\tau) = \tau (W\varphi) (\tau)\;.
\end{equation}
More generally,
\begin{equation}
(W_F [ A_0 + F(H_0) ] \varphi) (\tau) = \tau (W_F \varphi) (\tau)
\end{equation}
where 
\begin{equation}
W_F \varphi (\tau ) = (2 \pi)^{- 1/2} \int \lambda^{- 4i \tau -1/2}
 e^{if (\lambda)} \varphi (\lambda) d \lambda 
\end{equation}
and $\lambda f' (\lambda) = F(\lambda)$.

For example, if $\varphi (\lambda) = c (\lambda - z)^{-1}$ with 
$\,{\rm Im}\, z < 0$, a contour integral gives
\begin{equation}
(W \varphi) (\tau ) = \frac{icz^{- 4i\tau - \frac 12}}{1 + e^{-8\pi
\tau}} \;.
\end{equation}
If $z = |z| e^{- i \theta}$, the probability distribution is
\begin{equation}
|W\varphi (\tau)|^2 = \frac{|c|^2 e^{- 8 \theta \tau} |z|^{-1}}{(1 +
e^{- 8 \pi \tau})^2 } \;.
\end{equation}
For small $\theta$, this is very close to 0 for $\tau <0$, and to a
decaying exponential for $\tau >0$.

By the second expression for $W_\varphi (\tau)$ in (10), if $e^{s/8}
\varphi (e^{s/4})$ is close in $L^2 ({\mathbb R})$ to
\begin{equation}
c (\lambda - \lambda_0 + i \epsilon)^{-1} e^{i [ a + b (s-s_0)]},
\end{equation}
where $\lambda_0 = e^{s_0}$, then $(W \varphi (\tau))^2$ is close in
$L^1
({\mathbb R})$ to (13), translated by $b$.

For the case of short range $V \ne 0$, the operator (7) in a spectral
representation for $H$ can be diagonalized in exactly the same way.
Spectral representations of $H$ are given by expansions in
eigenfunctions $\psi (r, \lambda)$ corresponding to $\Omega_\pm$.  Let
\begin{equation}
\tilde{\varphi} (\lambda) = \int^\infty_0 \psi (r, \lambda) \varphi (r)
dr \;.
\end{equation}
In \cite{L1} a value $z$ was found so that if $\varphi (r) = X_{[0,R]}
(r) \psi (z, r)$, then $| \langle \varphi | e^{- iHt}\varphi \rangle |$
decays exponentially.  For such $\varphi$, integration by parts gives
\begin{equation}
\tilde{\varphi} (\lambda) = \frac{\varphi ' (R) \psi (\lambda, R) -
\varphi (R) \psi' (\lambda, R)}{\lambda -z} \;.
\end{equation}
Thus exponential decay of $| W \tilde{\varphi}|^2$ reduces to estimates
of the difference between (16) and (18).

\begin{thebibliography}{99}
\bibitem{L1}  Lavine, R., {\it Existence of almost exponentially
decaying state for barrier potentials}, preprint.
\bibitem{L2} Lavine, R., {\it Commutators and local decay}, 141--156, in
{\it Scattering Theory in Mathematical Physics}, LaVita, J. and
Marchand, J.-P. eds, Reidel, Boston (1974).
\bibitem{N} Narnhofer, H., {\it Time delay and dilation properties in
scattering theory}, J. Math. Phys. \textbf{25}, 987--991 (1984).
\end{thebibliography}
\end{document}