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# QS 10: Transition Rate

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$\newcommand{\kk}[1]{|#1\rangle} \newcommand{\bb}[1]{\langle #1} \newcommand{\dd}[1]{\delta_{#1}} \newcommand{\dydxt}[2]{\frac{d#1}{d#2}} \newcommand{\dydx}[2]{\frac{\partial#1}{\partial#2}} \newcommand{\molp}{\Omega^{(+)}}$

Let $\phi$ and $\xi$ be two free states at time $t=0$ orthogonal to each other, $(\phi,\xi)=0$.

Let us begin with a large number $N$ of identically prepared systems in the free state $\phi(-T_1)=U_0(-T_1)\phi,$ in remote past ($T_1$ a large positive number).
In scattering theory we are interested in a question of the following type. : how many of these $N$ systems will be found in a free state $\xi(+T_2)=U_0(+T_2)\xi$ in some remote future $t=+T_2$, where $T_2$ too is a large positive number?

If there was no interaction and evolution took place only with the free Hamiltonian $H_0$ then the answer would be zero, because for all values of time $t$ the  state $\phi(t)$ would be orthogonal to $\xi(t)$ both states evolving with $H_0$
so that $(\phi(T_2),\chi(T_2))=0$.

As it is, the state $\phi(-T_1)$ evolves to the scattering state $\psi(t)=\molp \phi(t)$ and may have a non-zero probability $P_{\xi\psi}(t)=|(\xi(t),\psi(t))|^2$ to be found in state $\xi(t)$ at time $t$. This probability is zero for large negative times, increases as interaction is switched on" and gradually saturates to a constant value $P_{\xi\psi}=|(\xi(T_2),\psi(T_2))|^2=|(\xi(T_2),\chi(T_2))|^2$ because for large positive times the state again evolves as a free state $\chi$.  Therefore the total number of transitions to state $\xi(T_2)$ at time $T_2$ are \begin{eqnarray*} NP_{\xi\psi}=N|(\xi(T_2),\psi(T_2))|^2 \end{eqnarray*} If we wait for a time $\Delta T_2$ more, the total number of transitions will be
\begin{eqnarray*} NP_{\xi\psi}=N|(\xi(T_2+\Delta T_2),\psi(T_2+\Delta T_2))|^2 \end{eqnarray*}
Therefore the number of transitions per unit time or the {\em transition rate} is given by
to be evaluated for large times.

The argument given above is based on the assumption that {\em once the system makes a transition to the free state $\xi(t)$ from $\psi(t)$ it continues to evolve as a free state from then on, and does nor revert back again to a scattering state}.

Therefore, transitions to $\xi$ occurring at different times keep accumulating and would be counted among the states which have already made transition to $\xi$.

At present we do not have a complete theory of measurement in quantum mechanics. These issues are not completely understood. But the formulas given below are based on these assumptions and they agree with experimental observations.

Let us put the transition rate in a more convenient form.

For any complex valued function $A(t)$ of $t$ such that $dA/dt=C/i$
\begin{eqnarray*} \dydxt{|A|^2}{t}=A^*\frac{C}{i}-\frac{C^*}{i}A=2\,{\rm Im}(A^*C) \end{eqnarray*}
Applying this simple identity to
\begin{eqnarray*} A=\bb{\xi(t)}\kk{\psi(t)}=\bb{\xi(t)}|\molp\kk{\phi(t)}\end{eqnarray*}
for which
\begin{eqnarray*} i\hbar\dydxt{A}{t}=\bb{\xi(t)}|V\kk{\psi(t)}=
\bb{\xi(t)}|V\molp\kk{\phi(t)}=C\hbar, \end{eqnarray*}
the transition rate is
\begin{eqnarray}n_\xi = N\dydxt{}{t}|(\xi(t),\psi(t))|^2=
N\frac{2}{\hbar}{\rm Im}\bb\phi(t)|B(t)\kk{\phi(t)}, \label{transition1}\end{eqnarray}
where
\begin{eqnarray}B(t)={\molp}^\dagger\kk{\xi(t)}\bb{\xi(t)}|V\molp.\label{transition2}\end{eqnarray}
Note that we have to evaluate this transition rate for $t\to\infty$. Note also the appearance of the projection operator to final states. In practice, both the free states $\phi(t)$ and $\xi(t)$ are stationary states with trivial time dependence $\exp(-iEt/\hbar)$.  Therefore $n_\xi$ is actually independent of time.

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