# The Transition Rate

In the next lecture, we will solve the quantum Liouville equation

$i\hbar{\partial \rho \over \partial t} = [H,\rho]$

perturbatively and derive quantum linear response theory. However, the transition rate can actually be determined directly within perturbation theory using the Fermi Golden Rule approximation, which states that the probability of a transition's occurring per unit time, $$R_{i\rightarrow f}$$, is given by

$R_{i\rightarrow f}(\omega) = {2\pi \over \hbar}\vert\langle ......gle f\vert B\vert i\rangle \vert^2\delta(E_f-E_i-\hbar\omega )$

The $$\delta$$-function expresses the fact that energy is conserved. This describes the rate of transitions between specific states $$\vert i\rangle$$ and $$\vert f\rangle$$. The transition rate between any initial and final states can be obtained by summing over both $$i$$ and $$f$$ and weighting the sum by the probability that the system is found in the initial state $$\vert i\rangle$$:

$P(\omega) = \sum_{i,f}R_{i\rightarrow f}(\omega) w_i$

where $$\underline {w_i}$$ is an eigenvalue of the density matrix, which we will take to be the canonical density matrix:

$w_i = {e^{-\beta E_i} \over {\rm Tr}\left(e^{-\beta H}\right)}$

Using the expression for $$R_{i\rightarrow f}(\omega)$$, we find

$P(\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i, f} w_i \vert \langle i \vert B \vert f \rangle \vert ^2 \delta(E_f-E_i-\hbar\omega )$

Note that

$P(-\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i, f } w_i \vert \langle i \vert B \vert f \rangle \vert^2\delta(E_f-E_i+\hbar\omega )$

This quantity corresponds to a time-reversed analog of the absorption process. Thus, it describes an emission event $$\vert i\rangle \rightarrow\vert f\rangle$$ with $$E_f = E_i-\hbar\omega$$, i.e., emission of a photon with energy $$\hbar \omega$$. If can also be expressed as a process $$\vert f\rangle \rightarrow \vert i\rangle$$ by recognizing that

$w_f = {e^{-\beta E_f} \over {\rm Tr}\left(e^{-\beta H}\right) } = {e^{\beta (E_i-\hbar\omega )} \over {\rm Tr}\left(e^{-\beta H}\right)}$

or

$w_f = e^{\beta\hbar\omega }w_i\;\;\;\;\Rightarrow\;\;\;\;w_i = e^{-\beta\hbar\omega }w_f$

Therefore

$P(-\omega) = {2\pi\over \hbar}\vert F_{\omega}\vert^2 e^{-\beta \hbar \omega } \sum _{i, f} w_f \vert \langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i+\hbar\omega )$

If we now interchange the summation indices, we find

 $$P(-\omega)$$ $${2\pi\over \hbar}\vert F_{\omega}\vert^2e^{-\beta\hbar\omega}\sum_{i,f}w_i\vert\langle i\vert B\vert f\rangle \vert^2\delta(E_i-E_f+\hbar\omega )$$ $${2\pi\over \hbar}\vert F_{\omega}\vert^2\sum_{i,f}w_i\vert\langle i\vert B\vert f\rangle \vert^2e^{-\beta\hbar\omega}\delta(E_i-E_f-\hbar\omega )$$

where the fact that $$\delta (x) = \delta (- x)$$ has been used. Comparing this expression for $$P (- \omega )$$ to that for $$P (\omega )$$, we find

$P(-\omega) = e^{-\beta\hbar\omega }P(\omega)$

which is the equation of detailed balance. We see from it that the probability of emission is less than that for absorption. The reason for this is that it is less likely to find the system in an excited state $$\vert f \rangle$$ initially, when it is in contact with a heat bath and hence thermally equilibrated. However, we must remember that the microscopic laws of motion (Newton's equations for classical systems and the Schrödinger equation for quantum systems) are reversible. This means that

$R_{i\rightarrow f}(\omega) = R_{f\rightarrow i}(-\omega)$

The conclusion is that, since $$P(\omega) > P(-\omega)$$, reversibility is lost when the system is placed in contact with a heat bath, i.e., the system is being driven irreversibly in time.

Define

 $$C_>(\omega)$$ $$\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i-\hbar\omega )$$ $$C_<(\omega)$$ $$\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2\delta(E_f-E_i+\hbar\omega )$$

then

$C_<(\omega) = e^{-\beta\hbar\omega }C_>(\omega)$

Now using the fact that the $$\delta$$-function can be written as

$\delta(E) = {1 \over 2\pi}\int_{-\infty}^{\infty}dt e^{-iEt}$

$$C_>(\omega)$$ becomes

 $$C_>(\omega)$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2 e^{-i(E_f-E_i-\hbar\omega )t/\hbar}$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i,f}w_i \vert\langle i\vert B\vert f\rangle \vert^2 e^{-i(E_f-E_i)t/\hbar}$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum _{i, f} w_i \langle i \vert B\vert f\rangle \langle f\vert B\vert i\rangle e^{-iE_f t/\hbar}e^{iE_it/\hbar}$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum _{i, f} w_i \langle i \vert e^{iH_0t/\hbar} B e^{-iH_0t/\hbar}\vert f\rangle\langle f\vert B\vert i\rangle$$

Recall that the evolution of an operator in the Heisenberg picture is given by

$B(t) = e^{iH_0t/\hbar}Be^{-iH_0t/\hbar}$

if the evolution is determined solely by $$H_0$$. Thus, the expression for $$C > (\omega )$$ becomes

 $$C_>(\omega)$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i,f}w_i \langle i\vert B(t)\vert f\rangle\langle f\vert B\vert i\rangle$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\sum_{i}w_i \langle i\vert B(t)B(0)\vert i\rangle$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;e^{i\omega t}{\rm Tr}\left[\rho B(t) B(0)\right]$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;e^{i\omega t}\langle B(t) B(0)\rangle$$

which involves the quantum autocorrelation function $$\langle B(t) B(0)\rangle$$.

In general, a quantum time correlation function in the canonical ensemble is defined by

$C_{AB}(t) = {{\rm Tr}\left[A(t)B(0)e^{-\beta H}\right] \over{\rm Tr}\left[e^{-\beta H}\right]}$

In a similar manner, we can show that

$C_<(\omega) = {1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle B(0)B(t)\rangle \neq C_>(\omega)$

since

$[B(0),B(t)]\neq 0$

in general. Also, the product $$B (0) B (t)$$ is not Hermitian. However, a hermitian combination occurs if we consider the energy difference between absorption and emission. The energy absorbed per unit of time by the system is $$P(\omega)\hbar\omega$$, while the emitted into the bath by the system per unit of time is $$P(-\omega)\hbar\omega$$. The energy difference $$Q ( \omega )$$ is just

 $$Q(\omega)$$ $$[P(\omega)-P(-\omega)]\hbar\omega$$ $$P(\omega)[1-e^{-\beta\hbar\omega }]\hbar\omega$$ $$2\pi\omega\vert F_{\omega}\vert^2C_>(\omega)[1-e^{-\beta\hbar\omega }]$$

But since

$C_<(\omega) = e^{-\beta\hbar\omega }C_>(\omega)$

it follows that

$C_>(\omega) + C_<(\omega) = \left(1+e^{-\beta\hbar\omega }\right)C_>(\omega)$

or

$C_>(\omega) = {C_>(\omega) + C_<(\omega) \over 1+e^{-\beta\hbar\omega}}$

Note, however, that

 $$C_>(\omega) + C_<(\omega)$$ $${1 \over 2\pi\hbar}\int_{-\infty}^{\infty}\;dt e^{i\omega t}\langle B(t)B(0) + B(0) B(t)\rangle$$ $${1 \over \pi\hbar} dt\;e^{i\omega t}\langle {1 \over 2}[B(0),B(t)]_+\rangle$$

where $$[...,...]_+$$ is known as the anticommutator: $$[A,B]_+ = AB+BA$$. The anticommutator between two operators is, itself, hermitian. Therefore, the energy difference is

 $$Q(\omega)$$ $$\underline { {2\omega \over \hbar}\vert F_{\omega}\vert^2 {1-e^{-\beta\hbar\omega} \over 1 + e^{-\beta \hbar \omega }}\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle [B(0),B(t)]_+\rangle }$$ $${2\omega \over \hbar}\vert F_{\omega}\vert^2 {\rm tanh}(\beta\hbar\omega /2)\int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle [B(0),B(t)]_+\rangle$$

The quantity $$\langle [B(0),B(t)]_+\rangle$$ is the symmetrized quantum autocorrelation function. The classical limit is now manifest ( $${\rm tanh}(\beta\hbar\omega /2)\rightarrow \beta\hbar\omega /2$$ ):

$Q(\omega)\rightarrow \vert F_{\omega}\vert^2 \int_{-\infty}^{\infty}\;dt\;e^{i\omega t}\langle B(0)B(t)\rangle$

The classically, the energy spectrum $$Q(\omega)$$ is directly related to the Fourier transform of a time correlation function.