The derivation of Fermi's golden rule uses the following approximations:

  1. Transition time is small.

  2. Photon frequency $\omega\approx\omega_f-\omega_i$.

The question is, are there any experiments where Fermi's golden rule cannot be used because either one or both approximations are not valid?


Just to clarify, this question is about the derivation of Fermi's golden rule, not the rule itself. Please see the following video series to learn more about the derivation of Fermi's golden rule and the approximations that are used.


My notes from the video series and Quantum Mechanics for Scientists and Engineers.

Step 1. Define $\Psi(x,t)$ for $\psi_1(x)$ the initial eigenstate, $\psi_2(x)$ the final eigenstate. \begin{equation*} \Psi(x,t)=c_1(t)\psi_1(x)\exp(-i\omega_1t)+c_2(t)\psi_2(x)\exp(-i\omega_2t) \end{equation*}

Step 2. Define the perturbing Hamiltonian. \begin{equation*} H'(x,t)=2V(x)\cos(\omega t+\phi) \end{equation*}

Step 3. From time dependent Schrodinger equation \begin{equation*} i\hbar\frac{\partial}{\partial t}\Psi(x,t)=H_0(x)\Psi(x,t)+H'(x,t)\Psi(x,t) \end{equation*}

Step 4. Result \begin{equation*} i\hbar\frac{\partial c_1(t)}{\partial t}\psi_1(x)\exp(-i\omega_1t) +i\hbar\frac{\partial c_2(t)}{\partial t}\psi_2(x)\exp(-i\omega_2t) =H'(x,t)\Psi(x,t) \end{equation*}

Step 5. Inner product with $\psi_2^*(x)$. \begin{equation*} i\hbar\frac{\partial c_2(t)}{\partial t}\exp(-i\omega_2t) =2\cos(\omega t+\phi) \Bigl( c_1(t)M_{21}\exp(-i\omega_1t)+ c_2(t)M_{22}\exp(-i\omega_2t) \Bigr) \end{equation*}

where \begin{align*} M_{21}&=\int\psi_2^*(x)V(x)\psi_1(x)\,dx \\ M_{22}&=\int\psi_2^*(x)V(x)\psi_2(x)\,dx \end{align*}

Step 6. Simplify. \begin{equation*} i\hbar\frac{\partial c_2(t)}{\partial t} =2\cos(\omega t+\phi) \Bigl( c_1(t)M_{21}\exp\bigl(i(\omega_2-\omega_1)t\bigr)+2c_2(t)M_{22} \Bigr) \end{equation*}

Step 7. Define initial conditions. \begin{equation*} c_1(0)=1,\quad c_2(0)=0 \end{equation*}

For small $t$ use approximations $c_1(t)=1$ and $c_2(t)=0$. \begin{equation*} i\hbar\frac{\partial c_2(t)}{\partial t} =2\cos(\omega t+\phi)M_{21}\exp\bigl(i(\omega_2-\omega_1)t\bigr) \end{equation*}

Step 8. Solve for $c_2(t)$ by integrating. \begin{equation*} c_2(t)=\frac{2M_{21}}{i\hbar} \int_0^t\cos(\omega t'+\phi)\exp\bigl(i(\omega_2-\omega_1)t'\bigr)\,dt' \end{equation*}

Result \begin{equation*} c_2(t) =-\frac{M_{21}}{\hbar} \left( \frac{\exp\bigl(i(\omega_2-\omega_1-\omega) t\bigr)-1}{\omega_2-\omega_1-\omega} \right)\exp(-i\phi) -\frac{M_{21}}{\hbar} \left( \frac{\exp\bigl(i(\omega_2-\omega_1+\omega) t\bigr)-1}{\omega_2-\omega_1+\omega} \right)\exp(i\phi) \end{equation*}

Step 9. For $\omega$ such that $\omega\approx\omega_2-\omega_1$ the first term dominates, discard the second term. \begin{equation*} c_2(t)=-\frac{M_{21}}{\hbar} \left( \frac{\exp\bigl(i(\omega_2-\omega_1-\omega) t\bigr)-1}{\omega_2-\omega_1-\omega} \right)\exp(-i\phi) \end{equation*}

Step 10. Change to sinc function. \begin{equation*} c_2(t)=-\frac{it}{\hbar}\,M_{21} \exp\left(i\,\frac{\omega_2-\omega_1-\omega}{2}\,t-i\phi\right) \operatorname{sinc}\left(\frac{\omega_2-\omega_1-\omega}{2}\,t\right) \end{equation*}

Step 11. Transition probability. \begin{equation*} P(1\rightarrow2)=|c_2(t)|^2=\frac{t^2}{\hbar^2}\,|M_{21}|^2 \operatorname{sinc}^2\left(\frac{\omega_2-\omega_1-\omega}{2}\,t\right) \end{equation*}

Step 12. Integrate to get total transition probability. \begin{equation*} P_{tot}(1\rightarrow2)=\frac{t^2}{\hbar^2}|M_{21}|^2 \int_{E-\epsilon}^{E+\epsilon} \operatorname{sinc}^2\left(\frac{E'/\hbar-\omega}{2}\,t\right) g(E')\,dE' \end{equation*}

where $E=\hbar(\omega_2-\omega_1)$ and $g(E')$ is joint density of states.

Step 13. Use the approximation $g(E')\approx g(\hbar\omega)$. \begin{equation*} P_{tot}(1\rightarrow2)=\frac{t^2}{\hbar^2}|M_{21}|^2g(\hbar\omega) \int_{E-\epsilon}^{E+\epsilon} \operatorname{sinc}^2\left(\frac{E'/\hbar-\omega}{2}\,t\right)\,dE' \end{equation*}

Step 14. Transform the integral. \begin{align*} y&=\frac{E'/\hbar-\omega}{2}\,t \\ E'&=\frac{2\hbar y}{t}+\hbar\omega \\ dE'&=\frac{2\hbar}{t}\,dy \end{align*}

New integration limits. \begin{equation*} E\pm\epsilon \rightarrow \frac{(E\pm\epsilon)/\hbar-\omega}{2}\,t =\frac{Et}{2\hbar}-\frac{\omega t}{2} \pm\frac{\epsilon t}{2\hbar} \approx \pm\frac{\epsilon t}{2\hbar} \end{equation*}

Transformed integral. \begin{equation*} P_{tot}(1\rightarrow2)=\frac{2t}{\hbar}|M_{21}|^2g(\hbar\omega) \int_{-\epsilon t/2\hbar}^{\epsilon t/2\hbar} \operatorname{sinc}^2y\,dy \end{equation*}

Step 15. The sinc squared function is very narrow so use the approximation \begin{equation*} \int_{-\epsilon t/2\hbar}^{\epsilon t/2\hbar}\operatorname{sinc}^2 y\,dy \approx \int_{-\infty}^\infty\operatorname{sinc}^2 y\,dy=\pi \end{equation*}

Result \begin{equation*} P_{tot}(1\rightarrow2)=\frac{2\pi t}{\hbar}|M_{21}|^2g(\hbar\omega) \end{equation*}

Step 16. The transition rate is the derivative of $P_{tot}(1\rightarrow2)$. \begin{equation*} \Gamma_{1\rightarrow2} =\frac{d}{dt}P_{tot}(1\rightarrow2)=\frac{2\pi}{\hbar}|M_{21}|^2g(\hbar\omega) \end{equation*}

  • $\begingroup$ The statement about "transition time" being "small" is also unclear. Small compared to what? What "transition time" are you even talking about? $\endgroup$
    – hft
    Commented Sep 21, 2023 at 4:16
  • $\begingroup$ A lot is hidden behind the apparent simplicity of the FGR. You may find some useful points in this answer and this one. $\endgroup$
    – Roger V.
    Commented Sep 22, 2023 at 7:48

2 Answers 2


In the paper...


the author addresses the issues of approximations used to derive Fermi's golden rule. In a remark on page 5 the author writes

In order to derive (8) we forced all of the coefficients of the states to remain virtually unchanged, at time t, from the values they initially had (t=0). For that we must pay a price. Equation (8) holds only for perturbations that last a very short period of time, i.e. that don't have enough time to significantly alter the state of the system.

Another interesting note is made on page 8 that the time is also bounded by the requirement that the transition probability cannot exceed unity.


The derivation of Fermi's golden rule (see below) uses the following approximations

Your derivation appears to use the following approximation.

Step 7. Transition time is small.

I can not make sense of an a priori requirement that the "transition time" be "small," since this requirement is not clear without further context. In particular, the word "small" does not have a meaning out of context: Small compared to what?

Given your definition of the matrix element $M$ as being linear in $V$, what is required is that the perturbation (which I will call $\hat V$) can be treated to first order in perturbation theory. This is not a general requirement, since it is possible to work to higher order in perturbation theory (see below), but in that case the matrix element $M$ will have higher order in $V$ contributions (and summations over intermediate states, as seen, e.g., in Raman scattering matrix elements).

My derivation, as opposed to your derivation or "the derivation," also requires that the perturbation is "turned on" in the very distant past. This is accomplished formally by letting: $$ \hat V \to e^{\delta t}\hat V\;, $$ where $$ \delta\to 0\;. $$

With these two assumptions, we can use ordinary first-order time-dependent perturbation theory to find the transition rate.

For example, when $\hat V$ has no explicit time dependence (in the Schrödinger picture), we find: $$ R_{i\to f} = \lim_{\delta\to 0}|\langle \phi_f|\hat V|\phi_i\rangle|^2 \frac{2\delta e^{2\delta t}}{|E_i - E_f + i\delta|^2} $$ $$ = 2\pi |\langle \phi_f|\hat V|\phi_i\rangle|^2\delta(E_i - E_j) $$

If you would like to work to higher order in $V$, you can still do so, but the definition of the matrix element is different. For example, at second order in $V$ we have $$ R_{i\to f} = 2\pi |\langle \phi_f| \left( \hat V+\sum_n\frac{\hat V|n\rangle\langle n|\hat V}{E_i-E_n+i\delta} \right)|\phi_i\rangle|^2\delta(E_i - E_j) $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.