# How does Fermi's golden rule yield a specific rate?

According to Fermi’s golden rule, the rate of an electronic transition is proportional to the magnitude squared of $\langle i|\hat{H}|f\rangle$. Since energy is always relative to some reference point, I could add an arbitrary constant to the Hamiltonian, and it would do nothing but shift all the energy eigenvalues by that constant. So I don’t understand how Fermi’s golden rule can yield a specific rate – changing the Hamiltonian by an arbitrary constant will (despite not changing the system) change the calculated rate. Can anyone explain?

• According to my quick read on wikipedia, the Hamilton operator to be put in is the perturbation. As such, it is not arbitrary and you may not add a constant. Furthermore, and someone may correct me on this, are constants not only applied to the diagonal? – TAR86 Apr 12 '17 at 19:07
• Good point, oops. thanks for pointing that out. What about the electronic coupling in Marcus Theory? – Maria Apr 12 '17 at 19:57


$$\braket{i|f} = \delta_{ij}$$

therefore if we define $\hat{V'} = \hat{V} + k$ where $\hat{V}$ is the perturbation Hamiltonian and $k$ is some arbitrary constant

\begin{align} \braket{i|\hat{V'}|f} &= \braket{i|\hat{V}|f} + \braket{i|k|f} \\ &= \braket{i|\hat{V}|f} + k\braket{i|f} \\ &= \braket{i|\hat{V}|f} \end{align}

if $\ket{i}$ and $\ket{f} \endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.

• Thanks for the reply. If |i⟩ and |f⟩ are both eigenstates of the system, though, then $⟨i|H|f⟩=⟨i|E|f⟩=0$, which isn't a case I'm interested in. So what about when |i⟩ and |f⟩ aren't orthonormal? – Maria Apr 12 '17 at 19:06
• @Felipe see TAR86's comment on your question; I forgot that $\hat{H}$ in your question refers to the perturbation. I'll remove my last sentence and update the notation. They have to be orthonormal because $\hat{H}_0$ is hermitian. – orthocresol Apr 12 '17 at 19:12
• In the case where the states are not eigenstates, you would need to expand them in the basis of eigenstates of the perturbation operator. – pentavalentcarbon Apr 13 '17 at 0:43