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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?

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    $\begingroup$ 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? $\endgroup$ – TAR86 Apr 12 '17 at 19:07
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    $\begingroup$ Good point, oops. thanks for pointing that out. What about the electronic coupling in Marcus Theory? $\endgroup$ – Maria Apr 12 '17 at 19:57
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I'm fairly sure the states $\require{\begingroup} \begingroup \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[1]{\langle #1 \rangle} \ket{i}$ and $\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\hat{H}_0$. Therefore, by virtue of the Hermiticity of $\hat{H}_0$, they are necessarily orthonormal

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

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.

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  • $\begingroup$ 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? $\endgroup$ – Maria Apr 12 '17 at 19:06
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    $\begingroup$ @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. $\endgroup$ – orthocresol Apr 12 '17 at 19:12
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    $\begingroup$ In the case where the states are not eigenstates, you would need to expand them in the basis of eigenstates of the perturbation operator. $\endgroup$ – pentavalentcarbon Apr 13 '17 at 0:43

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