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For a constant perturbation of the form $$\hat{H'}(t)=\hat{V}\theta(t)$$ to a time-independent Hamiltonian $\hat{H}_0$, the transition probability at time $t$ from an eigenstate $|i\rangle$ of $\hat{H}_0$ with energy $E_i$ at $t=0$ to a final state $|f\rangle$ ($\neq |i\rangle$) is given by $$P_{i\to f}(t;\omega_{fi})=|V_{fi}|^2\left[\frac{\sin(\omega_{fi}t/2)}{\omega_{fi}/2}\right]^2$$ where $\omega_{fi}=(E_f-E_i)/\hbar$. At a fixed $t$, the transition probability $P_{i\to f}(t;\omega_{fi})$ peaks at $\omega_{fi}=0$ i.e., for $E_f=E_i$. Why does a constant perturbation favour the transition at $\omega_{fi}=0$? Is it because the constant perturbation has only $\omega=0$ frequency component? But then why don't we see a delta function peak at $\omega=0$?

enter image description here

Solidification
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  • If perturbation is constant, it has zero matrix elements between any pait of eigenstates of the unperturbed Hamiltonian, hence no transitions. – Roger V. Jun 29 '21 at 15:43
  • Related: https://physics.stackexchange.com/q/89402/2451 – Qmechanic Jun 29 '21 at 15:44
  • @RogerVadim This is not true. There is no reason why $V$ should be diagonal in the eigenbasis of $H_0$. This is from Sakurai. – Solidification Jun 29 '21 at 16:02
  • @mithusengupta123 perhaps, you should clarify what constant means in your question: it is not constant in time (it is a sudden perturbation that turns on at $t=0$), and, according to your comment, it is not a constant. – Roger V. Jun 29 '21 at 16:05
  • I have defined the perturbation in the question. This is what Sakurai essentially calls a constant perturbation. You can also find it here: Zwiebach Lecture – Solidification Jun 29 '21 at 16:07
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    Chapter 5.6, Page 328 of Modern Quantum Mechanics Revised Edition, J.J. Sakurai, for those interested. – Lucas Baldo Jun 29 '21 at 16:15
  • @mithusengupta123 You didn't define $V$ in the question, it would be better to do so. In the book, Sakurai writes: "Even though the operator V has no explicit time dependence, it is in general made up of operators like $\mathbf{x}$, $\mathbf{p}$ and $\mathbf{s}$". From your question one could assume $V$ is just a real number (times identity). – Lucas Baldo Jun 29 '21 at 16:19
  • Agreed! Point taken. – Solidification Jun 29 '21 at 16:21

2 Answers2

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Your perturbation is by no means constant as it has a step function, which has all frequencies in its Fourier transformation. For such violent step-function perturbations you should use the "sudden" approximation in which the transition amplitude is simply the overlap between the eigenstates of $H_0$ and $H_0+V$.

mike stone
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  • This is in Sakurai, under a section titled constant perturbation. Your answer does not explain why the peak occurs at $\omega_{fi}=0$. – Solidification Jun 29 '21 at 16:42
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Note that your plot assumes that $V_{if}$ is non-zero. This is telling you that if the perturbation connects degenerate states (such that $\omega_{if} = E_f-E_i$ is zero and $V_{if}\neq 0$), then the transition between these states is favoured over the others.

The transition to other states would cost more energy (than zero) so it is less likely.

If the perturbation is diagonal in the degenerate subspace, or if there is no degeneracy, then there are no $\omega_{if}$ and $V_{if}$ that satisfy the conditions above simultaneously.

Lucas Baldo
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  • I didn't get it. In general, the given perturbation will connect both degenerate as well as nondegenerate states. What is so special about the perturbation that it favors transition between degenerate states? Not clear. – Solidification Jun 29 '21 at 16:54
  • It is not the perturbation per se that favors the degenerate states, it is conservation of energy. If the initial system is in a given energy eigenstate, it will need to absorb or give up energy in order to jump to states that are non-degenerate to it. This breaks conservation of energy, but that is ok because we are doing everything at finite times and the uncertainty relations warned us of this. As a consequence, though, the higher the energy jump the bigger the "breaking", so the less likely it should happen. – Lucas Baldo Jun 29 '21 at 17:12
  • Of course, this is a general behavior and the details of the perturbation can change this. That is why the matrix elements $V_{if}$ modulate the probability amplitude and if they are zero, that transition does not happen. – Lucas Baldo Jun 29 '21 at 17:14
  • "This breaks conservation of energy" I'm not sure if this is true. The perturbation is there to supply any energy. If the perturbation cannot supply adequate energy, there must be a reason. Moreover, although before the perturbation the system was in an eigenstate of $H_0$, after the perturbation acts, the system is not in an energy eigenstate. So statements about energy conservation are a bit shaky (because energy is not well-defined), in my opinion. – Solidification Jun 29 '21 at 17:29
  • Fair point. I will leave this answer here as is until I think of a counter-argument or someone comes up with a better explanation. – Lucas Baldo Jun 29 '21 at 17:58