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On page 257 of Griffiths' Quantum book, he says "Typically, the perturbation will break the degeneracy: As we increase $\lambda$, the common unperturbed energy $E^0$ splits into two. Going the other direction, when we turn off the perturbation, the "upper" state reduces down to one linear combination of $\psi^0_a$ and $\psi^0_b$ [the degenerate states], and the "lower" state reduces to some orthogonal linear combination, but we don't know a priori what these "good" linear combinations will be."

Griffiths provides no justification for the proposition that the perturbation splits degenerate states into degenerate linear combinations, he just states it which isn't very satisfying. It feels right, but I'm not sure how to prove that this must be true. In fact, I now realize that I don't think any justification was given for the stationary states of an unperturbed Hamiltonian being an orthogonal set.

Could you please show me why this must be the case? Thank you!

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In general a perturbation need not lift the degeneracy of the original Hamiltonian, Griffiths simply focuses on the case where it does because the case where it does not is essentially the same as non-degenerate perturbation theory and so relatively uninteresting.

When the perturbation does lift the degeneracy the eigenstates of the perturbed Hamiltonian are, by definition, eigenstates of a Hermitian operator with different eigenvalues and so must be orthogonal.

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  • $\begingroup$ Ok, but part of what I ask (at the end) is why eigenstates of a Hermitian operator with different eigenvalues must be orthogonal. Sorry if this seems trivial, but that means it's all the more important that I know it. Thanks for your answer $\endgroup$ – Mr. HelloBye May 2 '17 at 22:16
  • $\begingroup$ See this question. I don't have a copy of Griffiths to hand, but I would be shocked if he does not prove this somewhere as well $\endgroup$ – By Symmetry May 2 '17 at 22:24
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Degenerate states come from a symmetry in the Hamiltonian. For instance, some degree of freedom needed to describe a state of the system might not be present in the Hamiltonian at all. This is the case for the Hydrogen atom without corrections, the spin operator does not appear in the Hamiltonian at all.

If the perturbation you are considering also shares the same symmetry then it will not break the degeneracy of the states. However, again using the example of the hydrogen atom, the degeneracy for spin states is broken when you consider a perturbation that couples the spin to a magnetic field.

So in general, you need to see whether the symmetry of the Hamiltonian which causes the degeneracy is present in the perturbation.

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  • $\begingroup$ Thank you for the intuitive explanation of what determines whether a perturbation breaks the symmetry, but this doesn't answer my question of why the states must be orthogonal. Of course it's nice to work with an orthonormal basis, but I'm not clear why the basis of a Hermitian operator must be orthonormal. $\endgroup$ – Mr. HelloBye May 2 '17 at 22:20
  • $\begingroup$ I am not sure what you're asking. It's up to you to decide how you want to represent the states of the system, using a basis which is not orthogonal or normalized will just make doing calculations more difficult. I think you should ask "how do we know that there exists an orthonormal basis" rather than "why do we have to use one" if anything. Does that make sense? $\endgroup$ – Bobak Hashemi May 3 '17 at 0:36
  • $\begingroup$ That is what I'm asking. As I said, of course it's nice to work with an orthonormal basis, I just don't get why it must be so, or as you say how we know it must be so. $\endgroup$ – Mr. HelloBye May 6 '17 at 21:07
  • $\begingroup$ You don't need to, you just don't gain anything by working in a non-orthonormal basis. For instance, one can span the Cartesian plane with a pair of "oblique axes", which are straight lines that come together at an angle which is not 90 degs (so that the axes have non-zero projection on one another). The link shows you what it looks like to plot the standard function for a circle in oblique coordiantes, notice it is no longer a circle. So you see, standard formulas need to be rederived when working in another coordinate system. [cont...] $\endgroup$ – Bobak Hashemi May 6 '17 at 21:43
  • $\begingroup$ The equation for a circle can, of course, be found in the new coordinate system, but unless there is a good reason to use the oblique coordinates, why bother? Quantum mechanics has tons of dot products involved, using an orthonormal basis means that those dot products are easy to compute because for any pair of states $|x_i \rangle$ and $|x_j \rangle$, the simplest possible dot product surely is $\langle x_i | x_j \rangle = \delta_{i,j}$. In an arbitrary coordinate system you have $\langle x_i | x_j \rangle = f_{i,j}(x_1,...x_n)$, which makes computing answers more difficult in typical cases. $\endgroup$ – Bobak Hashemi May 6 '17 at 21:46

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