# The Φ equation in Griffiths' Introduction to QM

Why from Φ eq in page 136, he did not derive the solution in this form $Ae^{imφ}+Be^{-imφ}$

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Delta2
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The general solution is indeed as you say, ##Ae^{im\phi}+Be^{-im\phi}## however it seems (for reasons that are unclear to me) that he considers only the specific solution ##Ae^{im\phi}## and he says that the constant A is absorbed by ##\Theta## and that he also covers the other specific solution ##Be^{-im\phi}## by allowing m to become negative.

vela
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It's because in the end, he wants a solution which is an eigenfunction of ##\hat L_z##.

Anyway, it's just a choice of how to label the solutions. For ##m=1##, for example, you could say there are two solutions, ##e^{+i\phi}## and ##e^{-i\phi}##, or you could instead say there's one state labeled ##m=1## and one labeled ##m=-1##. Either way, you're accounting for all of the solutions.

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But why he didn't use the combination of two solutions simultaneously? I think exp(-1)+exp(1) is a different solution from either exp(-1) or exp(1)

George Jones
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The quoted passage comes from the second edition of "Introduction to Quantum Mechanics" by Griffiths (I just received my copy of the new third edition). @hoalacanhdk, have you covered section 3.3, in particular, the paragraph on page 102 that deals with degenerate eigenvalues?

PeroK
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But why he didn't use the combination of two solutions simultaneously? I think exp(-1)+exp(1) is a different solution from either exp(-1) or exp(1)
All Griffiths wants to do is prove that ##m## must be an integer. He does that for the simple solution ##e^{im\phi}##. It looks harder to prove this for the more general solution ##Ae^{im\phi} + Be^{-im \phi}##. And, I must admit, I don't immediately see a good argument that allows the simplification. (Perhaps someone else can?)

After that, the ##\theta## equation depends only on ##m^2##, hence you get the same spherical harmonics for ##m## and ##-m##. The general solution, therefore, ends up the same, whether you consider the solutions of the angular equation to be ##e^{im\phi}## for integer ##m##; or, ##Ae^{im\phi} + Be^{-im \phi}## for ##m \ge 0##.

Delta2
vanhees71
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Well, again it seems to me the best advice is to use another QM textbook ;-)). If you look for eigenvalues of ##\hat{L}_z=-\mathrm{i} \hbar \partial_{\varphi}## you get unique solutions ##u_m(\varphi)=\exp(\mathrm{i} m \varphi)/\sqrt{2 \pi}## with the eigenvalues being ##m \hbar##.

A short answer, why ##m## should be integer is that you want uniqueness of the wave functions under rotations by ##2 \pi## aroud the ##z## axis. This is an incomplete argument however, because in quantum mechanics the absolute phase doesn't count. A more complete answer is that you get a complete orthonormal set of functions by this set when ##m \in \mathbb{Z}##.

If you look for the eigenfunctions of ##\hat{L}_z^2## from this it's clear that for each eigenvalue (except 0) the eigenvalue is degenerate, i.e., for each possible eigenvalue you have two linearly independent solutions, which you can choose of course as the eigenstates of ##\hat{L}_z##, i.e., ##u_{\pm m}(\varphi)## for each ##m## (with the eigenvalue ##\hbar^2 m^2## for ##\hat{L}_z^2##). You can also choose ##\cos(m \varphi)## or ##\sin(m \varphi)##. Of course, all these sets form again a complete set of orthogonal functions on ##\mathrm{L}^2([0,2 \pi])##, but why you would bother the students with a case of degenerate eigenvalues where you can much simpler treat the problem by first looking for the eigenvalues and eigenstates of ##\hat{L}_z##, I can't tell :-((.

PeroK
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2020 Award
Well, again it seems to me the best advice is to use another QM textbook ;-)). If you look for eigenvalues of ##\hat{L}_z=-\mathrm{i} \hbar \partial_{\varphi}## you get unique solutions ##u_m(\varphi)=\exp(\mathrm{i} m \varphi)/\sqrt{2 \pi}## with the eigenvalues being ##m \hbar##.

A short answer, why ##m## should be integer is that you want uniqueness of the wave functions under rotations by ##2 \pi## aroud the ##z## axis. This is an incomplete argument however, because in quantum mechanics the absolute phase doesn't count. A more complete answer is that you get a complete orthonormal set of functions by this set when ##m \in \mathbb{Z}##.

If you look for the eigenfunctions of ##\hat{L}_z^2## from this it's clear that for each eigenvalue (except 0) the eigenvalue is degenerate, i.e., for each possible eigenvalue you have two linearly independent solutions, which you can choose of course as the eigenstates of ##\hat{L}_z##, i.e., ##u_{\pm m}(\varphi)## for each ##m## (with the eigenvalue ##\hbar^2 m^2## for ##\hat{L}_z^2##). You can also choose ##\cos(m \varphi)## or ##\sin(m \varphi)##. Of course, all these sets form again a complete set of orthogonal functions on ##\mathrm{L}^2([0,2 \pi])##, but why you would bother the students with a case of degenerate eigenvalues where you can much simpler treat the problem by first looking for the eigenvalues and eigenstates of ##\hat{L}_z##, I can't tell :-((.
In Griffiths' defence, the footnote in the above extract points this out as a flaw in the current argument and promises a more compelling proof when he gets to angular momentum.

vanhees71
vanhees71