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

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

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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.

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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George Jones

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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##.

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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 :-((.

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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.

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