The Φ equation in Griffiths' Introduction to QM

In summary, Griffiths is discussing the derivation of the general solution for the angular part of the Schrödinger equation in polar coordinates. He notes that while the general solution is expressed as ##Ae^{im\phi} + Be^{-im\phi}##, he only considers the specific solutions ##Ae^{im\phi}## and ##Be^{-im\phi}##, which can be accounted for by allowing ##m## to become negative. This choice of labeling the solutions is just a matter of convenience and does not affect the completeness of the solution set. The reason for only considering the specific solutions may be due to the difficulty in proving that ##m## must be an integer for the more general solution. The footnote
  • #1
Why from Φ eq in page 136, he did not derive the solution in this form $Ae^{imφ}+Be^{-imφ}$https://drive.google.com/open?id=1thokDDIVDytck1-2HI3qyiG07mAMO1nc
 
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  • #2
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  • #3
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.
 
  • #4
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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  • #5
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)
 
  • #6
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?
 
  • #7
hoalacanhdk said:
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##.
 
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  • #8
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 :-((.
 
  • #9
vanhees71 said:
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.
 
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  • #10
I'm so angry about Griffiths's QM text, because he's so brillant with his E&M textbook. Why wasn't he as careful with his QM book? :-(
 

1. What is the Φ equation in Griffiths' Introduction to QM?

The Φ equation in Griffiths' Introduction to QM is a mathematical expression that represents the wave function of a quantum mechanical system. It is a key equation in quantum mechanics and is used to describe the probability amplitude of finding a particle at a particular location and time.

2. Why is the Φ equation important in quantum mechanics?

The Φ equation is important in quantum mechanics because it is used to calculate the probability of finding a particle in a specific location and time. This equation helps us understand the behavior of particles on a microscopic level and is essential in predicting the outcomes of experiments in quantum physics.

3. How is the Φ equation derived?

The Φ equation is derived from the Schrödinger equation, which is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. The Φ equation is a solution to the Schrödinger equation and is obtained by applying certain mathematical techniques and boundary conditions.

4. What are the key components of the Φ equation?

The Φ equation has two key components: the wave function (Φ) and the Hamiltonian operator (H). The wave function represents the state of a quantum system, and the Hamiltonian operator describes the energy of the system. The equation also includes the time variable (t), which is used to calculate the time evolution of the system.

5. How is the Φ equation used in practical applications?

The Φ equation has many practical applications, including predicting the behavior of particles in quantum systems, calculating the energy levels of atoms and molecules, and understanding the properties of materials on a microscopic level. It is also used in developing technologies such as quantum computing and quantum cryptography.

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