Eigenvalue Problem of Quantum Mechanics

In summary, the conversation discusses the eigenvalue problem in quantum mechanics and the use of the "polar" form of the wavefunction in de Broglie-Bohm models. There is a question about how this form of the wavefunction, with the Action term, affects the eigenvalue problem and whether it is allowed. The experts explain that this form is not necessary for calculating momentum eigenstates, and the normalization constant is found using the Dirac delta distribution.
  • #1
Hello,

I hope you are doing well.

I had a question about the eigenvalue problem of quantum mechanics. In a past class, I remember it was strongly emphasized that the eigenvalues of an eigenvalue problem is what we measure in the laboratory.

##A\psi = a\psi##

where A would be the operator and the 'a' would be the eigenvalue.

I've also seen some books use the "polar" form of the wavefunction, where:

##\psi = Re^{iS/\hbar}##

This is often used in de Broglie-Bohm models of QM to get the "quantum" Hamilton-Jacobi equation. What I wanted to ask is this: If we assume that R is a function of position and time (as is done in the de Broglie-Bohm approach), doesn't this violate the eigenvalue problem?

Because then we get extra terms that come from the derivatives of the amplitude once acted upon by the momentum or energy operators. This is something I've been giving some thought, and it has been bothering me. If you have some insight to share, that would be much appreciated.

Many thanks.

All the best!
 
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  • #2
As for momentum operator P eigenstate
[tex]P \psi = p \psi [/tex]
[tex] \psi = e^{ipx/\hbar}[/tex]
aside from normalization. Is this a case you refer ?
CuriousLearner8 said:
doesn't this violate the eigenvalue problem?
What is the reason why your ##\psi## have to be an eigenstate ?
 
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  • #3
CuriousLearner8 said:
I've also seen some books use the "polar" form of the wavefunction, where:

ψ=ReiS/ℏ
Without constraints this means nothing. Choose S=0.
 
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  • #4
anuttarasammyak said:
As for momentum operator P eigenstate
[tex]P \psi = p \psi [/tex]
[tex] \psi = e^{ipx/\hbar}[/tex]
aside from normalization. Is this a case you refer ?

What is the reason why your ##\psi## have to be an eigenstate ?

Hi anuttarasammyak, thanks for the reply.

The eigenvalue problems look correct but I've seen the wavefunction written in terms of the Action. So I've seen some papers write:

[tex]\psi(x,t) = R(x,t)e^{iS(x,t)/\hbar}[/tex]

where the Action S can be written out as [tex]S(x,t) = px - Et[/tex]. This is especially done to recover the quantum Hamilton-Jacobi equation (the above wavefunction plugged into the Schrodinger Equation gives us the quantum Hamilton-Jacobi equation.)

But this seems to violate the above eigenvalue problem. For example:
[tex]P\psi = -i\hbar \partial_x \psi = -i\hbar (\partial_x R)e^{iS/\hbar} -i\hbar Re^{iS/\hbar}(\partial_x S)(i/\hbar) [/tex]
%
We know that [tex] \partial_x S = p_x [/tex], which is the momentum in the x direction. The p should be the eigenvalue, but the first term [tex] -i\hbar (\partial_x R)e^{iS/\hbar}[/tex] seems to violate the form of [tex]P\psi = p\psi[/tex].

I think my confusion comes from why is this allowed? I am probably missing some key insight that makes this look strange. Thank you for any insight you guys could provide.
 
  • #5
There's no need for this ansatz in calculating the (generalized) momentum eigenstates. This ansatz is needed to derive the WKB approximation as a series of powers in ##\hbar##.

The momentum eigenstates are rather simple to find, given that in the position representation ##\hat{p}=-\mathrm{i} \hbar \partial_x##:
$$\hat{p} u_{p}(x)=p u_p(x) \; \Rightarrow \; -\mathrm{i} \hbar u_p'(x)=p u_p(x)$$
The solution is very simple to find:
$$u_p(x)=N_p \exp(\mathrm{i} p x/\hbar).$$
It's unique up to a normalization constant, and this normalization is a bit tricky, because obviously this is not a proper wave function, because it's not square integrable, because ##|u_p(x)|^2=|N|^2=\text{const}##, i.e., the integral of ##|u_{p}(x)|^2## will diverge to infinity. This is, because the momentum operator is not bound and has a continuous spectrum, i.e., the "eigenvalue" (or rather its spectrum) is ##p \in \mathbb{R}##. In such cases you can only "normalize to a ##\delta## distribution", i.e., for two such "generalized eigenfunctions" you can make the "scalar product" a Dirac ##\delta## distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} \mathrm{d} x N_p^* N_{p'} \exp[\mathrm{i} x(p'-p)/\hbar] = |N_{p}|^2 2 \pi \delta[(p-p')/\hbar] = |N_p|^2 2 \pi \hbar \delta(p-p') \stackrel{!}{=} \delta(p-p'),$$
i.e., finally you get ##N_p=1/\sqrt{2 \pi \hbar}## and thus
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
 
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  • #6
vanhees71 said:
There's no need for this ansatz in calculating the (generalized) momentum eigenstates. This ansatz is needed to derive the WKB approximation as a series of powers in ##\hbar##.

The momentum eigenstates are rather simple to find, given that in the position representation ##\hat{p}=-\mathrm{i} \hbar \partial_x##:
$$\hat{p} u_{p}(x)=p u_p(x) \; \Rightarrow \; -\mathrm{i} \hbar u_p'(x)=p u_p(x)$$
The solution is very simple to find:
$$u_p(x)=N_p \exp(\mathrm{i} p x/\hbar).$$
It's unique up to a normalization constant, and this normalization is a bit tricky, because obviously this is not a proper wave function, because it's not square integrable, because ##|u_p(x)|^2=|N|^2=\text{const}##, i.e., the integral of ##|u_{p}(x)|^2## will diverge to infinity. This is, because the momentum operator is not bound and has a continuous spectrum, i.e., the "eigenvalue" (or rather its spectrum) is ##p \in \mathbb{R}##. In such cases you can only "normalize to a ##\delta## distribution", i.e., for two such "generalized eigenfunctions" you can make the "scalar product" a Dirac ##\delta## distribution,
$$\langle u_p|u_{p'} \rangle=\int_{\mathbb{R}} \mathrm{d} x N_p^* N_{p'} \exp[\mathrm{i} x(p'-p)/\hbar] = |N_{p}|^2 2 \pi \delta[(p-p')/\hbar] = |N_p|^2 2 \pi \hbar \delta(p-p') \stackrel{!}{=} \delta(p-p'),$$
i.e., finally you get ##N_p=1/\sqrt{2 \pi \hbar}## and thus
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
All of this makes sense to me, but I think what I'm trying to figure out is why the ansatz some use is not in contradiction to the eigenvalue problem? Isn't the eigenvalue problem a fundamental postulate of quantum mechanics, therefore should not be violated?
 
  • #7
Operators which commute allow choice of solutions which are eigenstates of both. The Ansatz you propose without explicit or implicit restriction on R (does that have a definitional reatriction?)) is not at all restrictive. With restriction one may still generate energy eigensatates (as @vanhees71 shows) and the form is very useful otherwise.
An ansatz is usually an educated guess whose deviation from generality will illuminate an exact solution (best case) or produce a well defined approximate form suitable for optimization with further work. Look up Bethe Ansatz for instance
 
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