# Eigenvalue Problem of Quantum Mechanics

CuriousLearner8
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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## Answers and Replies

Gold Member
As for momentum operator P eigenstate
$$P \psi = p \psi$$
$$\psi = e^{ipx/\hbar}$$
aside from normalization. Is this a case you refer ?
doesn't this violate the eigenvalue problem?
What is the reason why your ##\psi## have to be an eigenstate ?

Last edited:
topsquark
Homework Helper
2022 Award
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.

topsquark
CuriousLearner8
As for momentum operator P eigenstate
$$P \psi = p \psi$$
$$\psi = e^{ipx/\hbar}$$
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:

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

where the Action S can be written out as $$S(x,t) = px - Et$$. 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:
$$P\psi = -i\hbar \partial_x \psi = -i\hbar (\partial_x R)e^{iS/\hbar} -i\hbar Re^{iS/\hbar}(\partial_x S)(i/\hbar)$$
%
We know that $$\partial_x S = p_x$$, which is the momentum in the x direction. The p should be the eigenvalue, but the first term $$-i\hbar (\partial_x R)e^{iS/\hbar}$$ seems to violate the form of $$P\psi = p\psi$$.

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.

Gold Member
2022 Award
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
$$u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$

CuriousLearner8 and hutchphd
CuriousLearner8
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?