Now the eigenfunctions of the Hamiltonian clearly differ from one problem to another since the Hamiltonian depends on the potential and hence for a different potential we get a different eigenvalue equation for the Hamiltonian hence the eigenfunctions are different each time.

However, the eigenvalue equation for the position and momentum operators don't change (since they don't depends on the potential), so is it always the case that the eigenfunctions of momentum and position operators (which are continuous) are always the same? My gut tells me that no since the eigenfunctions must also satisfy boundary conditions which differ from one problem to another, but Griffiths book only solved this one case for the eigenfunctions of position and momentum operators and later used them as a standard eigenbasis of position/momentum for all types of problems. Even in the end of chapter problems he never asks us to solve the eigenvalue equation for the position and momentum operators for, say, the infinite well. Which left me with the aformentioned question: Are they always the same?


2 Answers 2


Yes they do vary. Say you're doing quantum mechanics on a circle. $\hat p\equiv -i\hbar \partial_x$ where $x\sim x+2\pi R$ is the coordinate along the circle, will then have discrete eigenvalues $k$ and complex exponential eigenfunctions: we have $$ \hat p e^{ikx/\hbar}=k e^{ikx/\hbar}\,, $$ but $e^{ikx/\hbar}$ will be a function on the circle iff it is periodic. Therefore $$ k\frac{R}{\hbar}\in\mathbb Z\,. $$

In contrast if you are considering quantum mechanics on say $\mathbb R^3$ then $\hat p$ will have continuous spectrum, so its "eigenvalues" $k\in\mathbb R^{3}$.

  • $\begingroup$ But the eigenfunctions with eigenvalues $k\in\mathbb R$ are still eigenfuncions of $\hat p$ whether or not you are on a circle or on $R$. The eigenfunctions are still eigenfunctions no matter the BCs $\endgroup$ Mar 3, 2020 at 18:33
  • 1
    $\begingroup$ I know that it is common to solve the Schroedinger differential equation for a free particle for example, and say that the general solution is an eigenfunction, no matter the boundary conditions. Just after that, there is the inevitable remark: it is not normalizable, and a valid function must vanish at infinity. It should be a better convention if eigenfunctions of an operator had to fulfill the boundary conditions as well. $\endgroup$ Mar 3, 2020 at 19:04
  • $\begingroup$ I don’t see how this is different from the case of say particle in a box. The restrictions on allowed 𝑘 comes from the boundary condition and not from the eigenstates of the operator. The eigenstates are like @AaronStevens pointed out, still all 𝑘 $\endgroup$ Mar 3, 2020 at 19:14
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    $\begingroup$ You have different Hilbert spaces for all these examples. Therefore, the operators are also different. We just do not write $\hat p_\text{circle}$ $\hat p_\text{line}$ and so on to save space. To further clarify @ClaudioSaspinski's remark: it is not just a convention that eigenfunctions have to satisfy the boundary conditions (i.e. lie inside the Hilbert space); we define an operator over some fixed space. There is no operator without a domain. $\endgroup$
    – user21299
    Mar 3, 2020 at 19:33
  • $\begingroup$ But can’t we consider those different Hilbert space as subspaces of an encompassing Hilbert space? $\endgroup$ Mar 3, 2020 at 19:35

The determination of eigenfunctions themselves do not depend on the boundary conditions. As you have said, the operators "don't change" from problem to problem, so the eigenfunctions will not change either. Where boundary conditions come into play is whether or not a quantum system can be in a specific eigenstate of some operator. If the eigenstate is "not allowed" due to inconsistent boundary conditions the system must have (or perhaps some other reason), then it stands to reason that the system will not be found to be in this state.

This is precisely what happens when solving for energy eigenstates of the electron in the Hydrogen atom.$^*$ We restrict the possible eigenfunctions to be periodic in the position basis (for the appropriate spatial coordinates). The eigenfunctions themselves aren't determined by periodic boundary conditions, but the possible eigenstates we can find the system in are determined by the boundary conditions. The "thrown out" eigenfunctions are still eigenfunctions of the energy operator, it's just that for this specific system we know we cannot use them.

In a more general sense, the reason why eigenfunctions are not determined by the boundary conditions is because the "eigenfunction equation" and the boundary conditions are separate things. You can write out general solutions to differential equations without specifying any boundary conditions. When you apply your boundary conditions to pick out the solutions that work for the specific system, the invalid solutions are still solutions to the "eigenfunction equation" even if you are not using them.

$^*$I know we recognized that these energy eigenstates "change" depending on the problem, but it still holds for an example of how the boundary conditions determine allowed eigenfunctions.

  • $\begingroup$ Ok I think I understand now. So just so we're clear: In the infinite square well on interval (0,a) the position eigenfunctions are still dirac deltas (technically incorrect I know, but besides the point) for all values of x, but the possible eigenfunction are dirac deltas which only range from 0 to a. Correct? $\endgroup$
    – Leonid
    Mar 3, 2020 at 17:52
  • $\begingroup$ @Leonid Yes, I think that is a valid understanding $\endgroup$ Mar 3, 2020 at 18:00

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