Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Are the possible momenta of a particle in an infinite square well on the interval [-½,½] countable or uncountable?

+ 2 like - 1 dislike
269 views

An operator $P$ in Hilbert space is called self-adjoint only if the domains of $P$ and its adjoint $P^\dagger$ coinside. Therefore, if we want to consider the momentum of a particle inside a square well on $I=[-½,½]$, the least boundary conditions one has to impose on the momentum operator $P=-i\partial_x$ are given by the domain $$D_p=\{\psi,P\psi\in L^2[I]:\psi(½)=e^{i\theta}\psi(-½)\},$$ for some real angle $\theta$.  In the following it might be sufficient to consider the special case $\theta=0.$ Then, the functions $$\phi_n(x)=e^{i 2 \pi nx},\qquad n=0,\pm1,\pm2,...$$ are eigenfunctions of $P$ with momentum eigenvalues
$$k_n=2\pi n$$ and form a complete orthonormal set. 

The momentum operator and the energy operator $H$ of the system do not commute and one can not expect them to have a common eigenbasis. Actually, the boundary conditions of the eigenvalue problem of $H$ is $\psi(-½)=\psi(½)=0$, which is different to that of the momentum above. Now, in wikipedia the possible values of the momentum of this problem are explained to be continuous (https://goo.gl/vrJxly). That is, on one hand there is the countable momentum spectrum $k_n$ above, but on the other hand there is the uncountable spectrum proposed in literature.

Question: Which one of them is realized in nature?

Note: From the experimental point of view one could refer to a single slit diffraction experiment and conclude that the momentum spectrum should be continuous with their maxima at the values of the point spectrum. However, we know that the diffraction pattern is just far field approximation of the time evolution of the position wave function, which can only be considered as a (continuous) Fourier integral transformation if the momentum is (heuristically defined) proportional to the position of the particle.  

asked Jan 1 in Theoretical Physics by kaffeeauf (45 points) [ revision history ]
edited Jan 4 by kaffeeauf

One should not confuse the problem of a particle in the real line \(\mathbb{R}\) subjected to an infinite square well potential to that of a particle confined in a circle \(S^1\). In the latter problem (particle in the circle), the Fourier modes \(\langle\theta|n\rangle=e^{inx/\hbar}/(2\pi\hbar)^{1/2}\) for \(n \in \mathbb{Z}\) are simultaneous eigenstates of the Hamiltonian and momentum operators, and they form a complete orthonormal set.

On the other hand, in the problem of a particle in the real line within an infinite square well potential, the momentum states do not need to satisfy yours boundary condition imposed on \(D_\boldsymbol{p}\), namely, that \(\psi(0)=e^{i\theta}\psi(L)\) (I have translated the boundaries of the box from \([-1/2,+1/2]\) to \([0,L]\)). The energy eigenfunctions need to have a support \(\subset [0,L]\). But the momentum states wave functions are still the ones from the whole line, namely, \(\langle x|p\rangle=e^{ipx/\hbar}/(2\pi \hbar)^{1/2} \forall p\in\mathbb{R}\), while the Hamiltonian eigenstates \(Hf_n=Ef_n\) are now given (in position representation) by the functions \(\langle x|f_n\rangle=(2/L)^{1/2}sin(2\pi n x/L)\forall x \in [0,L]\) and are zero otherwise, for \(n \in \mathbb{Z}\). In the momentum representation, these energy eigenstates reads

\[\tilde{f}_n(p)=\langle p|f_n\rangle = \int_{\mathbb{R}} dx\langle p|x\rangle\langle x|f_n\rangle\propto\int_0^L dx e^{-ipx/\hbar}sin(2\pi nx/L),\]

agreeing with Wikipedia's very much well-known result that these are just the Fourier transform of the position representation energy eigenfunctions. Just observe that now, generally, \(\tilde{f}_n(p) \neq 0\) for \(p \in \mathbb{R}\), which shows that the stationary states for yours Hamiltonian are now a superposition of uncountable many momentum states of the line. (This is needed to make the support of the energy eigenstates compact.)

From a physical viewpoint, you can understand qualitatively what is going on from the uncertainty principle. Since the particle is within \(x\in[0,L]\), the uncertainty in position is bounded from above, \(\Delta x \le L\). So, since \(\Delta p \Delta x \ge\hbar/2\) we cannot have \(\Delta p=0\), that is, we always must have an uncertainty in the momentum. Thus the energy eigenstates are indeed expected to be formed by a superposition of momentum waves.

Now you ask which one is realized in Nature. This sounds preposterous, since all this is very much idealized. The best one can do here is to confine a particle in a finite but very deep square well. You will have an amplitude for the particle leaking the well, described by an exponential decreasing term etc., but the above remarks still holds, namely, the available momentum are still a continuum. (I once again remark that this is not the same problem as that of a particle in the circle!)

Note. As an exercise, I recommend you to do the following: let \(k=2\pi n/L\) and, by taking the limit \(L \longrightarrow\infty\)\(k\) becomes a continuous variable. Then, using the exponential representation of the sine function prove that

\(\lim_{L \longrightarrow \infty}\tilde{f}_n(p)\propto \delta(k-p/\hbar)+\delta(k+p/\hbar).\)

So we recover the result that \(|p|=\hbar |k|\).

If one does not respect the "twisted" boundary conditions $\psi(0)=e^{i\theta}\psi(L)$ for a particle in the box, then the momentum operator would not be self-adjoined. And I'm not talking about a particle on the circle, but about the least conditon for self-adjointness of the momentum of a particle with a wave-function of compact support. 

Concerning the uncertainty principle, I did not state that the momentum dispersion does vanish. On the contrary, by a simple computation we can see that the momentum disperison satisfies the tight inequality $\sigma_p L\geq\pi\hbar$ in this approach. So, everyting seems to be consistent.

@IgorMol Great comment! But since it's quite long, could you post it as an answer instead?

@kaffeeauf The basic issue is that only the Hamiltonian has to be a self-adjoint operator in order to give a well-defined eigenvalue problem. It is not necessary for the momentum operator to be self-adjoint.

Thank you for your positive feedback Greg, I have just copied my comment to an answer.

2 Answers

+ 3 like - 0 dislike

One should not confuse the problem of a particle in the real line \(\mathbb{R}\) subjected to an infinite square well potential to that of a particle confined in a circle \(S^1\). In the latter problem (particle in the circle), the Fourier modes \(\langle\theta|n\rangle=e^{inx/\hbar}/(2\pi\hbar)^{1/2}\) for \(n \in \mathbb{Z}\) are simultaneous eigenstates of the Hamiltonian and momentum operators, and they form a complete orthonormal set.

On the other hand, in the problem of a particle in the real line within an infinite square well potential, the momentum states do not need to satisfy yours boundary condition imposed on \(D_\boldsymbol{p}\), namely, that \(\psi(0)=e^{i\theta}\psi(L)\) (I have translated the boundaries of the box from \([-1/2,+1/2]\) to \([0,L]\)). The energy eigenfunctions need to have a support \(\subset [0,L]\). But the momentum states wave functions are still the ones from the whole line, namely, \(\langle x|p\rangle=e^{ipx/\hbar}/(2\pi \hbar)^{1/2} \forall p\in\mathbb{R}\), while the Hamiltonian eigenstates \(Hf_n=Ef_n\) are now given (in position representation) by the functions \(\langle x|f_n\rangle=(2/L)^{1/2}sin(2\pi n x/L)\forall x \in [0,L]\) and are zero otherwise, for \(n \in \mathbb{Z}\). In the momentum representation, these energy eigenstates reads

\[\tilde{f}_n(p)=\langle p|f_n\rangle = \int_{\mathbb{R}} dx\langle p|x\rangle\langle x|f_n\rangle\propto\int_0^L dx e^{-ipx/\hbar}sin(2\pi nx/L),\]

agreeing with Wikipedia's very much well-known result that these are just the Fourier transform of the position representation energy eigenfunctions. Just observe that now, generally, \(\tilde{f}_n(p) \neq 0\) for \(p \in \mathbb{R}\), which shows that the stationary states for yours Hamiltonian are now a superposition of uncountable many momentum states of the line. (This is needed to make the support of the energy eigenstates compact.)

From a physical viewpoint, you can understand qualitatively what is going on from the uncertainty principle. Since the particle is within \(x\in[0,L]\), the uncertainty in position is bounded from above, \(\Delta x \le L\). So, since \(\Delta p \Delta x \ge\hbar/2\) we cannot have \(\Delta p=0\), that is, we always must have an uncertainty in the momentum. Thus the energy eigenstates are indeed expected to be formed by a superposition of momentum waves.

Now you ask which one is realized in Nature. This sounds preposterous, since all this is very much idealized. The best one can do here is to confine a particle in a finite but very deep square well. You will have an amplitude for the particle leaking the well, described by an exponential decreasing term etc., but the above remarks still holds, namely, the available momentum are still a continuum. (I once again remark that this is not the same problem as that of a particle in the circle!)

Note. As an exercise, I recommend you to do the following: let \(k=2\pi n/L\) and, by taking the limit \(L \longrightarrow\infty\)\(k\) becomes a continuous variable. Then, using the exponential representation of the sine function prove that

\(\lim_{L \longrightarrow \infty}\tilde{f}_n(p)\propto \delta(k-p/\hbar)+\delta(k+p/\hbar).\)

So we recover the result that \(|p|=\hbar |k|\).

answered Jan 8 by Igor Mol (420 points) [ no revision ]
+ 2 like - 0 dislike


There are two different problems to distinguish:

1) a free particle in a box with periodic boundary conditions, i.e. a free particle on a circle.

2) a particle on a line, in a infinite square well, i.e. a particle in a box.

In 1), the space of states is the space of periodic functions (i.e. functions on a circle) and the spectrum of the momentum is discrete. In this case, eigenstates of momentum are normalizable states and are also the eigenstates of the free Hamiltonian.

In 2), the space of states is the space of L^2 functions on a line. The spectrum of the momentum is continuous and eigenstates of the momentum are not normalizable. Eigenstates of the Hamiltonian are normalizable states (and are not eigenstates of the momentum).

1) and 2) being different problems, it does not make sense to say that  only one of them is relevant. The relevance of 1) or 2) for a particular experiment will depend on the details of the experiment.

Mathematically, 1) is much simpler than 2) and for a large box, 1) and 2) becomes close to each other (for a large box, what happens at the boundaries becomes less relevant) and so one can use results about 1) to derive approximate results about 2).

answered Jan 2 by 40227 (4,660 points) [ no revision ]

The particle under consideration has finite support $[−½,½]$ and the eigenvalue problem of the Hamiltonian is solved with respect to that support. Why should that not be the goal for the eigenvalue problem of the momentum operator - for instance, to ensure self-adjointness of the momentum operator with respect to $[−½,½]$, instead to the infinite real line? Neither in wikipedia nor in your (or Igor's) explanation I can see a justification why the momentum eigenvalue problem should make reference to positions outside $[−½,½]$. If you have an idea then please let me know. 

In general, the infinite square well is defined as the limit of a finite square well becoming deeper and deeper. For a finite square well, the wave function does not have finite support but has exponentially small contributions when going to infinity. Only in the infinite square well limit, the eigenfunctions of the Hamiltonian become of finite support: it is not something imposed from the beginning but it comes from solving the Schrödinger equation on a line.

But it seems that you want from the beginning restrict the space of wave functions to the space of functions with support on [-1/2,1/2] and vanishing at the boundaries. Solving the eigenvalue problem for the Hamiltonian on this space of functions will  indeed be equivalent to the above approach on the real line. But now the momentum operator is no longer defined: the derivative of a function vanishing at the boundaries of an interval has no reason to vanish there. It is physically clear: momentum is the generator of translations, so if your space of wave functions is not invariant under translation, there is no way to define a momentum operator on this space.

Sure, in my question I only consider the case where the square well is infinite just from the beginning. As far as I know, that seems to be the case im most standard contributions in literature. However, it would be interesting to know a reference where the limit process is performed afterwards.

I think, the vanishing boundary conditions of the energy eigenfunctions are only appropriate for the eigenvalue problem of the energy. For the momentum operator the eigenstates are expected to be different form the eigenfunctions of $H$ because of the incommensurability of $H$ and $P$. That releases us from the restriction that the momentum eigenstates or their derivatives have to vanish at the boundary. Anyway, general wave functions with vanishing boundary conditions (i.e. the ground state of $H$) can also be constructed in the Hilbert space generated by the eigenstates of the momentum operator. In my (humble) opinion the most important criterion is that the wave functions and their derivatives have to be members of $D_p$ such that self-adjointness of $P$ is ensured. However, what I miss is a justification of the superposition of the momentum eigenstates which goes beyond the theory of Fourier series at all. 

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...