Every now and then a question comes up about the status of the momentum operator in the infinite square well, and while we have two good answers on the topic here and here, I'm generally not satisfied by their level of detail and by how easy (not very) it is to find them when you need them. I therefore posit that it is high time we settle things and build a canonical Q&A thread for this, so, with that in mind:
What's the deal with momentum in the infinite square well?
- Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$ symmetric when restricted to the compact interval of the well? Are there any subtleties in its definition, via its domain or similar, that are not present for the real-line version?
- Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$ self-adjoint in these conditions? If not, why not, and what are the consequences in terms of the things we normally care about when doing one-dimensional QM?
- If it is not self-adjoint, does it admit a self-adjoint extension? If it does, is that extension unique? If the extension is not unique, what are the different possible choices, and what are their differences? Do those differences carry physical meaning / associations / consequences? And what is a self-adjoint extension anyways, and where can I read up about them?
- What is the spectrum and eigenvectors of the momentum operator and its extensions? How do they differ from each other? Is there such a thing as a momentum representation in this setting? If not, why not?
- What is the relationship between the momentum operator (and its possible extensions) and the hamiltonian? Do they commute? Do they share a basis? If not, why not?
- Do these problems have counterparts or explanations in classical mechanics? (nudge, nudge)
And, more importantly: what are good, complete and readable references where one can go and get more information about this?
Best Answer
(I henceforth assume the Hilbert space is $L^2([0,1],dx)$.)
It depends on the precise definition of the domain of $P$. A natural choice is $$D(P) = \{\psi \in C^2([0,1])\:|\: \psi(0)=\psi(1)=0\}\:.\tag{1}$$ With this definition $P$ is symmetric: (a) the domain is dense in $L^2([0,1], dx)$ and (b) the operator is Hermitian $$\langle P\psi| \phi \rangle = \langle \psi| P\phi \rangle\quad \mbox{for $\psi, \phi \in D(P)\:.$}\tag{2}$$ You can consider different definitions of the domain more or less equivalent. The point is that the self-adjoint extension are related with the closure of $P$ and not to $P$ itself, and you may have several possibilities to get the same closure stating from different domains. The situation is similar to what happens on the real line. There $P$ can be defined as a differential operator on $C_0^\infty(\mathbb R)$ or $\cal S(\mathbb R)$ (Schwartz' space), or $C^1_0(\mathbb R)$ and also interpreting the derivative in weak sense. In all cases the closure of $P$ is the same.
It is not self-adjoint with the said choice of its domain (or with every trivial modification of that domain). The consequence is that it does not admit a spectral decomposition as it stands and therefore it is not an observable since there is no PVM associated with it.
Defining $P= -i \hbar \frac{\mathrm d}{\mathrm dx}$ on the real line with one of the domain said above, the same problem arises.
The general fact is that differential operators are never self-adjoint because the adjoint of a differential operator is not a differential operator since it cannot distinguish between smooth and non-smooth functions, because elements of $L^2$ are functions up to zero-measure sets. At most a symmetric differential operator can be essentially self-adjoint, i.e., it admits a unique self-adjoint extension (which coincides to the closure of the initial operator). This unique self-adjoint operator is the true observable of the theory.
Yes. The canonical way is checking whether defect indices of $P$ with domain (1) are equal and they are. But the shortest way consists of invoking a theorem by von Neumann:
If a (densely defined) symmetric operator commutes with an antilinear operator $C$ defined on the whole Hilbert space and such that $CC=I$, then the operator admits self-adjoint extensions.
In this case $(C\psi)(x):= \overline{\psi(1-x)}$ satisfies the hypothesis.
NO it is not, the operator is not essentially self-adjoint.
There is a class of self-adjoint extensions parametrized by elements $\chi$ of $U(1)$. These extensions are defined on the corresponding extension of the domain $$D_\chi(P) := \{\psi \in L^2([0,1],dx)\:|\:\psi' \mbox{in weak sense exists in $L^2([0,1],dx)$ and $\psi(1) = \chi\psi(0)$} \}\:. $$ (It is possible to prove that with the said definition of $D_\chi$ the definition is consistent: $\psi$ is continuous so that $\psi(0)$ and $\psi(1)$ makes sense.) Next the self-adjoint extension of $P$ over $D_\chi(P)$ is again $-i\hbar \frac{d}{dx}$ where the derivative is interpreted in weak sense. The simplest case is $\chi=1$ and you have the standard momentum operator with periodic boundary conditions, that is self-adjoint. The other self-adjoint extensions are trivial changes of this definition. I do not know the physical meaning of these different choices (if any): the theory is too elementary at this stage to imagine some physical interpretation. Maybe with an improved model a physical interpretation arises.
You can easily compute the spectrum which is a pure-point spectrum and the eigenvectors are shifted exponentials. If $\chi = e^{i \alpha}$ where $\alpha \in \mathbb R$, and we denote by $P_\alpha$ the associated self-adjoint extension of $P$ a set of eigenvectors is $$\psi^{(\alpha)}_n(x) = e^{i(\alpha + 2\pi n)x}$$ with eigenvalues $$p^{(\alpha)}_n := \hbar(\alpha + 2\pi n)\quad n \in \mathbb Z\::$$ The set of the $\psi^{(\alpha)}_n$ is a Hilbert basis because it is connected with the standard basis of exponentials by means of the unitary operator $(U_\alpha \psi)(x) = e^{i\alpha x} \psi(x)$. Essentially Nelson's theorem and the spectral decomposition theorem prove that $P_\alpha$ has pure point spectrum made of the reals $p_n^{(\alpha)}$. So a momentum representation exists as you can immediately prove.
As you know, if you start form $H := -\hbar \frac{d^2}{dx^2}$ on $D(H):= \{\psi \in C^2([0,1]) \:|\: \psi(0)=\psi(1) =0\}$ (I assume $2m=1$) this is essentially self-adjoint, though the corresponding momentum operator with domain (1) is not. (The proof immediately arises from Nelson's theorem since $H$ is symmetric and admits a Hilbert basis of eigenfunctions.)
However there are also different candidates for the Hamiltonian operator arising by taking the second power of each self-adjoint extension of $P$ with domain (1). The spectrum is made of the second powers the elements of the spectrum of the corresponding seof-adjoint extension $\hbar^2(\alpha + 2\pi n)^2$
Momentum and associated Hamiltonian commute and a common basis is that written above for the momentum.
Different self-adjoint extensions and different Hamiltonians do not commute as you easily prove by direct inspection.
I do not know
I do not know, many results are spread in the literature. It is difficult to collect all them. A good reference is Reed and Simon's textbook: Vol I and II.
ADDENDUM. A technical point deserves a little discussion. Sometimes when introducing selfadjointness domains as above in the space $L^2(I)$, where $I\subset \mathbb{R}$ is a bounded interval, the functions $\psi$ are required to be absolutely continuous. This requirement is actually included in the condition that the weak derivative $\psi'$ exists and is included in $L^1$ (or $L^2$ since $I$ is bounded). In fact, a measurable function $\psi:I \to \mathbb{C}$ is absolutely continuous if and only if it admits weak derivative in $L^1(I)$. In this case, as the function is absolutely continuous, its derivative exists almost everywhere and coincides with $\psi'$.