In the context of mathematical quantum mechanics, a well known no-go theorem known as Hellinger-Töplitz tells us that an unbounded, symmetric operator cannot be defined everywhere on the Hilbert space $\mathcal H$. Thus, for many operators of interest in quantum mechanics, we must resort to restricting the domain on which they may act. Along with the Hamiltonian, the momentum operator $$ p:=-i\frac{\rm d}{\mathrm{d}x}$$ is the most prominent example of such an operator. Now, one of the most important questions is clearly: What is the correct domain of definition for the momentum operator?

In order to answer this question, it is of crucial importance to consider the question of self-adjointness: Only self-adjoint operators are admissible as observables of the theory, and there are certain vital mathematical results, such as the spectral theorem and Stone's theorem, which make it clear that any 'good' momentum operator must certainly be self-adjoint: $ p^\star= p$. So, the original question is refined slightly: What is the correct domain $\mathcal D( p)\subset \mathcal H$ that allows us to construct a self-adjoint momentum operator?

Recently, one of my lecturers treated a concrete incarnation of this (general and possibly vague) question: Consider a Hilbert space $\mathcal H=L^2(0,1)$ and two 'candidate momenta', $ p_0$ and $ p_\alpha$, with domains

$$\mathcal D(p_0)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=0=\psi(1)\} $$ $$ \mathcal D(p_\alpha)=\{\psi\in \mathcal H \,|\, \psi\in \text{AC}(0,1),\psi'\in\mathcal H, \psi(0)=\alpha\psi(1)\},\hspace{.5cm}|\alpha|=1 $$ Are these operators self-adjoint? As it turns out, it can be shown that $p_0$ is symmetric but not self-adjoint. However, $p_\alpha$ is self-adjoint. It seems that the domain of $p_0$ is 'too small'.

Now, my question is: What does the conclusion that $p_0\neq p_0^\star$ imply for the canonical freshman QM example of the infinite potential well? As far as I'm aware, it is conventional to take exactly the boundary conditions that we assume to define the domain of $p_0$ when solving the Schrödinger equation in this case. Can we conclude that the notion of momentum cannot be rigorously defined in this elementary example? Or, at the very least, do we have to admit some $\psi$ that obey non-physical boundary conditions?

For those interested, this is a related question, also touching upon domain issues of momentum operators.

  • $\begingroup$ Even for a free particle the domain of the momentum operator is larger than the domain of the Hamiltonian. So $p_1$ could be the right momentum operator but I haven't yet found a good argument for this. $\endgroup$
    – jjcale
    Oct 25, 2014 at 18:46
  • $\begingroup$ @jjcale I think this point is also touched upon in the answer by yuggib $\endgroup$
    – Danu
    Oct 25, 2014 at 18:52

2 Answers 2


Although we can define the momentum as a self-adjoint operator in $L^2[0,1]$ as you proposed, I think it's rather artificial to think about it as having relation to momentum in the case of $L^2(\Bbb{R})$. Realize that the operator $p_1$ with domain $D(p_1)=\{\psi\in\mathcal{H}^1[0,1]\,|\,\psi(1)=\psi(0)\}$, is related to spatial translations via the unitary group $U(t)=\exp(-itp_1)$, whose action is $$(U(t)\psi)(x)=\psi[x-t\pmod{1}],$$ so it's about a particle in a torus, not in an infinite square-well. Different values of $\alpha$, just give different phases to the wavefunction when it reaches the border and goes to the other side.

So, in my opinion, these operators are not actually related to the momentum as usually conceived. The idea of an infinite square-well does not allow spatial translations, so there's no self adjoint operator associated to a unitary translation group in this case. This happens for example in the case of a particle in the postive real line $\Bbb{R}_+$. In this case, the space $L^2[0,\infty)$ allows only translations to the right, not to the left, so you can not have a self-adjoint operator associated to a unitary group of translations. In this case, the operator $p=-i\dfrac{d}{dx}$ has no self-adjoint extensions, for any initial domain, although it is symmetric. For a particle in a box, we can think in the same way. There's no operator associated to spatial translations, because there is no spatial translations allowed.

It's also important to note that the hamiltonian $H$ in this case is given by the Friedrich extension of $$p_0^2=-\frac{d^2}{dx^2}\\ \mathcal{D}(p_0^2)=\{\psi\in\mathcal{H}^2[0,1]\,|\,\psi(0)=\psi'(0)=0=\psi'(1)=\psi(1)\}$$ $H$ cannot be the square of any $p_\alpha$, since the domains do not match.

Edit: As pointed out by @jjcale, one way to take the momentum in this case should be $p=\sqrt{H}$, but clearly, the action of $p$ can't be a derivative, because it has the same eigenfunctions of $H$, which are of the form $\psi_k(x)=\sin \pi kx$. This ilustrates the fact that it's not related to spatial translations as stated above.

Edit 2: There's is a proof that the Friedrich extension is the one with Dirichilet boundary conditions in Simon's Vol. II, section X.3.

The domains defined by the spectral theorem are indeed $\{\psi: p_\alpha\psi\in\mathcal{D}(p_\alpha)\}$. To see this, realize that in this case, since the spectrum is purely point, by the spectral theorem, we have $$p_\alpha=\sum_{n\in \Bbb{Z}}\lambda_{\alpha,n}P_n,$$ where $\lambda_{\alpha,n}$ are the eigenvalues associated to the normalized eigenvectors $\psi_{\alpha,n}$, and $P_n=\psi_n\langle\psi_n,\cdot\rangle$ are the projections in each eigenspace. The domain $\mathcal{D}(p_\alpha)$ is then given by the vectors $\xi$, such that $$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2\|P_n\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle\psi_n,\xi\rangle|^2<+\infty$$ Also, $\xi\in\mathcal{D}(p_\alpha^2)$ iff $$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4\|P_n\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4|\langle\psi_n,\xi\rangle|^2<+\infty$$ But then, $p_\alpha\xi$ is such $$\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2\|P_np_\alpha\xi\|^2=\sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle\psi_n,p_\alpha\xi\rangle|^2= \sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^2|\langle p_\alpha\psi_n,\xi\rangle|^2= \sum_{n\in \Bbb{Z}}|\lambda_{\alpha,n}|^4|\langle\psi_n,\xi\rangle|^2<+\infty$$ So, $\mathcal{D}(p_\alpha^2)= \{\psi: p_\alpha\psi\in\mathcal{D}(p_\alpha)\}$.

  • $\begingroup$ The last part is not precise. Why do you choose the hamiltonian to be the Friedrichs extension of $p_0^2$? This does not seem the best choice, and it is only one possible choice. Anyways, the domain of this extension is contained in the domain of any $H_\alpha=p_\alpha^2$ (defined by functional calculus). $\endgroup$
    – yuggib
    Oct 26, 2014 at 10:46
  • $\begingroup$ No, it's not in any extension. Self-adjoint operators do not admit proper extensions. The domain of the Friedrich extension $H_F$ will be $\mathcal{D}(H_F)=\{\psi \in \mathcal{H}^2[0,1]:\psi(0)=\psi(1)=0\}$. The domain of $H_\alpha$ put conditions in the derivatives of the functions and does not contain the domain cited above. For example, if $\alpha=1$, we have periodic conditions in $\psi$ and $\psi'$. The choice cited above is the natural one for Dirichilet boundary conditions, what is expected for the infinite well. $\endgroup$ Oct 26, 2014 at 14:14
  • $\begingroup$ I think you are making a mistake. The $p_0^2$ you have written is not a self-adjoint operator (since $p_0$ is not self-adjoint); it is a positive symmetric densely defined operator (on $D(p_0^2)$), hence it has a Friedrichs extension, $H_F$. However $H_F$ is not the only extension, there are infinitely many extensions (e.g. $H_\alpha$). The domain of definition of any of the extensions is bigger than the domain of the Friedrichs extension (it's a theorem, see e.g. here). $\endgroup$
    – yuggib
    Oct 26, 2014 at 14:42
  • $\begingroup$ I know it's not the only extension, but it's indeed the only one that has Dirichilet boundary conditions. There can't be extensions with a bigger domain than the Friedrich extension. Self-adjoint operators do not admit extensions. The link you posted says nothing about this. What you might be saying is about the form domain, which is indeed the smallest one for lower bounded symmetric operators. Realize for example, that the eigenfunction, $\psi_1=\sin \pi x$ is not in the domain of any $H_1$, since $p_1\psi_k \notin \mathcal{D}(p_1)$. Also $\psi_2 \notin \mathcal{D}(H_{-1})$. $\endgroup$ Oct 26, 2014 at 14:56
  • $\begingroup$ The operator domain is included in the form domain, and I was not saying that the other extensions include the Friedrichs extension, but simply that they are defined in a domain that is, in some sense, bigger (because it is included on a bigger form domain). Apart that you have to prove that the Friedrichs extension has the domain you claim (and I would be curious to see such proof), that does not tell you about the other domains $D(H_\alpha)$ defined by spectral theorem (they are not simply $\{\psi,p_\alpha\psi\in D(p_\alpha)\}$). $\endgroup$
    – yuggib
    Oct 26, 2014 at 15:17

The fact that $D(p_0)$ is not big enough to define a self-adjoint operator does not mean it is not included into the domain of the self-adjoint momentum. The choice of boundary conditions for the function is a specification of the vector, not of the operator.

Once you have fixed the self-adjoint extension you are considering (choosing the proper domain, usually of essential self-adjointness), you can take any state in the domain to apply the operator $p$. Obviously, to study the Schrödinger equation you need a vector in the domain of the Hamiltonian, not of the momentum (and there are vectors in $D(p_0)$, or also $D(p_\alpha)$, that does not belong to the domain of the Hamiltonian ;-) )

The problem, in general, is that the closure of an operator is taken in the graph topology, and that is not very explicit, so you have to be careful and be sure to end in the right closure, that have not to be too small to be equal to the domain of the adjoint.

  • $\begingroup$ Good point. I have slightly updated my question, taking your response into account. The remaining question is essentially: Do we need to allow non-physical $\psi$ in order to define momentum properly? Maybe this is already addressed by your second paragraph, though. $\endgroup$
    – Danu
    Oct 25, 2014 at 16:04
  • $\begingroup$ As I see it: physically you want states that have zero probability on the boundaries (or at the very least that have symmetric boundary conditions). On the other hand, you need an Hilbert space structure to do QM, so you need to consider the whole $L^2(0,1)$. Here the problem of boundary conditions affects the domains of operators; nevertheless once a (self-adjoint) operator is fixed (i.e. we fix its action and domain), it is ok. Then you may consider as physical only states with the desired boundary conditions, however in principle the action of an operator may change you to a state... $\endgroup$
    – yuggib
    Oct 25, 2014 at 16:22
  • $\begingroup$ with different boundary conditions. $\endgroup$
    – yuggib
    Oct 25, 2014 at 16:37

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