# Proof that the momentum operator is hermitian without assuming the wavefunction approaches zero at infinity?

I am currently taking my second semester of quantum mechanics. For a number of proofs in the course, we have used the assumption that the wavefunction goes to zero at infinity. We have simply used the normalization to justify this assumption, which seemed like a stretch to me, but I did not thoroughly question this assumption until recently. From my own examples and this question (Why are wave functions required to vanish at infinity?), I am now confident that this assumption obscures more complexity.

My question here is related to what we use that fact to prove. Most notably, we use it to prove that the momentum operator is hermitian. This seems to be one of the most fundamental things to me, as the momentum operator is used to define the hamiltonian.

Is there another proof that the momentum operator is hermitian that doesn't use the fact that the wavefunction disappears at infinity? Is there a more rigorous formulation of quantum mechanics not introduced in beginning courses that addresses this and related questions?

Consider the Hilbert space $$\mathcal H = L^2(\mathbb R)$$ - roughly, the set of square integrable functions on $$\mathbb R$$. The momentum operator $$P$$ is a linear map $$P : \mathcal D_P \rightarrow \mathcal H$$ $$\psi \mapsto -i \psi'$$ where $$\mathcal D_P \subseteq \mathcal H$$ is some densely-defined domain. First question - what is the appropriate choice for $$\mathcal D_P$$?

At a bare minimum we must have that if $$\psi\in \mathcal D_P$$, then $$-i\psi' \in \mathcal H$$. Therefore, a reasonable guess would be

$$\mathcal D_P := \{ \psi \in \mathcal H \ | \ \psi\in C^1(\mathbb R) , \psi' \in \mathcal H\}$$

where $$C^1(\mathbb R)$$ denotes the once differentiable functions on $$\mathbb R$$. We can now ask whether this $$P$$ is hermitian - that is, for any $$\psi,\phi \in \mathcal D_P$$, $$\langle \psi,P\phi\rangle = \langle P\psi,\phi\rangle$$.

$$\int dx \ \psi^*(x) \cdot (-i \phi'(x)) = -i\int dx \ \psi^*(x) \phi'(x) = -i \left[\left.\psi^*(x)\phi(x)\right|^\infty_{-\infty} - \int dx \ (\psi'(x))^* \phi(x)\right]$$

Now, it would be great if the first term in square brackets would vanish. You're right that it's perfectly possible to have square integrable functions which don't vanish at infinity (e.g. a rectangular "pulse train" with constant height but rapidly decreasing width). However, for all $$\psi\in\mathcal D_P$$ we made the additional demand that $$\psi' \in \mathcal H$$. This implies that $$\psi(\pm \infty)=0$$.

Note that $$\int_a^b dx \ \psi(x) \psi'(x) = \frac{1}{2}\int_a^b dx \ \frac{d}{dx}\left(\psi^2(x)\right) = \frac{1}{2}\left(\psi^2(b)-\psi^2(a)\right)$$

From the Cauchy-Schwarz inequality, $$\left|\int_a^b dx \ \psi(x) \psi'(x)\right|^2 = \frac{1}{4}\left|\psi^2(b)-\psi^2(a)\right|^2 \leq \left(\int_a^b dx \left|\psi(x)\right|^2\right)\cdot \left(\int_a^b dx \ \left|\psi'(x) \right|^2\right)$$

Because both $$\psi$$ and $$\psi'$$ are in $$L^2(\mathbb R)$$, it follows that if we let $$a$$ and $$b$$ both go to infinity, the integrals written above all vanish. This implies that

$$\psi^2(b)-\psi^2(a) \rightarrow 0 \text{ as } a,b\rightarrow \infty$$

Since $$a$$ and $$b$$ are independent, this means that $$\lim_{x\rightarrow \infty} \psi(x) = C$$ for some constant $$C$$; because $$\psi\in L^2(\mathbb R)$$, we must have that $$C=0$$. The same reasoning applies to $$\psi(-\infty)$$.

From here, the remainder of the analysis follows as usual, which demonstrates that $$P$$ is indeed hermitian on this domain.

The lesson to take away from this is that the domain of an operator is an absolutely essential part of its definition. Two operators which have the same "formula" (e.g. $$P_1\psi = -i\psi', P_2\psi = -i\psi'$$) but act on different domains are different operators. The domain of an operator must be chosen carefully if we want the operator to have certain properties, such as self-adjointness.

While we're on the subject, the fact that $$P$$ is hermitian does not imply that $$P$$ is self-adjoint. It is certainly a necessary condition, but it is not sufficient.

Given some operator $$P$$ with a densely-defined domain $$D_P\subseteq \mathcal H$$, the definition of the adjoint operator $$P^\dagger$$ goes as follows: Let $$\mathcal D_{P^\dagger}$$ be the set of all functions $$\eta\in\mathcal H$$ such that, for all $$\psi\in \mathcal D_P$$, there exists some $$\phi\in\mathcal H$$ such that $$\langle \eta, P\psi\rangle = \langle \phi,\psi\rangle$$. We then define $$P^\dagger \eta = \phi$$.

If you are not familiar with this definition, then it may take some time to understand. However, the main takeaway is that self-adjointness is usually extremely tedious and difficult to show. A self-adjoint operator is necessarily hermitian, but the reverse is not true - one needs to additionally show that $$\mathcal D_{P^\dagger} = \mathcal D_P$$ and this is not nearly as straightforward as it may sound. It's not too hard to show that $$\mathcal D_P \subseteq \mathcal D_{P^\dagger}$$, but the latter domain is generically bigger.

• Nice answer, I just remark that the initial domain must be dense, otherwise the adjoint operstor is not well defined. Oct 16 '19 at 5:53
• Thank you @ValterMoretti, I've updated my answer to clarify that point. Oct 16 '19 at 6:01
• Concerning the last part, it is indeed not easy in general to prove self-adjointness. For the momentum operator on the whole real line, however, the proof is not difficult (provided that one has some basic knowledge of Fourier analysis). Perhaps this could be included, for completeness, in your very good answer. Oct 16 '19 at 7:11
• Dear Murray, we know that ''the square-integrable functions no necessarily tends to zero at infinity'' but you used of this proposition in the end of your proof! I think we can not use from the mentioned proposition in the argument. Jan 2 '20 at 16:07
• @alamehhajimohamadi I did not use that proposition in the proof. Can you explain where you think I did? Jan 2 '20 at 16:15

I think that you should get familiar with Quantum Mechanics's groundings from functional analysis. I can suggest you lectures from Frederic Schuller, it will be easy for you to find it in internet. The key point is there a lot of ways how to determine definitions in Quauntum Mechanics and in your course you were introduced to nonrigorous way. Most rigorous and effective way is to learn functional analysis before Quantum Mechanics but few students can get through such hurdle. However, lectures which I suggested amazingly clear and easy to understand. You will find the answer on your question in corresponding lecture in that course. You will see why it is not easy to answer your question immediately: there are a lot of important formulations and theorems which you should know to make this answer clear to you. Good luck!