I am beginning to study quantum mechanics and I got stuck right at the beginning. I am trying to prove that the time derivative of the expected value of momentum of a particle is the (negative) expected value of the gradient of the potential, i.e., in one dimension we have \begin{equation} \frac{d\langle p\rangle}{dt}=-\left<\frac{dV}{dx}\right> \end{equation} which means \begin{equation} \frac{d\langle p\rangle}{dt}=-\int_{-\infty}^{\infty}\psi^*\frac{\partial V}{\partial x}\psi\ dx \end{equation} (here $\psi^*$ is the complex conjugate of $\psi$)

The wavefunction $\psi$ should be square summable, i.e. living in $L^2(\mathbb{R})$. I understand that we can quotient this space with the equivalence relation that associates two functions that are equal almost everywhere, so we can correctly state that $\psi(x,t)\longrightarrow 0$ when $x\longrightarrow\pm\infty$ and in this way we can get rid of the boundary terms when integrating by parts. But in the derivation of the above relation I got the following: \begin{equation} \frac{d\langle p\rangle}{dt}= \int_{-\infty}^\infty V\left(\psi^*\frac{\partial\psi}{\partial x}+\psi\frac{\partial\psi}{\partial x}^*\right)\ dx -\frac{\hbar^2}{2m}\left[\frac{\partial\psi}{\partial x}^*\frac{\partial\psi}{\partial x}\right]_{-\infty}^{\infty} \end{equation} Now, the boundary term should vanish for the relation to be verified, but why does it vanish? There are functions in $L^2$ whose derivatives are not in $L^2$, and that boundary term does not vanish in some of those cases. Should we be considering wave functions in some kind of Sobolev space instead? Is there a physical reason behind the vanishing of the boundary terms?


1 Answer 1


You're right that one technically has to be careful and Sobolev spaces etc. or some subset of $L^2$ with appropriate boundary conditions should be considered when one wants to be precise.

In introductory texts on quantum mechanics, however, these details concerning boundary terms are usually (succesfully) shoved under the rug by saying things like "the field and its derivatives fall off to zero near infinity". This approach allows one to focus more on the physics and less on the mathematical side of things. Of course, it's all just a matter of taste, as long as you get the right results ;-)

Vanishing of the wave function and its derivatives near infinity is typically motivated by stating that the system under consideration is somehow localized in space, and invoking the seemingly reasonable principle that this should not influence the physics far away: In mathematical terms, one assumes that all these fields have compact support.

From my personal experience with rigorous quantum mechanics, one typically ends up obtaining exactly the same results that one derives by more elementary---and to some, less satisfactory---means in "physical" QM.

  • $\begingroup$ Isn't this assumption of "locality" actually too anthropocentric? As far as I understand (which alas is not too much) some interesting facts about QM gravitate around the fall of local realism. $\endgroup$ Dec 4, 2015 at 9:55
  • $\begingroup$ But this may be a metaphysical question for philosophy.SE :) $\endgroup$ Dec 4, 2015 at 9:56
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    $\begingroup$ I assume you're alluding to things like Bell's theorem---such questions are not covered by this answer. However, I do think one fundamentally needs some kind of notion of locality to make physics meaningful. Without it, one can always "blame the aliens" for whatever is happening in the world around us. $\endgroup$
    – Danu
    Dec 4, 2015 at 9:58
  • $\begingroup$ Of course, this does include things like entanglement; Particles need to have interacted (i.e. they must have been in causal contact) for something like this to happen. In typical Bell-type setups, one first lets particles interact, then separates them. $\endgroup$
    – Danu
    Dec 4, 2015 at 9:59
  • $\begingroup$ The restriction to certain subspaces like Sobolev spaces is not a restriction of the space of physical states, imo, but merely a restriction of the domains of the operators (which of unbounded, cannot be defined everywhere to begin) $\endgroup$
    – ACuriousMind
    Dec 4, 2015 at 13:13

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