# In 1D wave mechanics, is there a counterexample to the relation $m \frac{d \langle x\rangle}{dt} = \langle p \rangle$?

The standard physicists' proof of the identity $m \frac{d\langle x\rangle}{dt} = \langle p \rangle$ involves integration by parts. For example, in Griffiths's "Introduction to Quantum Mechanics", the derivation goes as follows: $$\begin{split} m\frac{d\langle x \rangle}{dt} &= m\int x \frac{\partial|\psi|^2}{\partial t} dx\\ &= \frac{i\hbar}{2}\int x\frac{\partial}{\partial x}\left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) dx\\ &= -\frac{i\hbar}{2}\int \left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) dx\\ &= -i\hbar \int \psi^\ast\frac{\partial\psi}{\partial x} dx\\ & = \langle p \rangle, \end{split} \tag{1} \label{a}$$ Here, (among other things) one should integrate by parts to obtain the third line, where the associated boundary term is assumed to vanish, i.e., $$x\left(\psi^\ast\frac{\partial\psi}{\partial x}-\frac{\partial\psi^\ast}{\partial x}\psi\right) \Bigg|_{x=-\infty}^{\infty} = 0. \tag{2} \label{b}$$ But is it really OK to make such an assumption? In fact, for the normalizable wave function $$\psi_1(x) = \frac{e^{ix^4}}{x^2 + 1}, \tag{3}$$ the boundary term [Eq. $(\ref{b})$] does not vanish, making the whole derivation in Eq. $(\ref{a})$ invalid. Still, it is easy to see that $\langle p \rangle$ itself is ill-defined for the above wave function (i.e., the integral $\langle\psi_1|p|\psi_1\rangle$ is not convergent), so this counterexample is not very interesting.

Hence, my question is the following:

Is it possible to construct a counterexample to the relation $m \frac{d\langle x\rangle}{dt} = \langle p \rangle$, where both $\langle x\rangle$ and $\langle p\rangle$ are well-defined?

• No, it is not possible. If no answer comes along, I will write one in maximum two days. Commented Mar 12, 2018 at 21:41
• @DanielC Thank you. Before a full answer (provided by you or someone else), could you direct me to relevant materials or keywords? Commented Mar 12, 2018 at 21:48
• There are other ways in which things can be ill defined. 1) When wave functions are non-normalizable (e.g. free particle state in $\mathbb{R}$: take an eigenfunction of the Hamiltonian and hence these expectation values are time independent but then $\langle p\rangle$ is ill defined and $\langle x\rangle$ exists only as principal value). 2) When $x$ and/or $p$ are not well defined as self-adjoint operators (e.g. free particle in a circle: stationary states are normalizable and the expectation values of $x$ and $p$ can be "computed" and are time-independent, but $x$ is not self-adjoint). Commented Mar 12, 2018 at 23:08
• An extreme freak, even worse than what you are seeking, is Berry, M. V.; Balazs, N. L. (1979), "Nonspreading wave packets", Am J Phys (47): 264–267, doi:10.1119/1.11855 . Commented Mar 12, 2018 at 23:59
• @secavara You raise a good point, but I would like to restrict to normalizable wave functions and the case that both $x$ and $p$ are well-defined as self-adjoint operators. Commented Mar 13, 2018 at 4:33

The derivation in equation (1) assumes the following: in order that the boundary at infinity term to be 0 in the limit, it is enough/sufficient to consider that the wavefunctions are Schwartz test functions (i.e. they go to 0 to infinity quicker than any polynomial function). It is when this set of functions is used that the operators $$x$$ and $$p_x$$ are essentially self-adjoint, their matrix elements $$\langle \psi, x\psi\rangle$$ and $$\langle \psi, p_x\psi\rangle$$ are properly defined, well-behaved functions of the parameter $$t$$ by means of $$\psi(t)$$ and the space $$\mathcal{S}(\mathbb R)$$ is invariant under the uniparameter group $$e^{itH}$$ by the theorem of Hunziker. These are the sufficient conditions to make (1) into a valid derivation.

So the final question of the original post has this answer: "NO, because making the matrix elements as well-behaved functions of $$t$$ makes all manipulations in (1) mathematically valid".

Ehrenfest's theorem has been properly formulated and proven only as late as 2009 (i.e. more than 80 years after the original work by P. Ehrenfest). This can be found here: "On the Ehrenfest theorem of quantum mechanics", Gero Friesecke and Mario Koppen, Journal of Mathematical Physics 50, 082102 (2009); doi: 10.1063/1.3191679

• Thank you for your answer! Isn't there still a possibility of a counterexample that is not a Schwartz test function? Commented Mar 14, 2018 at 4:10
• Your equation (3) proves this, as this is not a Schwartz test function. Commented Mar 14, 2018 at 17:30
• The wave function in my equation (3) does not have a well-defined value of $\langle p \rangle$. Commented Mar 14, 2018 at 17:34

I would say the usual proof of this statement comes from Ehrenfest's theorem:

$$\frac{d \langle Q \rangle}{dt} = -\frac{i}{\hbar} [Q,H]$$

Then with the usual single particle Hamiltonian one has $$H=\frac{p^2}{2m} +V(x)$$ and so $$\frac{d \langle X \rangle}{dt}=\frac{1}{2m}[p^2,x]$$ This evaluate via standard commutation rules to your identity $$m \langle \dot{x} \rangle= \langle p \rangle$$.

At no point here did we invoke integration by parts. The proof of Ehrenfest's theorem, which does involves inner products (and so carries the risk of IBP), simply requires that all the inner products exist and this is equivalent to the statement that our wavefunctions are normalisable.

• I don't have enough reputation to have simply commented this. I think this purely algebraic approach doesn't have the flaw with boundary terms that Griffith's calculus based one has but I'm happy to be corrected on this. Commented Mar 13, 2018 at 0:11
• Thanks for your answer! But using the algebraic approach means that one implicitly restricts to the class of functions that are sufficiently well-behaved, on which various operator equations you wrote down make sense without additional boundary terms. Also, the requirement on the wave function is not just that ψ needs to be square-integrable, but objects like xψ should also be square-integrable. This make the problem much more complicated Commented Mar 13, 2018 at 4:44