While exploring the undergraduate Física Cuántica I course offered by the Pontificia Universidad Católica de Chile, I found a homework question asking to prove the momentum part of the theorem of Ehrenfest.
Although I know little to no quantum physics, I was curious whether, as a mathematician, I could understand and complete the proof.
I learned that the momentum part of the theorem of Ehrenfest relates the time derivative of the expectation value of the momentum $p$ to the expectation value of the force $F=-V'(x)$ on a massive particle moving in a scalar potential $V(x)$.
Mathematically,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\langle V'(x)\rangle.$$In simpler terms, as I understand it, this means that the average momentum of a quantum particle changes over time much like that of a classical particle under the average force from the potential (cf. second Newton law in momentum form); it connects quantum behavior with classical physics, showing that quantum averages obey familiar classical laws.
I also learned that this observation is often misused, e.g., by overlooking the assumptions on $V$, which is why I emphasize the importance of truly understanding the underlying principles rather than merely applying formulas.
I did not know how $\langle\cdot\rangle$ is defined in this context, so I looked it up: if $A$ is an operator such that $A\psi\in L^2(\mathbb R)$, where $\psi$ is the wavefunction of the particle, then$$\langle A\rangle=\underbrace{\langle \psi, A\psi \rangle}_{\text{inner product}}=\int_{\mathbb R}\psi^*(x,t)A\psi(x,t)\,\mathrm{d}x.$$In simpler terms, $\langle A \rangle$ is the average value one would get by measuring $A$ on a system described by $\psi$, weighted by the probabilities in $\psi$.
We actually study $L^p$ spaces in mathematics quite thoroughly since they are prototypical Banach spaces of Lebesgue-measurable function classes with $p$-integrable absolute value. In particular, $L^2$ is also a Hilbert space, so this quantum-mechanical setting is really nice.
By the way, a wavefunction $\psi(x,t)$ describes the state of a quantum particle: for each fixed time $t$, it is a function of position in $L^2(\mathbb R)$, meaning it is square-integrable. Moreover, physical wavefunctions are normalized so that total probability equals $1$, and we assume sufficient decay at infinity so we end up with nice equations. 😄
In order to proceed with the proof, one must know both the Schrödinger equation$$i\hbar\partial_t\psi(x,t)=\left(-\frac{\hbar^2}{2m}\partial_x^2+V(x)\right)\psi(x,t)$$and the definition of the momentum operator $p=-i\hbar\partial_x$.
I am not a physicist, so I am not concerned with how these were derived, but they seem like facts every physicist should know by heart.
Anyway, we proceed with the proof; from here I will abuse notation modestly to avoid clutter.$$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle&= \frac{\mathrm{d}}{\mathrm{d}t}\int \psi^*p\psi\\&= \int \partial_t\psi^*\,p\psi + \int \psi^*p\,\partial_t\psi\\&= \frac{i}{\hbar}\int (T+V)\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*p\,(T+V)\psi\\&= \underbrace{\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi}_{\text{potential energy}}+ \underbrace{\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi}_{\text{kinetic energy}}.\end{align}$$Here, $T=-(\hbar^2/2m)\partial_x^2$. This is often called the kinetic-energy operator.
We now tackle the potential-energy term:$$\begin{align}\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi&= \frac{i}{\hbar}\int V\psi^*(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_xV\psi)\\&= \int V\psi^*\partial_x\psi - \int \psi^*\partial_xV\psi\\&= \int V\psi^*\partial_x\psi - \int \psi^*(V'\psi+V\partial_x\psi)\\&= -\int V'|\psi|^2.\end{align}$$We now tackle the kinetic-energy term:$$\begin{align}\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi&= \frac{i}{\hbar}\int \Big(-\frac{\hbar^2}{2m}\partial_x^2\psi^*\Big)(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_x)\Big(-\frac{\hbar^2}{2m}\partial_x^2\psi\Big)\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \psi^*\partial_x^3\psi\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi\\&= 0.\end{align}$$Note that we integrated the rightmost term above by parts twice, still assuming sufficient decay so that the boundary terms vanish.
Finally,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\int V'|\psi|^2+0=-\langle V'(x)\rangle,$$which is what we wanted. This result extends analogously to $\psi(\cdot,t)\in L^2(\mathbb R^n)$.
Although I know little to no quantum physics, I was curious whether, as a mathematician, I could understand and complete the proof.
I learned that the momentum part of the theorem of Ehrenfest relates the time derivative of the expectation value of the momentum $p$ to the expectation value of the force $F=-V'(x)$ on a massive particle moving in a scalar potential $V(x)$.
Mathematically,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\langle V'(x)\rangle.$$In simpler terms, as I understand it, this means that the average momentum of a quantum particle changes over time much like that of a classical particle under the average force from the potential (cf. second Newton law in momentum form); it connects quantum behavior with classical physics, showing that quantum averages obey familiar classical laws.
I also learned that this observation is often misused, e.g., by overlooking the assumptions on $V$, which is why I emphasize the importance of truly understanding the underlying principles rather than merely applying formulas.
I did not know how $\langle\cdot\rangle$ is defined in this context, so I looked it up: if $A$ is an operator such that $A\psi\in L^2(\mathbb R)$, where $\psi$ is the wavefunction of the particle, then$$\langle A\rangle=\underbrace{\langle \psi, A\psi \rangle}_{\text{inner product}}=\int_{\mathbb R}\psi^*(x,t)A\psi(x,t)\,\mathrm{d}x.$$In simpler terms, $\langle A \rangle$ is the average value one would get by measuring $A$ on a system described by $\psi$, weighted by the probabilities in $\psi$.
We actually study $L^p$ spaces in mathematics quite thoroughly since they are prototypical Banach spaces of Lebesgue-measurable function classes with $p$-integrable absolute value. In particular, $L^2$ is also a Hilbert space, so this quantum-mechanical setting is really nice.
By the way, a wavefunction $\psi(x,t)$ describes the state of a quantum particle: for each fixed time $t$, it is a function of position in $L^2(\mathbb R)$, meaning it is square-integrable. Moreover, physical wavefunctions are normalized so that total probability equals $1$, and we assume sufficient decay at infinity so we end up with nice equations. 😄
In order to proceed with the proof, one must know both the Schrödinger equation$$i\hbar\partial_t\psi(x,t)=\left(-\frac{\hbar^2}{2m}\partial_x^2+V(x)\right)\psi(x,t)$$and the definition of the momentum operator $p=-i\hbar\partial_x$.
I am not a physicist, so I am not concerned with how these were derived, but they seem like facts every physicist should know by heart.
Anyway, we proceed with the proof; from here I will abuse notation modestly to avoid clutter.$$\begin{align}\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle&= \frac{\mathrm{d}}{\mathrm{d}t}\int \psi^*p\psi\\&= \int \partial_t\psi^*\,p\psi + \int \psi^*p\,\partial_t\psi\\&= \frac{i}{\hbar}\int (T+V)\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*p\,(T+V)\psi\\&= \underbrace{\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi}_{\text{potential energy}}+ \underbrace{\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi}_{\text{kinetic energy}}.\end{align}$$Here, $T=-(\hbar^2/2m)\partial_x^2$. This is often called the kinetic-energy operator.
We now tackle the potential-energy term:$$\begin{align}\frac{i}{\hbar}\int V\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pV\psi&= \frac{i}{\hbar}\int V\psi^*(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_xV\psi)\\&= \int V\psi^*\partial_x\psi - \int \psi^*\partial_xV\psi\\&= \int V\psi^*\partial_x\psi - \int \psi^*(V'\psi+V\partial_x\psi)\\&= -\int V'|\psi|^2.\end{align}$$We now tackle the kinetic-energy term:$$\begin{align}\frac{i}{\hbar}\int T\psi^*p\psi + \frac{1}{i\hbar}\int \psi^*pT\psi&= \frac{i}{\hbar}\int \Big(-\frac{\hbar^2}{2m}\partial_x^2\psi^*\Big)(-i\hbar\partial_x\psi) + \frac{1}{i\hbar}\int \psi^*(-i\hbar\partial_x)\Big(-\frac{\hbar^2}{2m}\partial_x^2\psi\Big)\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \psi^*\partial_x^3\psi\\&= -\frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi + \frac{\hbar^2}{2m}\int \partial_x^2\psi^*\partial_x\psi\\&= 0.\end{align}$$Note that we integrated the rightmost term above by parts twice, still assuming sufficient decay so that the boundary terms vanish.
Finally,$$\frac{\mathrm{d}}{\mathrm{d}t}\langle p\rangle=-\int V'|\psi|^2+0=-\langle V'(x)\rangle,$$which is what we wanted. This result extends analogously to $\psi(\cdot,t)\in L^2(\mathbb R^n)$.
This exercise humbled me. The mathematics of physics is not easy: it draws on a great deal of experimentally-derived knowledge, despite being modeled with elementary, yet nontrivial, tools like finite-dimensional Hilbert spaces.