Question 2 of 7: Hermitian Operators, Correspondence Principle and Ehrenfest's Theorem

Question 2: Hermitian Operators, Correspondence Principle and Ehrenfest's Theorem (20 marks: (a) 5, (b) 5, (c) 5, (d) 5)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The generalized Ehrenfest relation for the time-derivative of $\langle\hat A\rangle$, and the one-dimensional Hamiltonian $H=p_x^2/2m+V(x)$.

Find. (a) Why observables need Hermitian operators; (b) the correspondence principle in words; (c) $d\langle x\rangle/dt$ and $d\langle p_x\rangle/dt$; (d) whether (c) matches the correspondence principle.

Approach. (a)–(b) are conceptual; (c) evaluates the commutators $[H,x]$ and $[H,p_x]$ from the canonical relation $[x,p_x]=i\hbar$ and substitutes into the Ehrenfest relation.

1. Part (a) — why Hermitian. A measurement of an observable can only return a real number, so the operator representing it must have real eigenvalues (these are the only possible measured outcomes). Hermitian operators are guaranteed to have real eigenvalues and a complete, orthogonal set of eigenfunctions, so any physical state can be expanded in that basis and the Born rule ($|c_n|^2$) gives a consistent set of measurement probabilities. A non-Hermitian operator could return a complex "measured value," which has no physical meaning, and would not in general possess a complete eigenbasis to expand an arbitrary state in.
2. Part (b) — correspondence principle. The correspondence principle states that the predictions of quantum mechanics must reduce to those of classical mechanics in the appropriate limit — for large quantum numbers, macroscopic masses/energies, or equivalently $\hbar\to0$. Classical mechanics is recovered as the limiting case of the more fundamental quantum theory, not as an independent set of laws.
3. Part (c) — commutator $[H,x]$. Only the kinetic term fails to commute with $x$: $[H,x]=\frac{1}{2m}[p_x^2,x]$. Using $[p_x^2,x]=p_x[p_x,x]+[p_x,x]p_x$ and the canonical relation $[x,p_x]=i\hbar\Rightarrow[p_x,x]=-i\hbar$: $$ [p_x^2,x] = p_x(-i\hbar)+(-i\hbar)p_x = -2i\hbar p_x \;\Longrightarrow\; [H,x]=-\frac{i\hbar}{m}p_x. $$
4. Substitute into Ehrenfest's relation. Neither $H$ nor $x$ depends explicitly on $t$, and using the standard-sign form of the relation, $d\langle A\rangle/dt=\frac{i}{\hbar}\langle[H,A]\rangle$ (the printed sign in the question is the mirror convention $\frac{1}{i\hbar}\langle[A,H]\rangle$ of the same identity — both give the same physical result; see the check note below): $$ \frac{d\langle x\rangle}{dt}=\frac{i}{\hbar}\left\langle-\frac{i\hbar}{m}p_x\right\rangle = \frac{i\cdot(-i)}{m}\langle p_x\rangle \;\Longrightarrow\; \boxed{\frac{d\langle x\rangle}{dt}=\frac{\langle p_x\rangle}{m}} $$ — exactly the classical relation between velocity and momentum, now for expectation values.
5. Commutator $[H,p_x]$. Now $p_x^2$ commutes with itself, so only the potential term contributes: $[H,p_x]=[V(x),p_x]$. Using the identity supplied on the exam's formula page, $[f(x),p_x]=i\hbar\, df/dx$, with $f=V$: $$ [V(x),p_x] = i\hbar\frac{dV}{dx}. $$
6. Substitute again. $$ \frac{d\langle p_x\rangle}{dt}=\frac{i}{\hbar}\left\langle i\hbar\frac{dV}{dx}\right\rangle = i^2\left\langle\frac{dV}{dx}\right\rangle \;\Longrightarrow\; \boxed{\frac{d\langle p_x\rangle}{dt}=-\left\langle\frac{dV}{dx}\right\rangle} $$ — the quantum-mechanical statement of Newton's second law, force $=-dV/dx$, for expectation values.
7. Part (d) — consistency with correspondence. Yes: the pair $d\langle x\rangle/dt=\langle p_x\rangle/m$ and $d\langle p_x\rangle/dt=-\langle dV/dx\rangle$ are exactly Hamilton's classical equations of motion, but written for the expectation values of position and momentum rather than for definite classical trajectories. This is Ehrenfest's theorem: on average, quantum expectation values obey classical mechanics. The correspondence is exact for these two variables in any potential; it becomes an approximation only when $\langle dV/dx\rangle \ne dV/dx|_{\langle x\rangle}$, i.e. whenever $V(x)$ is not at most quadratic (a wavepacket must stay narrow compared to the curvature scale of $V$ for the classical trajectory of $\langle x\rangle$ itself to be recovered).