Harmonic Oscillator: General Solution, ⟨x⟩(t), and ⟨V⟩ = ½⟨E⟩
Problem Statement
(a) Write down the most general solution $\psi(x,t)$ of the time-dependent Schrödinger equation.
(b) Show that $\langle x \rangle(t) = A\cos(\omega t) + B\sin(\omega t)$ for real constants $A, B$ determined by the expansion coefficients.
(c) Prove that the time-averaged expectation value of the potential energy equals half the total energy: $\overline{\langle V \rangle} = \frac{1}{2}\langle E \rangle$.
Step-by-Step Solution
Part (a): General Solution
The energy eigenstates of the harmonic oscillator satisfy $H\phi_n = E_n\phi_n$ with $E_n = (n + \tfrac{1}{2})\hbar\omega$. The most general solution is a superposition:
where the coefficients $a_n = \langle \phi_n | \psi(x,0)\rangle$ are determined by the initial condition and satisfy $\sum_n |a_n|^2 = 1$.
Part (b): Expectation Value of Position
Using the ladder-operator representation $x = \sqrt{\hbar/(2m\omega)}\,(a + a^\dagger)$, the matrix elements of $x$ are nonzero only between adjacent states:
Therefore only $n \pm 1$ cross-terms contribute to $\langle x \rangle(t)$. The time-dependent phases give $e^{\pm i\omega t}$, so:
Writing $a_n^* a_{n+1} = |a_n^* a_{n+1}|e^{i\phi_n}$ and combining:
where $A$ and $B$ are real constants depending on the coefficients $a_n$. The expectation value oscillates at the classical frequency $\omega$—an instance of Ehrenfest's theorem.
Part (c): Virial Theorem — Time-Averaged Potential Energy
For a stationary state $\phi_n$, the virial theorem for the harmonic oscillator gives $\langle T \rangle_n = \langle V \rangle_n = \frac{1}{2}E_n$. For a general state:
Time-averaging kills all cross-terms ($m \neq n$) since $\overline{e^{i(E_m - E_n)t/\hbar}} = 0$ for $E_m \neq E_n$. Only diagonal terms survive:
This is the quantum virial theorem for the harmonic oscillator: on average, kinetic and potential energies each carry half the total energy.