Question

Consider a linear harmonic oscillator and let, ψ₀ and ψ₁ be its real, normalized ground and first excited state energy eigenfunctions respectively. Let Aψ₀ + Bψ₁ with A and B real numbers be the wave function of the oscillator at some instant of time. Show that the average value of x is in general different from zero. What values of A and B maximize (x) and what values minimize it?

Answer 1

The orthonormal condition

\(\int (A\psi_0 + B\psi_1)^2 dx = 1\)

gives A² + B² = 1.

Generally A and B are not zero, so the average value of 2,

\(\langle x\rangle = \int x(A\psi_0 + B\psi_1)^2 dx = 2AB\langle\psi_0|x|\psi_1\rangle\)

is not equal to zero. Rewriting the above as

\(\langle x\rangle = [1 - (A^2 + B^2 + 2AB)]\langle\psi_0|x|\psi_1\rangle\)

\(= [1 - (A - B)^2]\langle\psi_0|x|\psi_1\rangle\)

and considering f = AB = A (1 - A²)^1/2, which has extremums at A = \(\pm \frac 1{\sqrt 2}\), we see that if A = B = 1/√2, (x) is maximized; if A = -B = 1/√2, (x) is minimized.