Ehrenfest Theorem for a Linear Potential

Classical Mechanics Level 3

Consider the quantum-mechanical wavefunction of a particle of mass m m moving in a linear potential

V = α x . V = \alpha x.

The (normalized) wavefunction at t = 0 t=0 is

ψ ( x ) = π − 1 / 4 e − x 2 / 2 . \psi(x) = \pi^{-1/4} e^{-x^2/2}.

Find ⟨ x ⟩ \langle x\rangle as a function of time.

− t m α -\frac{t}{m\alpha} α t \alpha t α m t 2 \alpha m t^2 − α t 2 2 m -\frac{\alpha t^2}{2m}

Matt DeCross by Matt DeCross , 27, USA

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1 solution

Matt DeCross
May 10, 2016

The solution is by Ehrenfest's theorem. First compute the expectation value of momentum:

d d t ⟨ p ⟩ = − α ⟹ ⟨ p ⟩ = − α t . \frac{d}{dt} \langle p \rangle = -\alpha \implies \langle p \rangle = -\alpha t.

Then by the second half of the theorem,

d d t ⟨ x ⟩ = 1 m ⟨ p ⟩ = − α t m ⟹ ⟨ x ⟩ = − α t 2 2 m , \frac{d}{dt} \langle x \rangle = \frac{1}{m} \langle p \rangle = \frac{-\alpha t}{m} \implies \langle x \rangle = -\frac{\alpha t^2}{2m},

as claimed. Note that ⟨ x ⟩ = ⟨ p ⟩ = 0 \langle x \rangle = \langle p \rangle = 0 in the given wavefunction, so there are no constants of integration.

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