Ehrenfest Theorem

Given observable A^\hat{A}, recall the expectation value of A^\hat{A} is

E[A(t)]=⟨ψ(t)∣A(t)∣ψ(t)⟩\mathbb{E}[A(t)]=\left\langle \psi(t) \right\rvert A(t)\left\lvert \psi(t) \right\rangle

If we differentiate w.r.t time

ddtE[A(t)]=(ddt⟨ψ(t)∣)A(t)∣ψ(t)⟩+⟨ψ(t)∣A(t)(ddt∣ψ(t)⟩)\frac{d}{dt}\mathbb{E}[A(t)]=\left(\frac{d}{dt}\left\langle \psi(t) \right\rvert\right)A(t)\left\lvert \psi(t) \right\rangle+\left\langle \psi(t) \right\rvert A(t)\left(\frac{d}{dt}\left\lvert \psi(t) \right\rangle\right) =−1iℏ⟨ψ∣HA∣ψ⟩+⟨ψ∣A1iℏH∣ψ⟩= -\frac{1}{i\hbar}\left\langle \psi \right\rvert HA\left\lvert \psi \right\rangle+\left\langle \psi \right\rvert A\frac{1}{i\hbar}H\left\lvert \psi \right\rangle

because commutator,

=iℏ⟨ψ∣[H,A]∣ψ⟩=\frac{i}{\hbar}\left\langle \psi \right\rvert[H,A]\left\lvert \psi \right\rangle

Ehrenfest theorem

ddtE[A(t)]=iℏE[[H,A]]\boxed{\frac{d}{dt}\mathbb{E}[A(t)]=\frac{i}{\hbar}\mathbb{E}\left[[H,A]\right]}

Remarks

  1. If [H,A]=0[H,A]=0, then ddtE[A(t)]=0\frac{d}{dt}\mathbb{E}[A(t)]=0 The expected value of A is conserved if A and H are compatible. For example, energy H is always conserved in an Isolated System!

  2. Sometimes convientn to change basis to Heisenberg picture to study observable dynamics