QHO Quantum Dynamics

The Quantum Dynamics of a Quantum Harmonic Oscillator. Let all the variable from there

We know that from Heisenberg Picture (Heisenberg, 1925)

O^H(t)=U†(t) O^ U(t)\hat{O}_H(t) = U^\dagger(t)\,\hat{O}\,U(t)

Where O^\hat{O} is any observable, O^H\hat{O}_H is the Heisenberg picture operator.

Position Evolution

The position evolves like this

ddtx^H=iℏ[H^H,x^H]=iℏ[p^H22m,x^H]\frac{d}{dt}\hat{x}_H = \tfrac{i}{\hbar}[\hat{H}_H, \hat{x}_H] = \tfrac{i}{\hbar}\left[\tfrac{\hat{p}_H^2}{2m}, \hat{x}_H\right]

we know p^\hat{p} commutes with x^\hat{x}. use (via Commutator)

[p^2,x^]=2p^[p^,x^]=−2iℏp^[\hat{p}^2,\hat{x}]=2\hat{p}[\hat{p},\hat{x}]=-2i\hbar\hat{p}

so

ddtx^H=i2mℏ⋅(−2iℏp^H)=p^Hm\frac{d}{dt}\hat{x}_H = \tfrac{i}{2m\hbar}\cdot(-2i\hbar\hat{p}_H) = \tfrac{\hat{p}_H}{m}

this is already

Momentum Evolution

The momentum evolves like this

ddtp^H=iℏ[12mω2x^H2,p^H]\frac{d}{dt}\hat{p}_H = \tfrac{i}{\hbar}\left[\tfrac{1}{2}m\omega^2\hat{x}_H^2, \hat{p}_H\right]

We know [x^2,p^]=2iℏx^[\hat{x}^2,\hat{p}]=2i\hbar\hat{x} so

ddtp^H=−mω2x^H(t)\frac{d}{dt}\hat{p}_H = -m\omega^2\hat{x}_H(t)

Decouple

We know that

dx^Hdt=p^Hm,dp^Hdt=−mω2x^H\frac{d\hat{x}_H}{dt} = \tfrac{\hat{p}_H}{m}, \qquad \frac{d\hat{p}_H}{dt} = -m\omega^2\hat{x}_H

so

d2x^Hdt2=−ω2x^H(t)\frac{d^2\hat{x}_H}{dt^2} = -\omega^2\hat{x}_H(t)

Solving this ODE yields a form

x^H(t)=A^cos⁡(ωt)+B^sin⁡(ωt)\hat{x}_H(t) = \hat{A}\cos(\omega t) + \hat{B}\sin(\omega t) p^H(t)=mdx^Hdt=mω[−A^sin⁡(ωt)+B^cos⁡(ωt)]\hat{p}_H(t) = m\frac{d\hat{x}_H}{dt} = m\omega\left[-\hat{A}\sin(\omega t) + \hat{B}\cos(\omega t)\right]

so

x^H(0)=x^,p^H(0)=p^  ⟹  A^=x^,B^=p^mω\hat{x}_H(0) = \hat{x}, \quad \hat{p}_H(0) = \hat{p} \implies \hat{A} = \hat{x}, \quad \hat{B} = \tfrac{\hat{p}}{m\omega}

so

x^H(t)=x^cos⁡(ωt)+p^mωsin⁡(ωt)\boxed{\hat{x}_H(t) = \hat{x}\cos(\omega t) + \tfrac{\hat{p}}{m\omega}\sin(\omega t)} p^H(t)=p^cos⁡(ωt)−mω x^sin⁡(ωt)\boxed{\hat{p}_H(t) = \hat{p}\cos(\omega t) - m\omega\,\hat{x}\sin(\omega t)}