7 Quantum Harmonic Oscillator

Having shown an interconnection between the mathematics of classical mechanics and electromagnetism, let's look at the driven quantum harmonic oscillator too. Let's start with a one-dimensional quantum harmonic oscillator in its ground state at time t = 0 , and apply a force F(t) . The Hamiltonian is therefore

$${\frac{p^2}{2m}} {\textstyle\frac{1}{2}} \omega_{0}^{2}$$ (37)

and checking the classical equations of motion

$${\frac{dp}{dt}} {\frac{\partial H}{\partial x}} \omega_{0}^{2}$$ =

$${\frac{dx}{dt}} {\frac{\partial H}{\partial p}} {\frac{p}{m}}$$ = (38)

give Newton's equations of motion. The wave function satisfies Schrödinger's equation

$$\psi \rangle {\frac{\hbar}{i}} {\frac{\partial}{\partial t}} \psi \rangle$$ (39)

For short times, this can be integrated

$$\psi \Delta \rangle {\frac{i}{\hbar}} \Delta \psi \rangle$$ (40)

Let's assume that the force is weak, and let's calculate the expectation value of the operator x as a function of time. We write the Hamiltonian as

H(t) = H₀ - F(t)x (41)

and the short time solution is

$$\psi \Delta \rangle {\frac{i}{\hbar}} \Delta {\frac{i}{\hbar}} \Delta \psi \rangle \Delta$$ (42)

We can write

$$\psi \rangle \prod_{i=1}^{N} \left ( e^{- \frac{i}{\hbar} H_0 \Delta t} [1- \frac{i}{\hbar} F(t_i) x \Delta t ] \right)\vert \psi \rangle$$ (43)

where $\Delta$t = t/N , and the order of the operators must be kept since x and H₀ do not commute. Taking the limit that N goes to infinity and keeping only linear order terms, this becomes

$$\psi \rangle {\frac{i}{\hbar}} \psi \rangle {\frac{i}{\hbar}} \int_{0}^{t} {\frac{i}{\hbar}} {\frac{i}{\hbar}} \psi \rangle$$ (44)

The expectation value of x to linear order in F is then

$$\langle \rangle \langle \psi \psi \rangle$$ =

$$\langle \psi {\frac{i}{\hbar}} {\frac{i}{\hbar}} \psi \rangle$$ =

$${\frac{i}{\hbar}} \langle \psi {\frac{i}{\hbar}} \int_{0}^{t} {\frac{i}{\hbar}} {\frac{i}{\hbar}} \psi \rangle$$

$${\frac{i}{\hbar}} \int_{0}^{t} \langle \psi {\frac{i}{\hbar}} {\frac{i}{\hbar}} {\frac{i}{\hbar}} \psi \rangle$$ (45)

We now use the boundary condition that |$\psi$(0)$\rangle$ = |0$\rangle$ is the ground state with

$$\rangle {\textstyle\frac{1}{2}} \hbar \omega_{0}^{} \rangle$$ (46)

and x times the ground state is proportional to the first excited state,

$$\rangle \sqrt{\frac{\hbar}{2 m \omega_0}} \rangle$$ (47)

where

$$\rangle {\textstyle\frac{3}{2}} \hbar \omega_{0}^{} \rangle$$ (48)

Substituting these results, the expectation value of x for a weak force is

$$\langle \rangle {\frac{\hbar}{2m \omega_0}} {\frac{i}{\hbar}} \int_{0}^{t} \omega_{0}$$ =

$${\textstyle\frac{1}{m\omega_0}} \int_{0}^{t} \omega_{0}^{}$$ = (49)

which agrees with the classical result using the Green's function solution.