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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

H(t) = $\displaystyle{\frac{p^2}{2m}}$ + $\displaystyle{\textstyle\frac{1}{2}}$m$\displaystyle\omega_{0}^{2}$x 2 - F(t)x (37)

and checking the classical equations of motion
$\displaystyle{\frac{dp}{dt}}$ = - $\displaystyle{\frac{\partial H}{\partial x}}$ = - m$\displaystyle\omega_{0}^{2}$x + F(t)   
$\displaystyle{\frac{dx}{dt}}$ = $\displaystyle{\frac{\partial H}{\partial p}}$ = $\displaystyle{\frac{p}{m}}$ (38)
give Newton's equations of motion. The wave function satisfies Schrödinger's equation

H(t)|$\displaystyle\psi$(t)$\displaystyle\rangle$ = - $\displaystyle{\frac{\hbar}{i}}$$\displaystyle{\frac{\partial}{\partial t}}$|$\displaystyle\psi$(t)$\displaystyle\rangle$ . (39)

For short times, this can be integrated

|$\displaystyle\psi$(t + $\displaystyle\Delta$t)$\displaystyle\rangle$ = e - $\scriptstyle{\frac{i}{\hbar}}$H(t)$\scriptstyle\Delta$t|$\displaystyle\psi$(t)$\displaystyle\rangle$ . (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

|$\displaystyle\psi$(t + $\displaystyle\Delta$t)$\displaystyle\rangle$ = e - $\scriptstyle{\frac{i}{\hbar}}$H0$\scriptstyle\Delta$t[1 - $\displaystyle{\frac{i}{\hbar}}$F(t)x$\displaystyle\Delta$t]|$\displaystyle\psi$(t)$\displaystyle\rangle$ + O([F(t)$\displaystyle\Delta$t]2) . (42)

We can write

|$\displaystyle\psi$(t)$\displaystyle\rangle$ = $\displaystyle\prod_{i=1}^{N}$$\displaystyle\left (
e^{- \frac{i}{\hbar} H_0 \Delta t}
[1- \frac{i}{\hbar} F(t_i) x \Delta t ] \right)\vert$$\displaystyle\psi$(0)$\displaystyle\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

|$\displaystyle\psi$(t)$\displaystyle\rangle$ = e - $\scriptstyle{\frac{i}{\hbar}}$H0t|$\displaystyle\psi$(0)$\displaystyle\rangle$ - $\displaystyle{\frac{i}{\hbar}}$$\displaystyle\int_{0}^{t}$dt'e - $\scriptstyle{\frac{i}{\hbar}}$H0(t - t')F(t')xe - $\scriptstyle{\frac{i}{\hbar}}$H0t'|$\displaystyle\psi$(0)$\displaystyle\rangle$ (44)

The expectation value of x to linear order in F is then
$\displaystyle\langle$x(t)$\displaystyle\rangle$ = $\displaystyle\langle$$\displaystyle\psi$(t)|x|$\displaystyle\psi$(t)$\displaystyle\rangle$   
  = $\displaystyle\langle$$\displaystyle\psi$(0)|e $\scriptstyle{\frac{i}{\hbar}}$H0txe - $\scriptstyle{\frac{i}{\hbar}}$H0t|$\displaystyle\psi$(0)$\displaystyle\rangle$   
     - $\displaystyle{\frac{i}{\hbar}}$$\displaystyle\langle$$\displaystyle\psi$(0)|e $\scriptstyle{\frac{i}{\hbar}}$H0tx$\displaystyle\int_{0}^{t}$dt'e - $\scriptstyle{\frac{i}{\hbar}}$H0(t - t')F(t')xe - $\scriptstyle{\frac{i}{\hbar}}$H0t'|$\displaystyle\psi$(0)$\displaystyle\rangle$   
     + $\displaystyle{\frac{i}{\hbar}}$$\displaystyle\int_{0}^{t}$dt'$\displaystyle\langle$$\displaystyle\psi$(0)|e $\scriptstyle{\frac{i}{\hbar}}$H0t'F(t')xe $\scriptstyle{\frac{i}{\hbar}}$H0(t - t')xe - $\scriptstyle{\frac{i}{\hbar}}$H0t|$\displaystyle\psi$(0)$\displaystyle\rangle$ . (45)

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

H0|0$\displaystyle\rangle$ = $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\hbar$$\displaystyle\omega_{0}^{}$|0$\displaystyle\rangle$ . (46)

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

x|0$\displaystyle\rangle$ = $\displaystyle\sqrt{\frac{\hbar}{2 m \omega_0}}$|1$\displaystyle\rangle$ (47)

where

H0|1$\displaystyle\rangle$ = $\displaystyle{\textstyle\frac{3}{2}}$$\displaystyle\hbar$$\displaystyle\omega_{0}^{}$|1$\displaystyle\rangle$ . (48)

Substituting these results, the expectation value of x for a weak force is
$\displaystyle\langle$x(t)$\displaystyle\rangle$ = - 2Re $\displaystyle{\frac{\hbar}{2m \omega_0}}$$\displaystyle{\frac{i}{\hbar}}$$\displaystyle\int_{0}^{t}$dt'e - i$\scriptstyle\omega_{0}$(t - t')F(t')   
  = $\displaystyle{\textstyle\frac{1}{m\omega_0}}$$\displaystyle\int_{0}^{t}$dt'sin($\displaystyle\omega_{0}^{}$[t - t'])F(t') (49)
which agrees with the classical result using the Green's function solution.


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