Time Evolution Operator in Interaction Picture (Harmonic Oscillator with Time Dependent Perturbation)

Question

1. The problem statement, all variables and given/known data

Consider a time-dependent harmonic oscillator with Hamiltonian

$$\hat{H}(t)=\hat{H}_0+\hat{V}(t)$$

$$\hat{H}_0=\hbar \omega \left( \hat{a}^{\dagger}\hat{a}+\frac{1}{2} \right)$$

$$\hat{V}(t)=\lambda \left( e^{i\Omega t}\hat{a}^{\dagger}+e^{-i\Omega t}\hat{a} \right)$$

*(i) Compute $\hat{U}_S(t,0)$ using the interaction representation formula (Equation 1 in next section) to second order perturbation theory.

(ii) Compute $\hat{U}_S(t,0)$ using (Equation 2 in next section) to second order perturbation theory.

2. Relevant equations

EQUATION 1:

$$U_I(t,0)=1-\frac{i}{\hbar}\int_0^t dt' V_I(t')+\left( \frac{-i}{\hbar} \right)^2 \int_0^t dt' \int_0^{t'} V_I(t')V_I(t'') + \dots$$

EQUATION 2:

$$U(t,0)=1+\sum_{n=1}^{∞}\left( \frac{-i}{\hbar} \right)^n\int_0^t dt_1 \int_0^{t_1} dt_2 \dots \int_0^{t_{n-1}}dt_n H(t_1)H(t_2)\dots H(t_n)$$

3. The attempt at a solution

So I know that for the interaction picture the transformation of the operator $\hat{V}_I$ is..

$$\hat{V}_I=e^{\frac{i}{\hbar}\hat{H}_0 t} \hat{V} e^{\frac{-i}{\hbar}\hat{H}_0 t}$$

I also know that both operators and kets evolve in time. So I use the interaction picture equation of motion on the ladder operators so I can obtain an expression for them as a function of time.

$$\frac{d\hat{a}}{dt}=\frac{1}{i\hbar}\left[ \hat{a},\hbar \omega \left(\hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right) \right]$$

$$\frac{d\hat{a}^{\dagger}}{dt}=\frac{1}{i\hbar}\left[ \hat{a}^{\dagger},\hbar \omega \left( \hat{a}^{\dagger}\hat{a} + \frac{1}{2} \right) \right]$$

I then got..

$$\hat{a}(t)=\hat{a}(0)e^{-i\omega t}$$

$$\hat{a}^{\dagger}(t)=\hat{a}^{\dagger}(0)e^{i\omega t}$$

I plugged these into the expression for V to get,

$$\hat{V}=\lambda \left[ \hat{a}^{\dagger}(0)e^{i(\Omega + \omega)t} + \hat{a}(0)e^{-i(\Omega + \omega)t} \right]$$

So now what needs to be done, is to transform this into the interaction picture and then plug it into Equation 1 from above and integrate. But this seems very messy and I am having doubts if this is the correct way to I also know that both operators and kets evolve in time.

So I use the interaction picture equation of motion on the ladder operators so I can obtain an expression for them as a function of time. go about this problem. If anyone can shed some light onto this I would really appreciate it!

Answer 1

Let's start from your EQ1:

$\begin{eqnarray}U_I(t,0) = \mathbf{Id} + \frac{1}{i\hbar}\int_0^t dt_1V_H(t_1) +\cdots+\left(\frac{1}{i\hbar}\right)^k\int_0^tdt_1...\int_0^{t_{k-1} }V_H(t_1)\cdots V_H(t_k) \end{eqnarray}$

where $V_H$ means $V$ evolved by heisenberg. The operator is totally symmetric so we can adjust the integral extrema to write the well know path-order exponentail:

$\begin{eqnarray} U_I(t,0)=\mathbf{Id}+\sum_{k=1}^{+\infty}\frac{1}{k!}\int_0^tdt_1..\int_0^tdt_{k-1}V_H(t_1)...V(t_k) = \text{Texp}\left[\frac{1}{i\hbar}\int_0^tdt'V_H(t')\right] \end{eqnarray}$

So now you can use the form of potential that you fine in the path-order exponential, and with GellMann and Low theorem find the ground state of your hamiltonian.