Monday, 4 January 2016

quantum mechanics - Transformation of Jaynes-Cummings Hamiltonian


This is the Jaynes-Cummings Hamiltonian in the interaction picture: $H_i^n = \hbar g \sqrt{n+1}\begin{pmatrix} 0 & \exp(-i\delta t)\\ \exp(i\delta t) & 0 \end{pmatrix}$


I want to transform it into another basis, so that it looks like this:


$H_{i2}^n = \hbar \begin{pmatrix} -\delta/2 & g \sqrt{n+1}\\ g \sqrt{n+1} & \delta/2 \end{pmatrix}$



I present the solution that I found with help from DanielSank and march below.



Answer



Following DanielSank's suggestion, I tried writing both Hamiltonians in terms of Pauli operators: $$H_i^n=\hbar g \sqrt{n+1}\frac{1}{2}\left[(\sigma_x+i\sigma_y)e^{-i\delta t} + (\sigma_x-i\sigma_y)e^{i\delta t} \right],$$ and $$H_{i2}^n = -\hbar \sigma_z \delta/2 + \hbar g \sqrt{n+1}\sigma_x.$$


So I'm basically looking for a transformation that converts the content of the brackets into $\sigma_x$ and also makes this other term appear.


If I want to get rid of the time dependence, I can transform everything back into the Schroedinger picture by using $$U=\begin{pmatrix} \exp(i\delta t/2) &0 \\ 0 & \exp(-i\delta t /2) \end{pmatrix}.$$


This gives: $$UH_i^n U^+ = \hbar \sqrt{n+1}\sigma_x=H^*$$


I try to apply this $U$ to the Schrodinger equation from the left side:


$$UH_i^n|\Psi\rangle = i\hbar U \partial_t |\Psi\rangle $$


Then I insert $U^{-1}U$ before the state $|\Psi\rangle$ on both sides:


$$U H_i^n U^{-1}U |\Psi\rangle = i\hbar U\partial_t U^{-1}U |\Psi\rangle $$



Then I rename $$U|\Psi\rangle \equiv |\Psi'\rangle$$


Now product rule, as march suggested:


$$UHU^{-1} |\Psi'\rangle = i\hbar UU^{-1}\partial_t |\Psi'\rangle + i\hbar U \partial_t( U^{-1}) |\Psi'\rangle$$ which leads to $$\left(H^* - \hbar \begin{pmatrix} \delta/2 &0 \\ 0 &-\delta/2 \end{pmatrix}\right) |\Psi'\rangle =i\hbar\partial_t | \Psi' \rangle$$


The part of the left side is now exactly what I was looking for!




Update


A follow-up question generalizes this special case.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...