# Eigenstates/values of jaynes-Cummings Hamiltonian

1. Nov 27, 2012

### Millertron

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

The JCM has the Hamiltonian:

$\hat{H} = \hbar \omega \left(\hat{a}\hat{a}^{*} + 1/2 \right) + \frac{\hbar\omega_{0}\hat{\sigma}_{z}}{2} + \hbar g (\hat{\sigma}_{+}\hat{a} + \hat{\sigma}_{-}\hat{a}^{*}$

Find the eigenstates and energy eigenvalues in this non-resonant case assuming n excitations in the system.

3. The attempt at a solution

I initially thought this would be fairly simple (and I'm sure it is...). Firstly, I applied the Hamiltonian to the states |g,n> and |e,n-1>:

$\hat{H} \left|g,n\right\rangle = \hbar\omega(n+1/2) \left|g,n\right\rangle - \frac{\hbar\omega_{0}}{2} \left|g,n\right\rangle + \hbar g \sqrt{n} \left|e,n-1\right\rangle$

$\hat{H} \left|e,n-1\right\rangle = \hbar\omega(n-1/2) \left|e,n-1\right\rangle + \frac{\hbar\omega_{0}}{2} \left|e,n-1\right\rangle + \hbar g \sqrt{n} \left|g,n\right\rangle$

I also found the matrix elements

$\left\langle g,n \right| \hat{H} \left|g,n\right\rangle = \hbar\omega(n+1/2) - \frac{\hbar\omega_{0}}{2}$

$\left\langle e,n-1 \right| \hat{H} \left|e,n-1\right\rangle = \hbar\omega(n-1/2) + \frac{\hbar\omega_{0}}{2}$

$\left\langle e,n-1 \right| \hat{H} \left|g,n\right\rangle = \hbar g \sqrt{n}$

$\left\langle g,n \right| \hat{H} \left|e,n-1\right\rangle = \hbar g \sqrt{n}$

Here's where I start to get lost. I put the matrix elements into a matrix, and then my natural reaction would be to get the eigenvalues via det(H-Iλ) = 0. I think I need to diagonalise the matrix, only I'm very rusty at this... Find the eigenvalues, then put them into a 2x2 diagonal matrix? Also, I was told I may want to use the rotation matrix:

$\left(\stackrel{cos(\theta)}{-sin(\theta)} \stackrel{sin(\theta)}{cos(\theta)} \right)$

Which, to be honest, has just confused me... (Also how do I put a matrix into latek?!)

I know I'm looking for the dressed states |± n>, and I know these are related to the eigenvalues (obviously...). Argh. I know I'm close but can't quite think of the last few steps!

Any help would be VERY much appreciated! Assignment due in the near future and I'm in dire need of sleep...

Last edited: Nov 27, 2012
2. Nov 28, 2012

### Mute

Hm, it's been a while since I solved any JC Hamiltonian problems, but I'll try to help out.

To diagonalize a 2x2 matrix by the method you suggest, calculating the values for which $\mbox{det}(H-\lambda I) = 0$, you just need to remember that the determinant of a 2x2 determinant is

$$\left|\begin{array}{c c}a & b \\ c & d\end{array}\right| = ad-bc.$$

Applying this to your 2x2 matrix will give you a quadratic polynomial in $\lambda$ to solve for. Then, to find the eigenvectors, you plug each of the eigenvalues back into $H-\lambda I$ and find a vector which gives zero when you apply $H-\lambda I$ to it.

You can then diagonalize the hamiltonian as $H = P^{-1} D P$, where D is a diagonal matrix with the eigenvalues on the diagonal and P is a matrix whose columns are the eigenvectors you just calculated. (The order of the columns corresponds to the order of the eigenvalues in the columns in D).

The suggestion to use a rotation matrix comes in as a shortcut for some of the above: you know that your eigenvalues in the 2d space will be linear combinations of $|g,n\rangle$ and $|e,n-1\rangle$; you can parametrize the eigenvectors as

$$|\psi_1\rangle = \cos \theta |g,n\rangle + \sin\theta |e,n-1\rangle \\ |\psi_2\rangle = \cos \theta |g,n\rangle - \sin\theta |e,n-1\rangle.$$

You chose the angle $\theta$ by calculating the matrix elements of the Hamiltonian in this basis and choosing the angle such that the off-diagonal terms are zero.

This will ultimately get you the same result as the other diagonalization procedure, except that here you can skip the finding the eigenvalues and you can probably find the eigenvectors more quickly. Basically, you'll find that the matrix P in the first method can be written as a rotation matrix.

To write a matrix using LaTeX, I write

Code (Text):

\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)