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Diagonalizing a linearized optomechanical Hamiltonian

+ 1 like - 0 dislike
387 views

Hi all, 

in the paper Nonlinear Interaction Effects in a Strongly Driven Optomechanical Cavity, the authors diagonalize the Hamiltonian (equation (2) in the paper)

\[H_1=-\Delta d^\dagger d+\omega_Mb^\dagger b+G(d+d^\dagger)(b+b^\dagger)\]

in the appendix (equations S1-S3). First they define the vector \(\vec X=[b\,d\,b^\dagger\,d^\dagger]^T\), and then say it can be done "by standard means". I assumed they would mean writing the Hamiltonian as

\[H=\vec X^T M \vec X,\quad M=\left(\begin{matrix}\omega_M&0&0&0\\G&-\Delta&G&0\\0&0&0&0\\G&0&G&0\end{matrix}\right)\]

but $M$ is not diagonalisable according to WolframAlpha. 

What am I missing here?

asked Oct 27, 2015 in Theoretical Physics by danielm (5 points) [ no revision ]

You ignored the tensor product structure of the Hilbert space (assuming that $b$ and $d$ act on different components of a tensor product).

If done correctly, your $M$ should be Hermitian since $H_1$ is.

1 Answer

+ 2 like - 0 dislike

Watch out: Your problem arises because this second-quantized Hamiltonian contains terms that do not conserve the particle number (e.g. $d b$ in the "G" term). As a result, when "diagonalizing" such a Hamiltonian, you actually need a Bogoliubov transformation (instead of a simple unitary transformation) to obtain new normal modes. The requirement for this transformation, which mixes creation and annihilation operators, is that the new operators $d'$ and $b'$ fulfill the same bosonic commutation relations as the old ones (e.g. $[d',d'^{\dagger}]=1$).

While it is formally possible to write everything in terms of a matrix $M$ (as you have done), finding the Bogoliubov transformation is not equivalent to diagonalizing this matrix in the usual way. You need to fulfill the condition $[X_j,X_k^{\dagger}]=\delta_{jk} \sigma_k$, where $\sigma_k=+1$ for the $k=1,2$ and $\sigma_k=-1$ for $k=3,4$ in your example (the minus sign comes about because $X_3^{\dagger}=X_1$ and so on). Setting your Bogoliubov transformation to be $X_j=S_{jn} Y_n$ (summation over repeated indices), this implies $S \Sigma S^{\dagger} = \Sigma$, where $\Sigma$ would be the diagonal matrix with entries given by $\sigma_k$ ($1,1,-1,-1$). This is different from the usual unitary transformation, and indeed is known as a symplectic transformation.

answered Nov 1, 2015 by FlorianMarquardt (20 points) [ no revision ]

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