# Diagonalizing a linearized optomechanical Hamiltonian

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

$$H_1=−Δd^†d+ω_Mb^†b+G(d+d^†)(b+b^†)$$

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

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

but M is not diagonalisable according to WolframAlpha.

What am I missing here?

Actually the form you are looking for is
$$H = X^\dagger M X = \left[\begin{array}{cccc}b^\dagger & d^\dagger & b & d\end{array}\right]\left(\begin{array}{cccc}\alpha_1 & \beta_1 & \gamma_1 & \delta_1 \\ \alpha_2 & \beta_2 & \gamma_2 & \delta_2 \\ \alpha_3 & \beta_3 & \gamma_3 & \delta_3 \\ \alpha_4 & \beta_4 & \gamma_4 & \delta_4 \end{array}\right)\left[\begin{array}{c}b\\ d\\ b^\dagger\\d^\dagger\end{array}\right]$$ For the matrix $M$ try $$M = \frac{1}{2}\left(\begin{array}{cccc}\omega_M & G & 0 & G\\ G & -\Delta & G & 0 \\ 0 & G & \omega_M &G \\ G & 0 & G & -\Delta \end{array}\right)$$ Indeed, $$H = \frac{1}{2}\left[\begin{array}{cccc}b^\dagger & d^\dagger & b & d\end{array}\right] \left(\begin{array}{cccc}\omega_M & G & 0 & G\\ G & -\Delta & G & 0 \\ 0 & G & \omega_M &G \\ G & 0 & G & -\Delta \end{array}\right)\left[\begin{array}{c}b\\ d\\ b^\dagger\\d^\dagger\end{array}\right] = \\\; \\ = \frac{1}{2}\left[\begin{array}{cccc}b^\dagger & d^\dagger & b & d\end{array}\right]\left[\begin{array}{c}\omega_M b + Gd + Gd^\dagger\\ Gb - \Delta d + Gb^\dagger\\ Gd + \omega_Mb^\dagger + Gd^\dagger\\Gb + Gb^\dagger - \Delta d^\dagger \end{array}\right] = \\\; \\ = \frac{1}{2}\left[ \omega_Mb^\dagger b + Gb^\dagger d + Gb^\dagger d^\dagger + Gd^\dagger b - \Delta d^\dagger d + G d^\dagger b^\dagger +\\ Gbd + \omega_M bb^\dagger + Gbd^\dagger + Gdb + Gdb^\dagger - \Delta dd^\dagger + Gb^\dagger d \right]$$ Summing up and rearranging, $$H = \frac{\omega_M}{2}\left( b^\dagger b + bb^\dagger\right) - \frac{\Delta}{2}\left( d^\dagger d + dd^\dagger\right) + G\left( b^\dagger d + b^\dagger d^\dagger + bd + bd^\dagger \right)$$ and finally, $$H = \frac{\omega_M - \Delta}{2} - \Delta d^\dagger d + \omega_M b^\dagger b + G\left(b + b^\dagger \right)\left(d + d^\dagger \right)$$ where I assumed the canonical commutation relations (bosons).

All looks good and matrix $M$ is symmetric and diagonalizable.

• Thank you urdv, that was exactly what I was looking for! I wonder why I didn't have that idea myself.. Oct 29, 2015 at 8:24
• Welcome. As a general recipe, always look for a symmetric or hermitian matrix, unless you know the overall result is not self-adjoint. The matrix elements follow by identification when you expand the explicit expression.
– udrv
Oct 29, 2015 at 11:45
• I'm afraid I'm not convinced that this is the correct answer. The matrix $M$ here given has four distinct eigenvalues and so does not give a diagonalisation into two bosonic modes (two polaritons). Perhaps I misunderstand something. Oct 29, 2015 at 13:16
• @P.Plowman It is true that $M$ has 4 distinct eigenvalues. However, its form is unique if we demand both that $H = X^\dagger M X$ and that $H = Y^\dagger D Y$ for some $Y=[c_-\;c_+\;c_-^\dagger\;c_+^\dagger]^T$ and $D = (\Lambda \;\;O\;/ O\;\; \Lambda)$ a diagonal matrix with only two distinct eigenvalues ($\Lambda =(\lambda_1\;\;0\;/0\;\;\lambda_2$)). It is sufficient to assume $Y = UX$ for some invertible $U$ that is compatible with the structure of $X$ and $Y$, and the form in the answer follows.
– udrv
Oct 30, 2015 at 15:36
• @P.Plowman But since $M$ has 4 distinct eigenvalues, while $D$ has 2, $U$ cannot be unitary. Although we do get $M = U^\dagger D U$, I think this is a pseudo-diagonalization, not a regular diagonalization.
– udrv
Oct 30, 2015 at 15:37