• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,726 comments
1,470 users with positive rep
818 active unimported users
More ...

  Diagonalizing a linearized optomechanical Hamiltonian

+ 1 like - 0 dislike

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 ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights