Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the operator:

$$O = e^{\theta(a^\dagger b - b^\dagger a)}$$

where $\theta$ is a constant.

$O$ is a unitary operator.

$a$, $a^\dagger$, $b$, and $b^\dagger$ are ladder operators for two harmonic oscillators.

A normalized coherent state is defined as:

$$\lvert\alpha\rangle = e^{-\lvert\alpha\rvert^2/2} e^{\alpha a^\dagger} \lvert 0\rangle$$

where $\lvert0\rangle$ is the ground state of the harmonic oscillator.

I'm trying to see how $O$ acts on the coherent states by calculating $O \lvert\psi\rangle = O\lvert\alpha\rangle\lvert\beta\rangle$ in terms of coherent states.

Also, how does $O$ act on $\alpha$ and $\beta$?

I'm trying to use

$$O a O^\dagger = a \cos(\theta) + b \sin(\theta)$$


$$O b O^\dagger = -a \sin(\theta) + b \cos(\theta).$$

share|cite|improve this question

There are many ways to go around this. You can start from the coherent states and apply the unitary $\hat{O}$ directly on them. That will not be that simple because you will get a term $\hat{O}e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}$. Now, the typical approach would be to exchange the order of the operators to get something like $e^{\alpha\hat{a}^\dagger}e^{\beta\hat{b}^\dagger}\hat{O}$ (up to some extra term due to the commutator). This not really a simple task but once you are done, you can Taylor expand the operator $\hat{O}$ and keep only the zeroth order (all other terms contain annihilation operators acting on vacuum). I am not going to dig into the calculation in more detail, there are many ways to do it and none of them is really pleasant.

But there is a better way. You can define a displacement operator by the action $\hat{D}(\alpha)|0\rangle = |\alpha\rangle$ and then you have $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha) = \hat{a}+\alpha$. You can combine this with the formulas for $\hat{O}\hat{a}\hat{O}^\dagger$, $\hat{O}\hat{b}\hat{O}^\dagger$ to see how the annihilation operators are transformed. What you should get is a beam-splitting of the two coherent states, i.e., $$|\alpha,\beta\rangle\to|t\alpha+r\beta,t\beta-r\alpha\rangle$$, where $t = \cos\theta$, $r = \sin\theta$.

share|cite|improve this answer
Could you explain how to combine $\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha)$ with $\hat{O}\hat{a}\hat{O}^\dagger$ and $\hat{O}\hat{b}\hat{O}^\dagger$ to get the result? I don't see how to do this. – Randy May 7 '13 at 11:15
Hi, I would still like some clarification on the above step if possible. – Randy May 8 '13 at 4:24
I'll get to it in a while, having a busy week. Please be patient. – Ondřej Černotík May 8 '13 at 11:28
Okay, Thank you so much! – Randy May 8 '13 at 12:02

Let us change OP's notation $a\to a_1$ and $b \to a_2$. We write collectively the two annihilation operators as a column two-vector

$$ \tag{1} \vec{a}~:=~\begin{bmatrix} a_1 \\ a_2 \end{bmatrix}.$$

We have the Heisenberg algebra

$$ \tag{2} [a_i,a_j^{\dagger}] ~=~\delta_{ij} {\bf 1}\qquad i,j~\in~\{1,2\}. $$

and the vacuum state

$$ \tag{3} a_i | 0\rangle ~=~0, $$

Define un-normalized coherent states

$$ \tag{4} |\vec{\alpha} )_a ~:=~ e^{ a^{\dagger}_i \alpha_i} | 0\rangle . $$

The idea is now to diagonalize the ${\cal O}$ operator. Define unitary matrix

$$ \tag{5} U ~:=~\frac{\sqrt{2}}{2} \begin{bmatrix} 1 & i \\ i & 1 \end{bmatrix}~=~ \exp\left[i\frac{\pi}{4}\sigma_x \right] .$$

Define new operators

$$ \tag{6} b_i ~:=~ U_{ij} a_j, \qquad [b_i,b_j^{\dagger}] ~=~\delta_{ij} {\bf 1}, $$

and new coherent continuous labels

$$ \tag{7} \beta_i ~:=~ U_{ij} \alpha_j. $$

Define un-normalized coherent states

$$ \tag{8} |\vec{\beta} )_b ~:=~ e^{ b^{\dagger}_i \beta_i} | 0\rangle~=~ e^{ a^{\dagger}_i \alpha_i} | 0\rangle~=~|\vec{\alpha} )_a . $$

Note that the operator becomes diagonal

$$ \tag{9} {\cal O}~:=~ \exp\left[i\theta a^{\dagger}_i (\sigma_y)_{ij} a_j\right] ~=~ \exp\left[i\theta b^{\dagger}_i (\sigma_z)_{ij} b_j\right] ~=~ \exp\left[i\theta (n_1-n_2)\right],$$

where the number operators read

$$ \tag{10} n_i~:=~b^{\dagger}_i b_i \qquad\text{(no sum over $i$).} $$

Next deduce the commutation relations

$$ \tag{11} \exp\left[i\theta n_i\right]\exp\left[b^{\dagger}_i \beta_i \right]~=~\exp\left[b^{\dagger}_i \beta_i e^{i\theta} \right]\exp\left[i\theta n_i\right] \qquad\text{(no sum over $i$).} $$

We conclude from (11) that

$$ \tag{12} {\cal O}|\beta_1, \beta_2 )_b ~=~|\beta_1e^{i\theta}, \beta_2e^{-i\theta} )_b. $$

We leave it as an exercise to translate (12) back to normalized $a$-coherent states.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.