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

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


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


3 Answers 3


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 $t = \cos\theta$, $r = \sin\theta$.

  • $\begingroup$ 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. $\endgroup$
    – Randy
    May 7, 2013 at 11:15
  • $\begingroup$ Hi, I would still like some clarification on the above step if possible. $\endgroup$
    – Randy
    May 8, 2013 at 4:24
  • $\begingroup$ I'll get to it in a while, having a busy week. Please be patient. $\endgroup$ May 8, 2013 at 11:28

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

$$ \tag{2} [a_i,a_j^{\dagger}] ~=~\delta_{ij} {\bf 1}, \qquad [a_i,a_j] ~=~0, \qquad [a_i^{\dagger},a_j^{\dagger}] ~=~0,\qquad i,j~\in~\{1,2\}, $$

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.


Actually the operator you have is closely related to angular momentum. Indeed you can verify that the operators
$$ b^\dagger a \mapsto L_+\, ,\qquad a^\dagger b\mapsto L_-\, ,\qquad \frac{1}{2}(b^\dagger b-a^\dagger a) \mapsto L_z $$ satisfy the same commutation relations as the angular momentum operators. In this notation, the lowest state of angular momentum $s$ is the 2d harmonic oscillator state $$ \frac{(a^\dagger)^{2s}}{\sqrt{(2s)!}}\vert 0\rangle\to \vert s,-s\rangle $$ and in particular $$ a\vert 0\rangle \mapsto \vert \textstyle \frac{1}{2},-\frac{1}{2}\rangle\, ,\\ b\vert 0\rangle\mapsto \textstyle \frac{1}{2},\frac{1}{2}\rangle\, . $$ In general $$ \frac{(a^\dagger)^{s-m} (b^\dagger)^{s+m}}{\sqrt{(s-m)!(s+m)!}}\vert 0\rangle \mapsto \vert s,m\rangle\, . $$ Thus your operator $$ a^\dagger b-b^\dagger a\mapsto L_--L_+ =L_x-iL_y-(L_x+iLy)=-2iL_y $$ so that you can think of $O$ as the rotation $e^{-2i\theta L_y}$.

Now, the coherent states \begin{align} \vert\alpha\rangle \vert\beta\rangle&= \sum_{p,q}\frac{\alpha^p \beta^q}{p! q!} (a^\dagger)^p(b^\dagger)^q\vert 0\rangle \\ &\mapsto \sum_{p,q}\frac{\alpha^p \beta^q}{p! q!} \vert \textstyle\frac{1}{2}(p+q),\frac{1}{2}(p-q)\rangle \\ \end{align} where $\vert \textstyle\frac{1}{2}(p+q),\frac{1}{2}(p-q)\rangle$ is an angular momentum state with $s= \textstyle\frac{1}{2}(p+q)$ and $m_s=\frac{1}{2}(p-q)\rangle$. Thus \begin{align} O\vert\alpha\rangle \vert\beta\rangle &\mapsto \sum_{sm_s}\frac{\alpha^{s-m_s} \beta^{s+m_s}}{(s+m_s)! (s-m_s)!} e^{-i2\theta L_y}\vert \textstyle s,m_s\rangle\, ,\\ &=\sum_{sm_s}\frac{\alpha^{s-m_s} \beta^{s+m_s}}{(s+m_s)! (s-m_s)!} \sum_{m_s'} \vert s,m_s'\rangle d^s_{m_s',m_s}(2\theta) \end{align} where $d^s_{m_s',m_s}(2\theta)$ is a Wigner $d$-function. It remains to convert back $\vert s,m_s'\rangle$ to harmonic oscillator states.

Finally, the solution simplifies nicely if your initial state is either $\vert 0\rangle \vert\beta\rangle$ or $\vert \alpha \rangle \vert 0\rangle$. In such cases the functions $d^s_{m_s',\pm s}(2\theta)$ have a reasonably simple form, containing no summation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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