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.


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