# Displacement operator

The displacement operator for one mode in quantum optics is the shift operator

$\hat{D}(\alpha)=\exp \left ( \alpha \hat{a}^\dagger - \alpha^\ast \hat{a} \right )$,

where $\alpha$ is the amount of displacement in optical phase space, $\alpha^*$ is the complex conjugate of that displacement, and $\hat{a}$ and $\hat{a}^\dagger$ are the lowering and raising operators, respectively. The name of this operator is derived from its ability to displace a localized state in phase space by a magnitude $\alpha$. It may also act on the vacuum state by displacing it into a coherent state. Specifically, $\hat{D}(\alpha)|0\rangle=|\alpha\rangle$ where $|\alpha\rangle$ is a coherent state, which is the eigenstates of the annihilation (lowering) operator.

## Properties

The displacement operator is a unitary operator, and therefore obeys $\hat{D}(\alpha)\hat{D}^\dagger(\alpha)=\hat{D}^\dagger(\alpha)\hat{D}(\alpha)=\hat{1}$, where $\hat{1}$ is the identity operator. Since $\hat{D}^\dagger(\alpha)=\hat{D}(-\alpha)$, the hermitian conjugate of the displacement operator can also be interpreted as a displacement of opposite magnitude ($-\alpha$). The effect of applying this operator in a similarity transformation of the ladder operators results in their displacement.

$\hat{D}^\dagger(\alpha) \hat{a} \hat{D}(\alpha)=\hat{a}+\alpha$
$\hat{D}(\alpha) \hat{a} \hat{D}^\dagger(\alpha)=\hat{a}-\alpha$

The product of two displacement operators is another displacement operator, apart from a phase factor, has the total displacement as the sum of the two individual displacements. This can be seen by utilizing the Baker-Campbell-Hausdorff formula.

$e^{\alpha \hat{a}^{\dagger} - \alpha^*\hat{a}} e^{\beta\hat{a}^{\dagger} - \beta^*\hat{a}} = e^{(\alpha + \beta)\hat{a}^{\dagger} - (\beta^*+\alpha^*)\hat{a}} e^{(\alpha\beta^*-\alpha^*\beta)/2}.$

which shows us that:

$\hat{D}(\alpha)\hat{D}(\beta)= e^{(\alpha\beta^*-\alpha^*\beta)/2} \hat{D}(\alpha + \beta)$

When acting on an eigenket, the phase factor $e^{(\alpha\beta^*-\alpha^*\beta)/2}$ appears in each term of the resulting state, which makes it physically irrelevant.[1]

## Alternative expressions

Two alternative ways to express the displacement operator are:

$\hat{D}(\alpha) = e^{ -\frac{1}{2} | \alpha |^2 } e^{+\alpha \hat{a}^{\dagger}} e^{-\alpha^{*} \hat{a} }$
$\hat{D}(\alpha) = e^{ +\frac{1}{2} | \alpha |^2 } e^{-\alpha^{*} \hat{a} }e^{+\alpha \hat{a}^{\dagger}}$

## Multimode displacement

The displacement operator can also be generalized to multimode displacement. A multimode creation operator can be defined as

$\hat A_{\psi}^{\dagger}=\int d\mathbf{k}\psi(\mathbf{k})\hat a^{\dagger}(\mathbf{k})$,

where $\mathbf{k}$ is the wave vector and its magnitude is related to the frequency $\omega_{\mathbf{k}}$ according to $|\mathbf{k}|=\omega_{\mathbf{k}}/c$. Using this definition, we can write the multimode displacement operator as

$\hat{D}_{\psi}(\alpha)=\exp \left ( \alpha \hat A_{\psi}^{\dagger} - \alpha^\ast \hat A_{\psi} \right )$,

and define the multimode coherent state as

$|\alpha_{\psi}\rangle\equiv\hat{D}_{\psi}(\alpha)|0\rangle$.

## References

1. ^ Christopher Gerry and Peter Knight: Introductory Quantum Optics. Cambridge (England): Cambridge UP, 2005.