Displacement operator

In the quantum mechanics study of optical phase space, the displacement operator for one mode is the shift operator  in quantum optics,
 * $$\hat{D}(\alpha)=\exp \left ( \alpha \hat{a}^\dagger - \alpha^\ast \hat{a} \right ) $$,

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 an eigenstate 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 whose total displacement, up to a phase factor, is 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.

It further leads to the braiding relation
 * $$\hat{D}(\alpha)\hat{D}(\beta)=e^{\alpha\beta^*-\alpha^*\beta} \hat{D}(\beta)\hat{D}(\alpha)$$

Alternative expressions
The Kermack-McCrae identity gives two alternative ways to express the displacement operator:
 * $$\hat{D}(\alpha) = e^{ -\frac{1}{2} | \alpha |^2  } e^{+\alpha \hat{a}^{\dagger}} e^{-\alpha^{*} \hat{a} } $$


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

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


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