Displacement operator

In the quantum mechanics study of optical phase space, the displacement operator for one mode is the shift operator in quantum optics,
:$\hat(\alpha)=\exp \left ( \alpha \hat^\dagger - \alpha^\ast \hat \right )$,
where $\alpha$ is the amount of displacement in optical phase space, $\alpha^*$ is the complex conjugate of that displacement, and $\hat$ and $\hat^\dagger$ 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,
$\hat(\alpha), 0\rangle=, \alpha\rangle$ 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
$\hat(\alpha)\hat^\dagger(\alpha)=\hat^\dagger(\alpha)\hat(\alpha)=\hat$,
where $\hat$ is the identity operator. Since $\hat^\dagger(\alpha)=\hat(-\alpha)$, 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.

:$\hat^\dagger(\alpha) \hat \hat(\alpha)=\hat+\alpha$
:$\hat(\alpha) \hat \hat^\dagger(\alpha)=\hat-\alpha$

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.

:$e^ e^ = e^ e^.$

which shows us that:

:$\hat(\alpha)\hat(\beta)= e^ \hat(\alpha + \beta)$

When acting on an eigenket, the phase factor $e^$ appears in each term of the resulting state, which makes it physically irrelevant.Christopher Gerry and Peter Knight: ''Introductory Quantum Optics''. Cambridge (England): Cambridge UP, 2005.

It further leads to the braiding relation
:$\hat(\alpha)\hat(\beta)=e^ \hat(\beta)\hat(\alpha)$

Alternative expressions

The Kermack-McCrae identity gives two alternative ways to express the displacement operator:
:$\hat(\alpha) = e^ e^ e^$

:$\hat(\alpha) = e^ e^e^$

Multimode displacement

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

:$\hat A_^=\int d\mathbf\psi(\mathbf)\hat a^(\mathbf)$,

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

:$\hat_(\alpha)=\exp \left ( \alpha \hat A_^ - \alpha^\ast \hat A_ \right )$,

and define the multimode coherent state as

:$, \alpha_\rangle\equiv\hat_(\alpha), 0\rangle$.

See also

* Optical phase space

References

{{Physics operators

Quantum optics