Displacement operator

Quantum optics operators
Ladder operators
Creation and annihilation operators
Displacement operator
Rotation operator (quantum mechanics)
Squeeze operator
Anti-symmetric operator

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

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

where $α$ is the amount of displacement in optical phase space, $\alpha^\ast$ 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 $α$ . 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. Displaced states are eigenfunctions of the annihilation (lowering) operator.

1 Properties
2 Alternative expressions
3 Multimode displacement
4 References
5 Notes
6 See also

Properties

The displacement operator is a unitary operator, and therefore obeys $\hat{D}(\alpha)\hat{D}^\dagger(\alpha)=\hat{D}^\dagger(\alpha)\hat{D}(\alpha)=I$ , where I is the identity matrix. 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 ( $- α$ ). 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.

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

which shows us that:

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

When acting on an eigenket, the phase factor $e^{(\beta\alpha^*-\alpha\beta^*)/2}$ 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.

References

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