Displacement operator

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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.


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[edit]

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[edit]

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



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


See also[edit]