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<math>\hat{D}(\alpha)|0\rangle=|\alpha\rangle</math> where <math>|\alpha\rangle</math> is a coherent state.
Displaced states are [[eigenfunctions]] of the annihilation (lowering) operator.
== Properties ==
The displacement operator is a [[unitary operator]], and therefore obeys
<math>\hat{D}(\alpha)\hat{D}^\dagger(\alpha)=\hat{D}^\dagger(\alpha)\hat{D}(\alpha)=I</math>,
where I is the identity matrix. Since <math> \hat{D}^\dagger(\alpha)=\hat{D}(-\alpha)</math>, the [[hermitian conjugate]] of the displacement operator can also be interpreted as a displacement of opposite magnitude (<math>-\alpha</math>). The effect of applying this operator in a [[similarity transformation]] of the ladder operators results in their displacement.
<math>\hat{D}^\dagger(\alpha) \hat{a} \hat{D}(\alpha)=\hat{a}+\alpha</math><br>
<math>\hat{D}(\alpha) \hat{a} \hat{D}^\dagger(\alpha)=\hat{a}-\alpha</math>
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#The Hadamard lemma|Baker-Campbell-Hausdorff formula]].
:<math> 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}. </math>
which shows us that:
<math>\hat{D}(\alpha)\hat{D}(\beta)= e^{(\beta\alpha^*-\alpha\beta^*)/2} \hat{D}(\alpha + \beta)</math>
When acting on an eigenket, the phase factor <math>e^{(\beta\alpha^*-\alpha\beta^*)/2}</math> appears in each term of the resulting state, which makes it physically irrelevant.<ref>Gerry, Christopher, and Peter Knight: ''Introductory Quantum Optics''. Cambridge (England): Cambridge UP, 2005.</ref>
== Alternative expressions ==
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