Displacement operator: Difference between revisions

Revision as of 06:16, 3 January 2011

Revision as of 07:23, 6 January 2012

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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]].

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which shows us that:

:<math>\hat{D}(\alpha)\hat{D}(\beta)= e^{(\alpha\beta^*-\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>