Displacement operator: Difference between revisions

Revision as of 01:16, 26 June 2012

Revision as of 12:40, 16 July 2012

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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\beta^*-\alpha\beta^*\beta)/2}. </math>

which shows us that:

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:<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\beta^*-\alpha\beta^*\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 ==