Displacement operator: Difference between revisions

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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{Da}(^{\dagger} - \alpha)^*\hat{Da}(} e^{\beta)=e\hat{a}^{\mathrmdagger} - i\cdotbeta^*\operatornamehat{Ima}} \left= e^{(\alpha + \beta)\hat{a}^{\astdagger} - (\rightbeta^*+\alpha^*)}\hat{Da}} e^{(\beta\alpha^*-\alpha + \beta^*)/2}. </math>

which shows us that:

When acting on an eigenket, the phase factor <math>e^{\mathrm i\cdot\operatorname{Im} \left(\alpha \beta^\ast \right)}</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>

<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^{(\mathrm ibeta\cdot\operatorname{Im} \left(alpha^*-\alpha \beta^\ast \right*)/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>

== Multimode displacement ==