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I'm confused by a line in the following wikipedia article on the squeeze operator in deriving the action of the squeeze operator on Heisenberg basis, the article seems to imply that $$[(\hat{a}^{\dagger})^2, \hat{a}] = -2\hat{a}^{\dagger}.$$

I'm confused by this, because as I expand $$[(\hat{a}^{\dagger})^2, \hat{a}] = (\hat{a}^{\dagger})^2\hat{a} - \hat{a}(\hat{a}^{\dagger})^2 = \hat{a}^{\dagger}[\hat{a}^{\dagger}, \hat{a}]\hat{a}^{\dagger}=-(\hat{a}^{\dagger})^{2}.$$

I'm wondering if there is typo in the article?

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2 Answers 2

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There is an error in your equation, namely $$\hat{a}^{\dagger 2}\hat{a} - \hat{a} \hat{a}^{\dagger 2} \neq \hat{a}^\dagger [\hat{a}^\dagger, \hat{a}]\hat{a}^\dagger \,.$$

The correct equation is $$\hat{a}^{\dagger 2}\hat{a} - \hat{a} \hat{a}^{\dagger 2} = \hat{a}^\dagger [\hat{a}^\dagger, \hat{a}] +[\hat{a}^\dagger, \hat{a}] \hat{a}^\dagger \,.$$

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With $\left[\hat{a}, \hat{a}^{\dagger}\right] = \hat{a} \hat{a}^{\dagger} -\hat{a}^{\dagger} \hat{a} = 1$, $$ \begin{eqnarray} [(\hat{a}^{\dagger})^2, \hat{a}] &=& \hat{a}^{\dagger} \hat{a}^{\dagger} \hat{a} - \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger} \\ &=& \hat{a}^{\dagger} \left(\hat{a} \hat{a}^{\dagger} - 1\right) - \left(\hat{a}^{\dagger} \hat{a} + 1\right) \hat{a}^{\dagger} \\ &=& \hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} \\ &=& -2\hat{a}^{\dagger} \end{eqnarray} $$

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  • $\begingroup$ ...don't give it away so easily... $\endgroup$ Commented May 17, 2022 at 20:16
  • $\begingroup$ Meh, I appreciate the clear response. I know it's important to struggle generally. $\endgroup$
    – user135520
    Commented May 17, 2022 at 20:22

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