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I'm not clear why, in case $\hat A$ and $\hat B$ are self-adjoint, then $(\hat A \hat B + \hat B \hat A )$ is also self-adjoint. In order to show that $(\hat A \hat B + \hat B \hat A )=(\hat A \hat B + \hat B \hat A )^\dagger$, I have seen that the property $(\hat a \hat b)^\dagger=\hat b^\dagger \hat a^\dagger$ is applied, so $(\hat A \hat B + \hat B \hat A )^\dagger=\hat B^\dagger \hat A^\dagger + \hat A^\dagger \hat B^\dagger$. However, I don't get with this last is equal to $\hat A \hat B + \hat B \hat A $. Could someone give me a hint?

Invenietis
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1 Answers1

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$$\begin{align} \left( \hat A\hat B+\hat B\hat A\right)^\dagger &=\left(\hat A\hat B\right)^\dagger+\left( \hat B\hat A\right)^\dagger \tag{1 }\\ \tag{2} &= \hat B^\dagger\hat A^\dagger+\hat A^\dagger\hat B^\dagger \\ \tag{3} &= \hat A^\dagger\hat B^\dagger+ \hat B^\dagger\hat A^\dagger \\ \tag{4 }&= \hat A\hat B + \hat B\hat A.\end{align}$$

  1. Distribute the Hermitian conjugate between terms.

  2. $(\hat X\hat Y)^\dagger = \hat Y^\dagger \hat X^\dagger$ as you have correctly said.

  3. Matrix addition is commutative.

  4. $\hat A^\dagger=\hat A$ and $\hat B^\dagger =\hat B$ as they are both self-adjoint.

Charlie
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  • I was stuck with something so silly (point 4). Thank you very much! – Invenietis Aug 31 '20 at 09:42
  • No problem, easy enough to overlook when you're starting out. – Charlie Aug 31 '20 at 10:07
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    Just to add a word of caution: If $\hat A,\hat B$ are general unbounded (self-adjoint) operators one has to ensure that the expression $\hat A\hat B+\hat B\hat A$ is defined at least on a dense subspace of the underlying Hilbert space. For example it could happen that $\operatorname{im}(\hat A)\cap D(\hat B)={0}$ in which case $\hat B\hat A$ does not have a meaning (cf. Chapter VIII.5 in Reed & Simon, "Functional Analysis"). – Frederik vom Ende Sep 13 '20 at 15:25
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    @FrederikvomEnde A good point to add, but probably quite a lot more advanced than is needed for the target audience of OP's question. – Charlie Sep 13 '20 at 15:46