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I'm pretty new to quantum mechanics (I would like to understand it broadly as an hobbyist). I'm trying to reading Principles of Quantum Mechanics by Dirac. I've found difficult to understand a particular piece on p. 28 in the II chapter. It says that:

the conjugate complex of the product of two linear operators equals the product of the conjugate complexes of the factors in reverse order: $$\overline{\beta} \overline{\alpha} = \overline{\alpha \beta}\tag{$*$} $$

As simple examples of this result, it should be noted that, if $\xi$ and $\eta$ are real, in general $\xi \eta$ is not real. This is an important difference from classical mechanics. However, $\xi \eta + \eta \xi$ is real, and so is $i(\xi \eta - \eta \xi)$. Only when $\xi$ and $\eta$ commute is $\xi \eta$ itself also real.

I didn't understand why $\xi \eta$ is not real in general and why $\xi \eta + \eta \xi$ and $i(\xi \eta - \eta \xi)$ are, even if $\xi$ and $\eta$ are real.

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I hope this will help you to understand some of the statements. I will slightly modify the notation. We will denote the adjoint of an operator $A$ by $A^{\dagger}$ (in your post this corresponds to $\overline{A}$). If $A=A^{\dagger}$, then we call $A$ hermitian (='real'). We further say that two operators $A$ and $B$ commute if $[A,B]\equiv AB-BA = 0$. In this notation, your equation $(*)$ reads: $(A\,B)^{\dagger} = B^{\dagger}\, A^{\dagger}$.

We want to prove the following: For two hermitian operators, $A$ and $B$, it holds that

$$ (A\,B)^{\dagger}= A\,B \Longleftrightarrow [A,B]=0 \quad $$

Let us start with the $'\Leftarrow'$ direction: If $[A,B]=0$, then by definition it holds that $A\,B=B\,A$. It follows that $$ (AB)^{\dagger} = (BA)^{\dagger} \quad .$$ But since $A=A^{\dagger}$ and $B=B^{\dagger}$ by assumption, we obtain (recall equation $(*)$): $$(AB)^{\dagger} = A^{\dagger}\, B^{\dagger} = A\,B \quad .$$

Now the direction $'\Rightarrow'$, so let $A\,B= (A\,B)^{\dagger} \overset{(*)}{=} B^{\dagger}\,A^{\dagger}$. Then again by our assumption that $A$ and $B$ are hermitian, it follows that $A\,B = B\,A$ and thus $[A,B]=0$.

So, we see that indeed the product of two hermitian (real) operators is hermitian (real) if and only if both operators commute.

As a last point, let us take a look on
$$ A\,B + B\,A \quad .$$ If both operators are hermitian, then it follows that $$(A\,B + B\,A)^{\dagger} =(A\,B)^{\dagger} + (B\,A)^{\dagger} \overset{(*)}{=} B^{\dagger}\,A^{\dagger} + A^{\dagger} \, B^{\dagger} = B\,A+ A\,B = A\,B+ B\,A \quad. $$ Can you show that $i\,(\xi \,\eta - \eta\,\xi)$ is real for real $\xi$ and $\eta$?

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  • $\begingroup$ Thank you a lot for your time and effort, now I think I understood. I also tried to demonstrate that $i(\xi \eta - \eta \xi)$ is real for $\xi$ and $\eta$ real. My reasoning is the following (using your notation): $i(AB - BA)* = i(AB)* - i(BA)* = iB*A* - iA*B* =>$ since $A$ and $B$ real $=> iBA - iAB = i(BA - AB) => i(AB - BA) = i(BA - AB) => i(AB - BA) is real$. (Where * is your dagger symbol, I didn't know how to type it in...) $\endgroup$ – Luke__ Jan 10 at 20:19
  • $\begingroup$ Use \dagger to get $\dagger$. $\endgroup$ – G. Smith Jan 10 at 20:34
  • $\begingroup$ No problem! I think there is an error in your derivation: You have to use $(i\,C)^{\dagger} = -i \, C^{\dagger}$, i.e. also consider the $i$ in the process of taking the adjoint. Thus, for example, it reads $(i\,A\,B)^{\dagger} = -i\, (A\,B)^{\dagger}$. BTW: You can use the dagger symbol by using ^{\dagger}. $\endgroup$ – Jakob Jan 10 at 20:36
  • $\begingroup$ In fact I noticed the error and I was going to correct it. As usual, thank you! $\endgroup$ – Luke__ Jan 10 at 20:51
  • $\begingroup$ @Luke__ I just saw that you asked for the reason why the order of the operators in equation $(*)$ is changed. I think you can find an answer here $\endgroup$ – Jakob Jan 11 at 14:56

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