Conjugate complex of linear operators in quantum mechanics

Question

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.

Answer 1

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 $\eta$?