# Conjugate complex of linear operators in quantum mechanics

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.

I hope this will help you to understand some of the statements. In the following, we will use a slightly different notation:

We 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: $$(A\,B)^{\dagger} = B^{\dagger}\, A^{\dagger} \quad \tag{*}$$

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

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

To start, assume that both operators commute: $$[A,B]=0$$. This in turn implies $$A\,B=B\,A$$ and thus $$(AB)^{\dagger} = (BA)^{\dagger} \quad .$$ Since $$A=A^{\dagger}$$ and $$B=B^{\dagger}$$ by assumption, we obtain: $$(AB)^{\dagger} = (BA)^{\dagger}\overset{(*)}{=} A^{\dagger}\, B^{\dagger} = A\,B \quad .$$

Now suppose $$A\,B= (A\,B)^{\dagger}$$. We then find that $$A\,B \overset{(*)}{=} B^{\dagger}\,A^{\dagger}$$ and since $$A$$ and $$B$$ are hermitian, it follows that $$A\,B = B\,A$$ and hence $$[A,B]=0$$.

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 $$(A\,B + B\,A)^{\dagger} =(A\,B)^{\dagger} + (B\,A)^{\dagger} \overset{(*)}{=} B^{\dagger}\,A^{\dagger} + A^{\dagger} \, B^{\dagger} = B\,A+ A\,B = A\,B+ B\,A \quad,$$ where we have used that the adjoint of a sum is the sum of the adjoints.

Can you show that $$i\,(\xi \,\eta - \eta\,\xi)$$ is real for real $$\xi$$ and $$\eta$$?

• 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...) Jan 10 at 20:19
• Use \dagger to get $\dagger$. Jan 10 at 20:34
• 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}. Jan 10 at 20:36
• In fact I noticed the error and I was going to correct it. As usual, thank you! Jan 10 at 20:51
• @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 Jan 11 at 14:56