Parity operator and change of variable question

[andresordonez]

Hi, while reading the section about the parity operator from the QM book by Cohen-Tannoudji (complement F II, page 192), I found this:

"
Consider an arbitrary vector $$|\psi\rangle$$ of $$\mathcal{E}_\vec{r}$$:

$$|\psi\rangle = \int d^3 r \psi(\vec{r})|\vec{r} \rangle$$

If the variable change $$\vec{r'}=-\vec{r}$$ is performed, $$|\psi \rangle$$ can be written:

$$|\psi \rangle = \int d^3 r' \psi (-\vec{r'}) |-\vec{r'} \rangle$$
"

But $$d^3 r = dx dy dz$$ and after the variable change I get $$d^3 r' = dx' dy' dz' = - dx dy dz$$, so I don't understand what happened to that minus sign. It should be:

$$|\psi \rangle = -\int d^3 r' \psi (-\vec{r'}) |-\vec{r'} \rangle$$

right??

Someone told me it had to do something with the meaning of the differential volume, but I'm not sure about that.

Thanks.

 

[schieghoven]

The limits on the integral also got swapped. When you change them back around, there's another three factors of -1.

 

[andresordonez]

Right. Thanks schieghoven!