- #1
andresordonez
- 65
- 0
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.
"
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.