I Parity Selection Rules: I'm Confused

[ergospherical]

I'm confused by the discussion in section §30 (Parity of a state), page 98 of Landau's QM. The functions $\psi_u$ and $\psi_g$ are odd an even states respectively. If $f$ is a true scalar, then it should remain unchanged by inversion of the co-ordinates. Writing $q' = -q$, then its matrix element at position $(u,g)$ is\begin{align*}
f_{ug} = \int \psi_u^*(q) \hat{f}(q) \psi_g(q) dq =-\int [-\psi_u^*(q')] \hat{f}(q') \psi_g(q') dq' = \int \psi_u^*(q') \hat{f}(q') \psi_g(q') dq'
\end{align*}however it is written in the text that $f_{ug} = -f_{ug}$. What did I mis-understand?

 

[Gaussian97]

What is $q$? What are the limits of all those integrals?

 

[ergospherical]

What is $q$? What are the limits of all those integrals?

Oh yes, thanks, there should be an extra minus sign due to inverting the limits of the third integral.

 

[vanhees71]

I'm confused by the discussion in section §30 (Parity of a state), page 98 of Landau's QM. The functions $\psi_u$ and $\psi_g$ are odd an even states respectively. If $f$ is a true scalar, then it should remain unchanged by inversion of the co-ordinates. Writing $q' = -q$, then its matrix element at position $(u,g)$ is\begin{align*}
f_{ug} = \int \psi_u^*(q) \hat{f}(q) \psi_g(q) dq =-\int [-\psi_u^*(q')] \hat{f}(q') \psi_g(q') dq' = \int \psi_u^*(q') \hat{f}(q') \psi_g(q') dq'
\end{align*}however it is written in the text that $f_{ug} = -f_{ug}$. What did I mis-understand?

Where does the additional $-$ sign after the 2nd equality sign come from? This should be absent since $f(q)=f(-q)$ by assumption, if I understand right what you mean by "true scalar", i.e., a scalar under rotations AND parity.

 

[vanhees71]

Oh yes, thanks, there should be an extra minus sign due to inverting the limits of the third integral.

No, why? You have
$$\mathrm{d}^3 q' =\left | \mathrm{det} \frac{\partial(q')}{\partial q} \right| \mathrm{d}^3 q = |-1| \mathrm{d}^3 q=\mathrm{d}^3 q.$$

 

[ergospherical]

I have taken $dq’ = -dq$, but in changing variables must also reverse the limits of the integral
\begin{align*}
f_{ug} = \int_{-\infty}^{\infty} \psi_u^*(q) \hat{f}(q) \psi_g(q) dq &=-\int_{\infty}^{-\infty} [-\psi_u^*(q')] \hat{f}(q') \psi_g(q') dq' \\
&= \int_{\infty}^{-\infty} \psi_u^*(q') \hat{f}(q') \psi_g(q') dq' \\
&= -f_{ug}
\end{align*}