Proof of parity formula

Question

The parity formula is given by $p=n(-1)^l$ where $n$ is the intrinsic parity and $l$ the orbital angular number. How can we prove this formula?

Answer 1

I will give you an example maybe it will be clear to you.

Under space reflection $\textbf{x}\rightarrow \textbf{x}'=-\textbf{x}$ the relativistic wave equation $\psi(\textbf{x},t)$ transform as
$\psi(\textbf{x},t)\rightarrow\psi'(\textbf{x}',t)=e^{i\phi}\gamma^0\psi(\textbf{x},t) $

See Greiner "Relativistic quantum Mechanics"

Now we define the intrinsinc parity $n$ as the parity in rest frame system that is , if $\psi$ is eigenfunction of the pariy operator than in the rest frame we have

$P\psi=n\psi$

For example given

$ u_1=\sqrt{2m}\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$

we have that

$Pu_1=+u$

That is for a particle who his wave function in the rest frame is $u_1$ his intrinsic parity $n$ is $+1$

In the same way we have particles whose their wave function in the rest frame $v_1$ have the propertie

$Pv_1=-v_1$ that is their intrinsic parity $n$ is -1.

Now suppose that in the non relativistic regime that we have a particle whose his eingenfunction is $R(r)Y_m^l(\theta,\phi)$ Under parity transformation we have that

$PR(r)Y_m^l(\theta,\phi)=(-1)^lR(r)Y_m^l(\theta,\phi)$

Now you would be tentated to define parity as $(-1)^l$, but remember in the rest frame we have that the orbital number $l$ is zero so for a particle whose intrinsic parity $n$ is $-1$ we would have a parity 1 which is wrong. So if we define parity as $n(-1)^l$ we would avoid this problem