In B.R.Martin's textbook, 'Nuclear and Particle Physics: An Introduction', he writes the parity operator as $$\hat{P}\psi(x,t)\equiv P \psi(-x,t)$$ where $\hat{P}$ is the parity operator, $\psi$ is the wavefunction of some particle and $P$ is the 'intrinsic parity'. However, is the parity operator not defined as $x \rightarrow -x$, and as such $\hat{P}\psi(x,t)=\psi(-x,t)$?

If this is the case does that not mean that the initial equation can be written as $$\psi(-x,t)=P\psi(-x,t)$$ and hence $P\equiv1$ always? How can $P=-1$?

  • $\begingroup$ $P$ is the eigen value of the parity operator $\hat{P}$. To see that, apply parity operator again on $\hat{P}\psi=P\psi$. That will give $P^{2}=\pm{1}$. $\endgroup$ – AMS Oct 23 '16 at 12:27
  • $\begingroup$ I understand that, but what I'm saying is the left hand of the top equation seems to be equivalent to $\psi(-x,t)$ and hence how can the equality hold if $P\neq 1$? $\endgroup$ – R L W Oct 23 '16 at 12:30
  • $\begingroup$ Perhaps you do not understand that. Your rhetorical question "However, is the parity operator not defined..." has, of course the answer "it is not". Martin's definition is correct. $\endgroup$ – Cosmas Zachos Oct 23 '16 at 13:16
  • $\begingroup$ Ah okay, so are you saying that $x\rightarrow -x$ doesn't mean that under the same transform doesn't also act such that $\psi(x) \rightarrow \psi(-x)$? If so, how can this be the case? $\endgroup$ – R L W Oct 23 '16 at 17:09

When the parity operation is performed on a particle (its wavefunction), there are two possibilities

$\hat{P}\psi(x,t)\equiv P \psi(-x,t)$ = $\psi(x,t)$ (meaning the wavefunction is symmetric wrt spatial co-ordinates inversion; the wavefunction has same value at both $\pm x$)


$\hat{P}\psi(x,t)\equiv P \psi(-x,t)$ = $-\psi(x,t)$.......this is also possible (it is antisymmetric wrt co-ordinate inversion; the wavefunction's value at $-x$ is the negative of the wavefunction's value at $+x$)

Hence, P can have both the values $\pm1$

For more info, check out this link

  • 1
    $\begingroup$ There's a third possibility, which is that $\psi$ isn't an eigenstate of parity and $\hat P\psi = +\psi_\text{even} - \psi_\text{odd}$. However, all states must be unchanged under application of $\hat P^2$, which just takes you back to your original coordinate system. $\endgroup$ – rob Oct 23 '16 at 15:03

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