# Problem with Parity for photons

1. Oct 18, 2012

### LayMuon

I am trying to understand how it can be shown that under parity transformation we have to have $\hat{P} \hat{a}_{\mathbf{p},\lambda} \hat{P} = - \hat{a}_{-\mathbf{p},\lambda}$, I mean the negative sign (negative intrinsic parity of photon). So I am trying to prove that from the vector nature of four potential it follows that $\eta = -1$ in the relation $\hat{P} \hat{a}_{\mathbf{p},\lambda} \hat{P} = \eta \hat{a}_{-\mathbf{p},\lambda}$

In radiation gauge the second quantized four-potential is:
$$\mathbf{A}(x) = \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 \omega_p}} \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p}, \lambda) \hat{a}_{\mathbf{p},\lambda} e^{-ipx} +\boldsymbol{\epsilon}^*(\mathbf{p}, \lambda) \hat{a}_{\mathbf{p},\lambda}^* e^{ipx} \right ]$$
From vector nature of $\mathbf{A}$ we should have:
$$\hat{P} \mathbf{A}(t,\mathbf{x}) \hat{P} = - \mathbf{A}(t,\mathbf{x'}=-\mathbf{x})$$

Applying parity operator:
$$\hat{P} \mathbf{A} \hat{P}= \int \frac{d^3p}{(2 \pi)^3 \sqrt{2 \omega_p}} \eta \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p}, \lambda) \hat{a}_{-\mathbf{p},\lambda} e^{-ipx} +\boldsymbol{\epsilon}^*(\mathbf{p}, \lambda) \hat{a}_{-\mathbf{p},\lambda}^* e^{ipx} \right ] \\ \hat{P} \mathbf{A}(x) \hat{P}= \int \frac{d^3p'}{(2 \pi)^3 \sqrt{2 \omega_p}} \eta \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(-\mathbf{p'}, \lambda) \hat{a}_{\mathbf{p'},\lambda} e^{-ip'x'} +\boldsymbol{\epsilon}^*(\mathbf{-p'}, \lambda) \hat{a}_{\mathbf{p'},\lambda}^* e^{ip'x'} \right ]$$

On the other hand for circular polarization vectors the flipping of the momentum sign ammouts to rotation of the direction of momentum by 180, e.g. around the axes $x$:
$$\boldsymbol{\epsilon}(\mathbf{p},1) = \frac{1}{\sqrt{2}}\{1,i,0\} \Rightarrow \boldsymbol{\epsilon}(-\mathbf{p},1) = \frac{1}{\sqrt{2}}\{1,-i,0\} \\ \boldsymbol{\epsilon}(\mathbf{p},2) = \frac{1}{\sqrt{2}}\{1,-i,0\} \Rightarrow \boldsymbol{\epsilon}(-\mathbf{p},2) = \frac{1}{\sqrt{2}}\{1,i,0\}$$

We see that polarizations exchange their places in the equation. Under spatial inversion all coordinates would be reversed and a minus sign would appear in the equation (4). Hence

$$\hat{P} \mathbf{A}(x) \hat{P}= \int \frac{d^3p'}{(2 \pi)^3 \sqrt{2 \omega_p}} (-\eta) \sum_{\lambda =1,2} \left [ \boldsymbol{\epsilon}(\mathbf{p'}, \mp) \hat{a}_{\mathbf{p'},\pm} e^{-ip'x'} +\boldsymbol{\epsilon}^*(\mathbf{-p'}, \mp) \hat{a}_{\mathbf{p'},\pm}^* e^{ip'x'} \right ]$$

I don't understand how can this equation (7) be compared with equation (2) to deduce that $\eta=-1$, the polarizations echanged their places!!!!

2. Oct 18, 2012

### Bill_K

Remember that, although momentum is a normal (polar) vector, spin is an axial vector. So that under a parity operation, p → -p but ss. The components of the polarization vectors remain unchanged: ε(-p,1) is still (1/√2){1, i, 0}.

3. Oct 18, 2012

### LayMuon

That I intuitively realize but I need mathematically sound proof. Which part of my logic fails mathematically?

Last edited: Oct 18, 2012