- #1

Prologue

- 185

- 1

## Homework Statement

Show that the anticommutator of parity and boost is zero.

## Homework Equations

[tex]\{\mathcal{P},K^{i}\}=0[/tex]

## The Attempt at a Solution

Let the anti commutator act on a state

[tex]\{\mathcal{P},K^{i}\}\Psi(t,\vec{x})=\mathcal{P}K^{i}\Psi(t,\vec{x})+K^{i}\mathcal{P}\Psi(t,\vec{x})[/tex]

[tex]\{\mathcal{P},K^{i}\}\Psi(t,\vec{x})=\mathcal{P}K^{i}\Psi(t,\vec{x})+K^{i}\Psi(t,-\vec{x})[/tex]

Now, I could go off on a presumably wrong tangent, here goes. There is some definition that the K operating on Psi is the same as inverting the boost and popping it into the argument of Psi, i.e.

[tex]K^{i}\Psi(x^{\mu})=\Psi((K^{i})^{-1}x^{\mu})[/tex]

Now, multiplying that last equation on both sides by [tex]\mathcal{P}[/tex], we get

[tex]\mathcal{P}K^{i}\Psi(x^{\mu})=\mathcal{P}\Psi((K^{i})^{-1}x^{\mu})[/tex]

And since The parity worked in a somewhat easy way, basically [tex]\mathcal{P}\Psi(x^{\mu})=\Psi(\mathcal{P}x^{\mu})[/tex] (although it wasn't defined this way I don't think - can't recall now), why not continue the trend to get

[tex]\mathcal{P}K^{i}\Psi(x^{\mu})=\Psi(\mathcal{P}(K^{i})^{-1}x^{\mu})[/tex]Of course this is a convoluted mess and I am sure something is wrong by now. I do realize that I could write down the 4x4 representation of the boost matrices along with the parity matrix, and use the obvious fact that parity flips the sign of the three bottom rows when it acts on the left and flips the sign of the three rightmost columns when acting from the right - but I am sure that isn't the point.

Last edited: