- #1
copernicus1
- 99
- 0
In the operation $$U(\Lambda)|{\bf p}\rangle=|{\Lambda\bf p}\rangle,$$ if we define the state covariantly, $$|{\bf p}\rangle=\sqrt{2E_{\bf p}}a_{\bf p}^\dagger|0\rangle,$$ then does the unitary operator [itex]U(\Lambda)[/itex] affect the factor of [itex]\sqrt{2E_{\bf p}}[/itex]? In other words, can we write $$U(\Lambda)|{\bf p}\rangle=U(\Lambda)\sqrt{2E_{\bf p}}a_{\bf p}^\dagger|0\rangle=\sqrt{2E_{\Lambda\bf p}}U(\Lambda)a_{\bf p}^\dagger|0\rangle,$$ or does [itex]\sqrt{2E_{\bf p}}[/itex] remain unaffected?