# Unitary operator acting on state

1. May 24, 2014

### copernicus1

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 $U(\Lambda)$ affect the factor of $\sqrt{2E_{\bf p}}$? 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 $\sqrt{2E_{\bf p}}$ remain unaffected?

2. May 24, 2014

### Matterwave

The unitary operator is linear and so cannot "affect" numbers, only states (and operators).

3. May 24, 2014

### Bill_K

Yes. As you say, the covariant states are created by the covariant form of the creation operator, namely $b_{\bf p}^\dagger = \sqrt{2E_{\bf p}}a_{\bf p}$, and $U(\Lambda)b_{\bf p}^\dagger U(\Lambda)^{-1}= b_{\bf \Lambda p}^\dagger$

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