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Unitary operator acting on state

  1. May 24, 2014 #1
    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?
  2. jcsd
  3. May 24, 2014 #2


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    The unitary operator is linear and so cannot "affect" numbers, only states (and operators).
  4. May 24, 2014 #3


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