Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

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

    Bill_K

    User Avatar
    Science Advisor

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Unitary operator acting on state
  1. Unitary time operator (Replies: 20)

  2. Unitary operators (Replies: 14)

Loading...