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Trouble with Hermitian operators?

  1. Nov 17, 2014 #1
    I am looking at the derivation of the Heisenberg Uncertainty Principle presented here: http://socrates.berkeley.edu/~jemoore/p137a/uncertaintynotes.pdf [Broken]

    and am confused about line (21)...

    I do not understand why [itex]AB[/itex] and [itex]BA[/itex] are complex conjugates of each other... (I'm still in high school so I don't really have much of a background in the algebra of operators).

    If I assume what is written in line 21, I think I get the following lines- if [itex]AB[/itex] and [itex]BA[/itex] are complex conjugates, then [itex]|\langle AB-BA \rangle |^2 = |2\langle AB\rangle |^2 = 4|\langle AB\rangle |^2[/itex] I think... The expected value signs are what is confusing me most...

    Thank you for any help!
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 17, 2014 #2


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    Staff: Mentor

    A hermitian operator has the property ##A^\dagger = A##, where the ##\dagger## symbol refers to hermitian conjugation, which is the extension to operators (or matrices) of complex conjugation. Also, when applying hermitian conjugation to a product of operators, you have to invert their order:
    (A B)^\dagger = B^\dagger A^\dagger
    So, if ##A## and ##B## are hermitian, you get
    (A B)^\dagger = B^\dagger A^\dagger = BA
    The complex conjugate of a bracket is given by
    \left( \langle \psi | A | \phi \rangle \right)^* = \langle \phi | A^\dagger | \psi \rangle
    When you combine these things together, you find
    \left( \langle AB \rangle \right)^* &= \left( \langle \psi | AB | \psi \rangle \right)^* \\
    &= \langle \psi | (A B)^\dagger | \psi \rangle \\
    &= \langle \psi | BA | \psi \rangle \\
    &= \langle B A \rangle
    So, as for all complex numbers ##|z| = |z^*|##, you have that ##|\langle AB \rangle| = |\langle BA \rangle|##.

    You can only write an inequality here, not equalities.
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