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Expectation value of an anti-Hermitian operator

  1. Apr 30, 2005 #1
    Hi, could anyone tell me how one would show that the expectation value of a anti-Hermitian operator is a pure imaginary number? Thanks.
  2. jcsd
  3. Apr 30, 2005 #2


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    I solved this problem once right on this site...(Dunno if in the QM forum,or college homework).Use the site's search engine,or google to find it...

  4. Apr 30, 2005 #3
  5. Apr 30, 2005 #4


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    Yes,post #2 of that thread.Consider an antiself-adjoint linear operator [itex] \hat{A} [/itex] for which u wish to prove that it has a spectrum made up of 0 & purely imaginary #-s...

    [tex] \hat{A}|\psi\rangle=\lambda|\psi\rangle [/tex]

    for an arbitrary eigenvector [itex] |\psi\rangle [/itex] corresponding to an eigenvalue [itex] \lambda[/itex]

    Then,the expectation value for this eigenstate is the eigenvalue,because

    [tex]\langle\hat{A}\rangle_{|\psi\rangle}=\langle\psi|\hat{A}|\psi\rangle=\lambda [/tex](1)

    The matrix element involved in (1) has the property

    [tex] \langle\psi|\hat{A}|\psi\rangle=\left(\langle\psi|\hat{A}^{\dagger}|\psi\rangle\right)^{*}=\left(-\lambda\right)^{*} [/tex] (2)

    Equating (1) & (2),you get that

    [tex]\lambda=-\lambda^{*} [/tex] (3)

    which means [itex] \mbox{Re}(\lambda) =0 [/itex],Q.e.d.

    Last edited: Apr 30, 2005
  6. May 1, 2005 #5
    In another form:

    A anti hermitian => i.A is hermitian
    => eigenvalues of A= (eigenvalues of i.A)/i= -i.(real number)= imaginary number.

  7. May 1, 2005 #6
    Ok thanks Daniel and Seratend.
  8. May 1, 2005 #7


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    By the way, "expectation" is a noun. "Expected" is an adjective.
    The "expected value" is the "expectation".
  9. May 1, 2005 #8
    I believe that used in this context, "expectation" becomes genitive, i.e. "value of an expectation". Other examples: economics textbook, price theory, etc.

    Also - "expectation value" seems to be normal usage:

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