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Is the following operator hermitian? C|Phi> = |Phi>*

  1. Oct 2, 2003 #1
    -hey everyone,

    this one might be a little too math based for this forum, but I ran across it studying for one of my quantum exams and it seemed like an interesting problem. Haven't figured it out completly.
    We all know hermitian operators play a central role in quantum and so being able to tell if an operator is hermitian or not is important. Is the following operator hermitian?

    C|Phi> = |Phi>*

    (Takes a state function and gives the complex conjugate)

    All the methods to show that an operator is hermitian (that I've seen) rely on the eigen-value equation

    A|Phi> = a|Phi>

    But the way this operator is defined, I'm not sure what the eigen values are (Or if it is an eigen value problem since the original function is not returned just its conjugate).

    All I got was that for a Phi that is real, the only eigen value is one and it has infinite degeneracy. Not as exciting as faster than light travel, but dig in if you'd like...
  2. jcsd
  3. Oct 3, 2003 #2


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    Have you tried writing down the complete definition of a hermitian operator and try to prove that C satisfies each of the properties.
    Last edited: Oct 3, 2003
  4. Oct 4, 2003 #3


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    Below is another hint. (don't peek until you've tried the above!)

    C is not linear.
  5. Oct 7, 2003 #4

    Thanks Hurkyl,

    Appreciate the push in the right direction.

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