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Eigenvalues of an operator

  1. Dec 14, 2008 #1
    Hi, everyone!

    While I was studying for my midterm, I encountered this question.
    Consider the hermitian operator H that has the property that
    H4 = 1
    What are the eigenvalues of the operator H?
    What are the eigenvalues if H is not restricted to being Hermitian?

    What I am going to ask is that does it matter this operator have a different power like H6 or H5? I mean what is the role of the power in this question? Also, I couldn't figure out that how I can find the eigenvalues of such operator when it is not Hermitian?

    Thank you.
  2. jcsd
  3. Dec 14, 2008 #2


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    Assume a finite dimensional vector space on which H acts. Can you write the characteristic equation for H ?
  4. Dec 14, 2008 #3
    I think you can do it simple like this:

    Let E be an eigenvector of H with eigenvalue e:

    H E = e E

    Then H^4 E = e H^3 E = e^2 H^2 E = ....

    And use H^4 E = E since H^4 = 1.

    This gives you an general equation for the eigenvalues. Now when H is hermitian the eigenvalues must be real, and this restricts the solutions to the equation obtained.
  5. Dec 15, 2008 #4
    Isn't it H*[tex]\Psi[/tex] = E * [tex]\Psi[/tex]?
  6. Dec 15, 2008 #5
    Also, can I say that since H^4 = 1, then one eigenvalue of it is 1 and if H were not a Hermitian operator, one eigenvalue would be i, since H^2 = -1 or 1?
  7. Dec 15, 2008 #6
    Let me just post the solution, u are very close anyway:

    Let E be an eigenvector of H with eigenvalue e:

    H E = e E

    Then we have:

    H^4 E = e H^3 E = e^2 H^2 E = e^3 H E = e^4 E.

    You simply let one H act on E in each step.

    Since you have H^4 = 1 we have:

    H^4 E = E = e^4 E.

    So all you need to solve is:

    e^4 = 1.

    If H is hermitian it has 4 solutions. But if H is not Hermitian the complex eigenvalues are ruled out and you only have the real solutions.
  8. Dec 15, 2008 #7
    Student111, thanks for your answer, but I guess H has 2 values if H is hermitian, if it isn't, it has 4?
  9. Dec 15, 2008 #8
    exactly, i mistyped.

    Nonhermitian: 1, -1, i, -i.

    And only 1,-1 if it is hermitian.
  10. Dec 15, 2008 #9


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    That's a good thing. Not all operators are guarunteed to be diagnonalizable, but Hermitian is an example that is. Of course, since a power of the operator is unity (i.e. the identity operator), they are hinting that H is at least unitary, which is also diagonalizable.
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