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Homework Help: Unitary and self-adjoint operators

  1. Sep 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Let U and A be two linear maps related by U=e^iA. Show that U is unitary if A is self-adjoint. Give a counterexample to show that U can be unitary if A is not self-adjoint.

    2. Relevant equations
    Self-adjoint: A*=A

    3. The attempt at a solution
    OK, so I had no problem with the first part. It's easy to show that U dagger=U inverse. However, I'm having some trouble coming up with a counter example for the second part. I tried something anti-Hermitian, but that just shows that U is self-adjoint, not that it is unitary.
  2. jcsd
  3. Sep 19, 2012 #2
    Write out what it means that [itex]UU^*=I=U^*U[/itex]. What could go wrong?
  4. Sep 19, 2012 #3
    Wait. What do you mean "what could go wrong"? I need to show that U can be unitary even if A isn't self-adjoint by way of an example.
  5. Sep 19, 2012 #4
    I didn't phrase that the best way as I could.

    So, indeed, you need to find a non-self-adjoint matrix A such that [itex]U=e^{iA}[/itex] is unitary. So, assume that U is unitary, then [itex]UU^*=I=U^*U[/itex]. What does that mean for A??
  6. Sep 19, 2012 #5

    I feel like this is really simple; I wonder why I'm not seeing it.
  7. Sep 19, 2012 #6
    OK, so you want [itex]UU^*=I=UU^*[/itex]. You indeed say that this is the same as


    Now, do you know what [itex](e^{iA})^+[/itex] is?? Can you write it in another way??
  8. Sep 19, 2012 #7
  9. Sep 19, 2012 #8
    OK, so now you have


    Now, what happens if you use [itex]e^{A+B}=e^A e^B[/itex]?? (I know the rule isn't always valid for matrices, but let's assume [itex]AA^+=A^+A[/itex], in that case the rule is valid).
  10. Sep 19, 2012 #9
    You get [tex]e^{i(-A^++A)}=I=e^{i(A-A^+)}[/tex].
  11. Sep 19, 2012 #10
    OK, so what did we find now? We found out that if A isn't self-adjoint, then there are two ways that [itex]e^{iA}[/itex] could still be unitary.

    Either, [itex]B=A+A^+[/itex] is a matrix such that [itex]e^{iB}=I[/itex].
    Or [itex]AA^+\neq A^+A[/itex].

    One of those two things must hold for A.

    Let's focus on the second one, can you find a matrix such that [itex]AA^+\neq A^+A[/itex].
  12. Sep 19, 2012 #11
    I think I can come up with a matrix, but can't we just treat these as operators, too? How about the ladder operator in QM. It is non-Hermitian; i.e. [itex]AA^+\neq A^+A[/itex].
  13. Sep 19, 2012 #12
    The condition [itex]AA^+=A^+A[/itex] is known as "normal".
    But anyway, what do you mean with the ladder operator??
    Matrices are fine examples of operators. Matrices are exactly the operators on a finite dimensional space. So any counterexamples you find as matrices will be a counterexample of operators.
  14. Sep 19, 2012 #13

    I am terrible at Latex, so I'm just linking to Wikipedia. I think this makes sense, right?
  15. Sep 19, 2012 #14
    Oh, those. I call them shift operators :smile:

    Anyway, maybe those ladder operators work, but for that you got to calculate [itex]e^{iA}[/itex] and that seems a little tedious.
    The easiest calculations are, in my opinion, with matrices.
  16. Sep 19, 2012 #15
    OK, how about let A=({2,3}, {-2,1})? I think that works.
  17. Sep 19, 2012 #16
    OK, but the calculations are a bit annoying. Can't we find a matrix whose exponential can easily be calculated? For example, try a nilpotent matrix.
  18. Sep 19, 2012 #17
    OK, how about ({0,1}, {0,0})?
  19. Sep 19, 2012 #18
    Yes, try that one! What do you get?
  20. Sep 19, 2012 #19
    You get A dagger does not commute with A, as we wanted. What else do I need to show?
  21. Sep 19, 2012 #20
    What do I need to show from there?
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