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How unitary change of basis related to Trace?

  1. Nov 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Shanker 1.7.1
    3.)Show that the trace of an operator is unaffected by a unitary change of basis (Equivalently, show [itex]TrΩ=TrU^{\dagger}ΩU[/itex]


    2. Relevant equations

    I can show that via Shanker's hint, but I however can't see how a unitary change of basis links to [itex]TrΩ=TrU^{\dagger}ΩU[/itex], (and it really giving me a headache!) Would anyone be kind enough to explain to me?
     
    Last edited: Nov 2, 2012
  2. jcsd
  3. Nov 2, 2012 #2
    There is some identity which tells you that [itex]\mathrm{Tr}(AB) = \mathrm{Tr}(BA)[/itex] (more generally one could state that the trace is invariant under cyclic permutations). Use it and your problem should be as good as solved.
     
  4. Nov 2, 2012 #3
    Forgive my ignorance,
    But shouldn't "unaffected by a unitary change of basis" be expressed as[itex]TrUΩ=TrΩ[/itex]
     
    Last edited: Nov 2, 2012
  5. Nov 2, 2012 #4

    Dick

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    No. A matrix A is a change of basis of a matrix B if A=T^(-1)BT for a nonsingular matrix T.
     
  6. Nov 2, 2012 #5
    Thanks for reply,
    So here is my understanding: we map A to B, A and B represent the same thing in different bases, and their mathematical relation is [itex]A=T^{-1}AT[/itex]
    am I right?
    I have a rough picture now, but Sadly, I still can't see the physical picture, neither the mathematics.
    , and shanker surely was telling the truth! One should know a bit more of linear algebra before embrace into it!

    Thanks guys, anyway :)
     
  7. Nov 2, 2012 #6
    Now I got it a bit!
    For column vector (1,0) -> (0 ,1)
    And T is {(0,1),(1,0)} :)
    Did I get it right?
     
  8. Nov 2, 2012 #7

    Dick

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    If T is unitary then sure that's an option. So T takes (1,0) -> (0,1) and (0,1) -> (1,0). T^(-1) (which happens to be the same as T, but that's usually not the case) does the opposite. So to figure out what A is 'equivalent' (not equal!) to B, you use T to rotate a vector to the basis of B, then let B act on it, then undo the rotation with T^(-1), so A=T^(-1)BT. I know this is vague. But none of this vagueness should stop you from being able to show Tr(B)=Tr(A)=Tr(T^(-1)BT). That change of bases don't change the trace.
     
    Last edited: Nov 2, 2012
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