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Hermitian operator represented as a unitary operator

  1. Dec 30, 2014 #1
    1. The problem statement, all variables and given/known data
    I know that any unitary operator U can be realized in terms of some Hermitian operator K (see equation in #2), and it seems to me that it should also be true that, starting from any Hermitian operator K, the operator defined from that equation exists and is unitary.

    2. Relevant equations
    U=exp(iK)
    Letting M# be the adjoint of M (I can't find a dagger), and * be multiplication,
    unitary: M#*M = M*M# = I (Identity operator)
    Hermitian:M#=M

    3. The attempt at a solution
    Given a Hermitian operator K, define U as above, and then one must prove that exp(iK)*(exp(iK))# = I. That is, that (exp(iK))# =exp(-iK)
    Obviously I must use the fact that K=K# to do so, but this only makes the problem to show that exp(-iK#) = (exp(iK))# , which doesn't seem to get me very far. The fact that (M#)-1=(M-1)# also seems tempting to apply somewhere, but I do not see where. Oh, of course there is also the possibility that my conjecture is wrong.

    Side note: I presume this should go in some Homework rubrik, because although this is not a Homework problem, it seems like something that would be given for homework in an algebra course. Therefore this should probably go under a rubrik for Algebra, but there was no Algebra listing in the Homework menu.
     
  2. jcsd
  3. Dec 31, 2014 #2

    DEvens

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    " ... exp(-iK#) = (exp(iK))# ..."

    Can you show that? Can you do some operations on exp(iK) that will allow you to bring the # operator into the ()? It's hard to know how much of a hint is considered acceptable in a homework forum, but how about Taylor? So, given that K# = K, what is ( i K * K )#?
     
  4. Dec 31, 2014 #3
    Thanks, DEvens. Let me see if I see where your hints are leading. First, since if K is Hermitian so is iK, so
    (iK2)#=(iK*K)#=(K#(iK)#)=-i(K#)2.
    Similarly
    (iK3)#=(iK2*K)#=(iK2)#*K#=
    =(-i(K#)2)K#=-i(K#)3
    and so forth, so
    expanding (exp(iK))# via Taylor, we get (∑ (iK)n/n!)# = (∑ (-iK#)n/n! = exp(-iK#)
    Is this correct?
    PS I found the dagger M
     
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