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Conjugation of operators

  1. Nov 6, 2013 #1
    Why do we multiply some operator A both on the left and on the right with, say, A and A^(-1) in order to perform some kind of conjugation?

    If it helps, the example I'm thinking of is the relationship between Schrodinger and Heisenberg operators in QFT.

    Thanks.
     
  2. jcsd
  3. Nov 6, 2013 #2
    Hi. I'm sorry but I don't understand what you are saying. Could you elaborate on the example?
     
  4. Nov 6, 2013 #3
    Suppose we have some states ##| \psi \rangle##, ##| \phi \rangle## and an operator ##A##. We can compute the matrix element ##\langle \phi | A | \psi \rangle##. Then we can apply some transformation operator ##T## to the states to get new states ##T | \psi \rangle##, ##T | \phi \rangle##, and we can compute the matrix element of A between the new states, namely ##\langle \phi | T^\dagger A T | \psi \rangle##. We can see that this is equivalent to computing the matrix element of the transformed operator ##T^\dagger A T## between the original states ##| \psi \rangle## and ##| \phi \rangle##. So performing this sort of conjugation on operators is equivalent to performing a certain transformation on states. That's why this sort of conjugation appears so much.
     
  5. Nov 6, 2013 #4
    The_Duck, thanks, great answer.
     
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