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About Schrödinger-Heisenberg connection

  1. Mar 3, 2013 #1
    I stumbled over something reading Green's Matrix Mechanics (1965) this afternoon. There was an equation very similar to one I saw in Dirac's Lectures on Quantum Field Theory (1966), where he talks about the equivalence (or near equivalence) of the Schrödinger and Heisenberg formulations of ordinary quantum mechanics:

    [tex]

    U_{S} = e ^ {-iHt/ \hbar} U_{H} e^ {iHt/ \hbar }

    [/tex]

    I take it that the U's are matrices, but what are the exponential terms? Vectors? Other matrices? If H is a matrix, what does it mean to raise a real number (e) to the power of a matrix? If instead they are real numbers, wouldn't the two exponential terms then simply cancel each other out?
     
  2. jcsd
  3. Mar 3, 2013 #2
  4. Mar 3, 2013 #3

    dextercioby

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    Homework Helper

    They are operators. H is an operator, the exponential of an operator is defined through the series expansion as an infinite sum. Of course, technicalities such as convergence, domain mustn't be overlooked.
     
  5. Mar 3, 2013 #4
    Hey thanks, I had no idea!
     
  6. Mar 3, 2013 #5

    tom.stoer

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    Usually a function of an (hermitean) operator is defined using the spectral decomposition. Let A be an hermitean operator, let a be its eigenvalues and let |a> define the eigenvectors = an orthonormal basis.

    Then

    [tex]f(A) = f(A) \sum_a|a\rangle\langle a| = \sum_a f(a)\,|a\rangle\langle a|[/tex]
     
  7. Mar 3, 2013 #6
    Interesting . . Will try this on a few very simple examples and see what happens.
     
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