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Exponential of operator

  1. Jul 31, 2012 #1


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    Hi there,
    I learn from the text that the exponential of an operator could be expanded with a series such that

    [itex]e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}[/itex]
    So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]

    will be a matrix with diagonal elements given as [tex]\exp(a_i)[/tex], is that right?

    So I am wondering what happen if [itex]\hat{A}[/itex] is now written as a power form, i.e. [itex]\hat{A}^n[/itex], can we conclude that
    So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]


    gives a matrix with diagonal elements as [itex]\exp(a_i^n)[/itex] ?
    Last edited: Jul 31, 2012
  2. jcsd
  3. Jul 31, 2012 #2


    Staff: Mentor

    That's correct - for operators that can be diagonalised.

    Its basically how functions of diagonalisable operators are defined - when diagonalised its a function of the eigenvalues.

    Non diagonalisable operators are however a bit trickier - there are a few definitions I have seen and even a paper showing they are basically all equivalent - but the point is they are a more difficult proposition.

  4. Jul 31, 2012 #3
    If [itex]\hat{A}|\psi\rangle = a_i|\psi\rangle[/itex], then [itex]e^{\hat{A}}|\psi\rangle= \exp(a_i)|\psi\rangle[/itex] and [itex]e^{\hat{A}^n}|\psi\rangle = \exp(a_i^n)|\psi\rangle[/itex]. You should be able to prove this pretty easily using the Taylor series definition you mentioned.
  5. Aug 1, 2012 #4


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    Science Advisor

    afaik the proof of

    [tex]f(A) = \sum_n f_n A^n[/tex]

    where fn are the Taylor coefficients is rather involved mathematically. What you need is the spectral theorem for self-adjoint operators. For us physicists it reads

    [tex]f(A) = f(A) \sum_n |a\rangle\langle a| = \sum_n f(a) |a\rangle\langle a| [/tex]

    where the states |a> are the eigenstates of A and form a complete set.
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