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Operators in denominators

  1. Jun 18, 2010 #1
    I am led to believe (because it is in a paper I am reading) that
    [tex] \frac{1}{H - z} \left|\phi\rangle = \frac{1}{E - z}\left|\phi\rangle[/tex]
    where [tex]H[/tex] is the hamiltonian, [tex]\left|\phi\rangle[/tex] is an energy eigenstate with energy [tex]E[/tex], and [tex]z[/tex] is a complex variable.
    In attempting to understand this expression, I have realized I do not know what is meant by
    [tex]\frac{1}{A}\left|\phi\rangle[/tex]
    for some operator [tex]A[/tex]. Is this the same thing as the inverse of A?

    Thanks.
     
  2. jcsd
  3. Jun 18, 2010 #2

    strangerep

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    Science Advisor

    Welcome to the wonderful world of the spectral theorem(s)! :-)

    The basic idea is this: let A is a self-adjoint operator on a Hilbert space,
    and suppose it has a (continuous) spectrum from 0 to infinity.
    Then it is possible to find a basis in the Hilbert space where each basis
    state corresponds to a particular value in the spectrum, and in fact
    a complex-analytic function of A, written "f(A)", can be expressed as

    [tex]
    f(A) ~\leftrightarrow~ \int_0^\infty\! dk\; f(k) \; |k\rangle\langle k|
    [/tex]

    where each |k> is one of the eigenstates of A, with eigenvalue k.
    (The set of all such k is called the "spectrum" of A.)

    In your case, we're dealing with the Hamiltonian operator H, and your
    particular complex-analytic function is f(H) := 1/(H-z), so we can
    express it as

    [tex]
    \frac{1}{H - z} ~\leftrightarrow~ \int_0^\infty\! dk\; \frac{ |k\rangle\langle k|}{k - z}
    [/tex]

    where now each |k>is an eigenstate of H with eigenvalue k, and these states are
    "normalized" according to [itex]\langle k|k'\rangle = \delta(k - k')[/itex].

    Now apply this to your [itex]|\phi\rangle[/itex], but first let's rewrite [itex]|\phi\rangle[/itex]
    as [itex]|E\rangle[/itex], since this is a more helpful notation. This is ok because
    [itex]|\phi\rangle[/itex] is (by definition) the eigenstate of H with eigenvalue E. We get:

    [tex]
    \int_0^\infty\! dk\; \frac{ |k\rangle\langle k|}{k - z} ~ |E\rangle
    ~=~ \int_0^\infty\! dk\; \frac{ |k\rangle}{k - z} ~ \delta(k - E)
    ~=~ \frac{1}{E - z} \; |E\rangle
    [/tex]

    To understand the spectral theorem(s) in more detail, it's probably best
    to start with the finite-dimensional matrix versions in linear algebra
    textbooks, and then progress to the functional-analytic versions,
    then to the versions for rigged Hilbert space which is what I've been
    using above. Or, for a more QM-flavored overview, try Ballentine ch1,
    in particular sections 1.3 and 1.4.

    HTH.
     
    Last edited: Jun 18, 2010
  4. Jun 19, 2010 #3

    Hurkyl

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    Staff Emeritus
    Science Advisor
    Gold Member

    While it's good to know there's a generality, this particular case can be handled by lower-powered means -- so long as z ranges over the non-eigenvalues* of H, the operator (H-z) has a multiplicative inverse. 1/(H-z) is the unique operator that you can substitute for T in
    [tex]T \left( (H-z) | \psi \rangle \right) = | \psi \rangle[/tex]​
    which makes the equation hold for all kets.

    If you restrict to the space generated by [itex]|\phi\rangle[/itex], then z merely has to avoid the value E. (Because, on this subspace, E is the only eigenvalue of H)

    *: Really, I mean the points that aren't in the spectrum. But I think your H has a discrete spectrum... and so they are all eigenvalues too.
     
  5. Jun 20, 2010 #4
    Thanks to both of you! That's a big help.
     
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