Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Densely defined operators of Fock space

  1. Sep 9, 2008 #1


    User Avatar
    Gold Member

    Could someone prove the following (if it is precisely correct):

    Since Fock space is the closure of the finite linear span of finite excitations of the vacuum state $\Omega$, then the operator $\hat{O}$ is densely defined if and only if $\hat{O} \Omega$ has finite norm.

    Or more generally, an operator is densely defined on a separable space if it is defined on a given single vector in that space.

  2. jcsd
  3. Sep 10, 2008 #2


    User Avatar
    Science Advisor

    I'm not an expert on Functional Analysis, but since no one else has
    answered yet I'll have a go...

    (BTW, $....$ doesn't work here for latex strings. I've fixed them
    below. Click on any one of them to see the latex source, and a link
    to some latex info relevant to PhysicsForums.)

    I think you can use the axiom that Fock space is generated by
    cyclic operations of the annihilation and creation operators
    upon a distinguished vacuum vector (and a subsequent
    closure). Under this construction the a/c operators form an
    irreducible set, meaning that *any* operator on the Fock
    space can be expressed as a sum of polynomials of these
    operators. (Since you only want "densely defined",
    polynomials are enough -- you don't need to worry about
    infinite series and associated existence issues.)

    Then you take an arbitrary product of a/c operators.
    Eg, for a quadratic in the creation operators it might
    be something like:
    O_2 ~:=~ \int dp dq ~ f(p,q) a^*(p) a^*(q)
    More generally, there might also be annihilation operators
    but you can use the CCRs to normal-order them, and if
    the result is any naked delta fns (which cannot be
    integrated away) then such an operator doesn't give a
    finite-norm vector when acting on the vacuum.

    I think that's enough to do the proof in one direction
    (though you'll have to clean it up and fill in more details).
    Once you've seen it in one direction you can probably figure
    out the reverse direction.

    And if I've got anything wrong in the above, maybe someone
    else will enjoy correcting it. :-)

    I don't think that's true in general without further
    assumptions (such as the cyclicity of the vacuum and
    maybe some other stuff). "Separability" in this context
    only means the space has a countable orthonormal basis,
    so you've got to show that the operator acting on any
    basis vector results in a finite-norm vector. (Actually,
    that's kinda what I did above in a clumsy way for
    the Fock, since one gets a countable basis there by
    acting on the vacuum with products of creation

  4. Oct 5, 2008 #3


    User Avatar
    Gold Member

    Thanks for the response

    sorry i've been a bit busy to reply.

    Yes, a lot of operators have the same action on a given vector (in a separable space) but some will give non-normalizable vectors when acting on other vectors - unless addition assumptions are made.

    I didn't read the paper carefully anough, te separable space in question is isomorphic to Fock space


Share this great discussion with others via Reddit, Google+, Twitter, or Facebook