Densely defined operators of Fock space

1. Sep 9, 2008

julian

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.

thanks

2. Sep 10, 2008

strangerep

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
operators.)

HTH.

3. Oct 5, 2008

julian

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

Thanks

julian