# Confusion about Fock subspace

I'm currently reading Folland's book on quantum field theory and came along some definitions. On p.90 of his book, Folland defines the symmetric Fock space as $$\mathcal{F}_s(\mathcal{H})=\bigoplus_{k=0}^\infty \overset{k}{\bigotimes}_s\mathcal{H} \tag{1}$$ where $\otimes_s$ is to be understood as the symmetric tensor product and $\mathcal{H}$ as Hilbert space.

On the next page he talks about finite-particle subspace (of the symmetric Fock space) of states in which the total number of particles is bounded above defined as

$$\mathcal{F}_s^0(\mathcal{H})=\text{the algebraic direct sum of the spaces}\,\overset{k}{\bigotimes}_s\mathcal{H} \tag{2}$$

Maybe its the language barrier, but doesn't algebraic direct sum mean exactly the direct sum which is in $(1)$? This would make them the same, which is probably not what he means. So I don't really understand the difference between $(1)$ and $(2)$.

• (1) is the completion of (2) in the metric induced by the inner product on (2). That is the standard way of constructing a Hilbert space out of a countable direct sum of Hilbert spaces. – Phoenix87 Aug 1 '15 at 14:04

What you have in (1) is not an algebraic direct sum as "An element of the direct sum is zero for all but a finite number of entries" (http://mathworld.wolfram.com/DirectSum.html ), and that is not true for (1).

• Ok, so notationwise, would it be correct to use $\oplus_{k=0}^n$ instead of using infinity? – Apogee Aug 1 '15 at 12:52
• @Apogee: No. Whatever $n$ you choose, there can be more non-zero elements in a point from $F^0_s$. What's important, there must be a finite number of such non-zero elements for each individual point. – akhmeteli Aug 1 '15 at 13:16

Ask yourself if you want your Fock space to itself be a Hilbert space.

Now look at the so called "finite-particle space" which is ... by definition ... a subset of the set of all functions from $\mathbb N$ into the union of all the finite products of 1d particle spaces. Specifically it is the subset where each function sends each n into the space of products of n single particle states and such that the function gives zero for all but finitely many $n\in \mathbb N.$ That's just what the words mean (and you cab restrict to where the finite products are symmetric or antisymmetric as needed).

So we clearly know what this "finite-particle space" is and it is clearly a vector space, and we can give it an inner product, but it is most definitely not a Hilbert space. Can we think of this vector space as a subspace of a larger space that is a Hilbert Space?

Yes. The Fock space. You can think of the Fock space concretely as the topological/metric closure of the "finite-particle space" as an inner product space.

Fine. But why was this so confusing? And what was going on with the notation? The notation and terminology can be understood from the perspective of category theory. If you take the algebraic category of vector spaces and take the direct sum you got the "finite-particle space" and if instead you took the Hilbert Spaces in the topological category of Hilbert Spaces and take the sum you get the Fock space.

In both cases you can think of them as the smallest algebraic type vector space capable of holding any of the spaces (the algebraic sum) versus the smallest topological type vector space (i.e. Hilbert Space) that is capable of holding any of the multiparticle spaces.

So if you think of them as vector spaces and take the sum you get the "finite-particle space" and if you think of then as Hilbert spaces and take the sum you get the Fock space.

It just comes down to whether you want the sum to be a Hilbert space or just a vector space. If you want it to be a Hilbert space it has to be a bit bigger to fill in the holes the "finite-particle space" leaves

The space of possible states of a single particle is a Hilbert space $H$. A state with $n$ particles is an element of $H^{\otimes n}$ or the tensor product of $H$ with itself $n$ times. However if we add a splash of relativity to the mix the number of particles isn't constant so the state is now an element of the direct sum, $$F:= \bigoplus^\infty_{n=0} H^{\otimes n}$$ This is called the Fock space $F$. Since you talk of symmetry we will also mention that it has symmetric subspaces $F^+$; $$F^+:=\bigoplus ^\infty_{n=0}Sym ^nH^{\otimes n}$$ And antisymmetric subspaces $F^-$; $$F^-:=\bigoplus^\infty _{n=0}\bigwedge ^nH$$ These represent the space of states of bosons and Fermions respectively. That is my take on it at least.

• It's true that the full Fock space has these subspaces, but from what I understand the $\mathcal{F}_s^0$ is a (dense?) subspace of the already symmetric Fock space $\mathcal{F}_s$ – Apogee Aug 1 '15 at 12:22
• @AngusTheMan Your answer doesn't really answer the question; it's just recapping the definition of the Fock space which the OP already gave. – Jonas Greitemann Aug 1 '15 at 13:07
• @Jonas Yeah I agree with your point. I am hoping to edit it to account for this in the future, but I don't think I will delete it as it may help someone reading the post who isn't familiar with Fock spaces. Thanks for your comment. – AngusTheMan Aug 1 '15 at 13:16
• What's finite is the number of particle described by each subspace. This doesn't imply that the subspaces are finite dimensional. This happens for the fermionic case, where the single particle Hilbert space is $\mathbb C^2$, but it is infinite dimensional, however separable, for the bosonic case. – Phoenix87 Aug 1 '15 at 14:07