# Fock Space: I don't understand the definition

My question is really close to this one Meaning of Fock Space

In fact I don't understand why we define Fock space as :

$$\Gamma(H):=\mathbb C\oplus H\oplus H^2\oplus\cdots\oplus H^N\oplus\cdots$$

Indeed, I would define the Fock space as a Hilbert space which is an infinite tensorial product of Hilbert space.

Because for me to write :

$$|01\rangle = |0\rangle|1\rangle$$ is physically the same state as $$|1\rangle$$

Indeed (for me but I may be wrong), to say "first state is vacuum and second one is with one particle" is physically the same as saying "we have one particle"

Where is my mistake?

• Because the infinite tensorial product simply does not exist. $||\otimes_{n=1}^{+\infty} a_n\psi_n || = |a_n|^{\infty} = \infty$ if $||\psi_n||=1$ and $a_n = a>1$ even if $||\otimes_{n=1}^{+\infty} \psi_n || = 1$. You should exlude certain vectors from the space against the natural structure of tensor product and tensor space. – Valter Moretti Jan 24 '17 at 13:00
• Roughly dealing, you may handle the Fock space as if it were an infinite tensor product, but you cannot if you want to be mathematically precise. – Valter Moretti Jan 24 '17 at 13:04
• So, is it only for pure mathematical reasons ? If we would'nt have this problem of convergence, then we would use the "more simple" definition of infinite tensor product ? – StarBucK Jan 24 '17 at 15:31
• That is a question with a very strange status, I cannot answer. physical tools in physics must be logically consistent, first of all (and physically sound obviously). An infinite tensor product is not logically consistent, at least within a natural approach. This is enough in my view to rule out it. Also because a Fock space is not an absurdly complicated concept...This is different form other tools think of the Dirac's delta, where the mathematical formalization is a bit more complicated... – Valter Moretti Jan 24 '17 at 15:48
• To my experience when physics is correct mathematics can be fixed accordingly. Fock space is just the mathematical notion arising from the physical idea of a quantum system made of an undefined number of elementary subsystems... – Valter Moretti Jan 24 '17 at 15:51

One has to take into account all the subspaces of definite particle number of the full Hilbert space in order to describe interacting theory, because in QFT particles can be created from the vacuum by operators in the interaction term $H_{int}$ in the Hamiltonian, and annihilated. How are you going to define the action of the creating operator $a_1^{\dagger}$ on your two-particle Hilbert space, if you "absorbed" the state space of the first particle into the second one? The operators (observables) in QFT are defined to act on the full Hilbert space.
• "if you "absorbed" the state space of the first particle into the second one?" I don't understand what you mean ? I just work in a larger space that contains all number of particle ever possible. For me it is the same as studying an object that can move through a line $y=2$ but then decide to go on all the plane. I can describe its movement from the beginning by a movement in $\mathbb{R}^2$. I'm not forced to work in $\mathbb{R}\oplus\mathbb{R}^2$ to study it. – StarBucK Jan 24 '17 at 15:36