# Tensor product of Hilbert space and Fock space

If we have system consist of two or more than two subsystems, we can write hilbert space of a system in terms of tensor product of Hilbert spaces of subsystems and we can write state of a system. But also in Fock space for same system, when we write state of a system, Fock space is direct sum of hilbert spaces of subsystems. So can we use both formalisms for a system at the same time?

There are no different formalisms here. When you have two subsystems $$S_1$$ and $$S_2$$, the Hilbert space is the tensor product $$\mathcal H = \mathcal H_1 \otimes \mathcal H_2$$.
Let $$\mathcal H_1$$ be the Hilbert space of a single particle. The Hilbert space describing $$N$$ (distinguishable) particles is then $$\mathcal H_N = \mathcal H_1 \otimes \cdots \otimes \mathcal H_1 = \mathcal H_1^{\otimes N} .$$ If you can have any number of particles, then the total state lies in the direct sum $$\mathcal H = \mathcal H_1 \oplus \cdots \oplus \mathcal H_N \oplus \cdots .$$ (To get the Fock space, you now have to take the symmetrization / antisymmetrization, because the particles are indistinguishable.)
• @glS Maybe you should ask this as a separate question. In short: the tensor product describes a system made up of different subsystems, the direct sum describes different states the same system can be in. Note that e.g. $\lambda |+\rangle + |-\rangle$ is different from $|+\rangle + \lambda |-\rangle$ which does actually make sense :) – Noiralef Jan 14 '19 at 22:16
• @Noiralef ah, right, I see what you meant now. I still think it's worded a bit weirdly though. In your second equation you are essentially decomposing the Hilbert spaces of the first equation then? Wouldn't it be clearer to write it as something like $\mathcal H_k=\bigoplus_{j=1}^\infty \mathcal H^{(1)}$ then? Or even better just $\mathcal H_k=\bigoplus_{j=1}^\infty \mathbb C$, as you are simply saying that the space is infinite dimensional and nothing more here – glS Jan 15 '19 at 9:13