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I am beginner in QFT. I would like to know why there is a need of constructing Fock space for a $N$-particle system? Why can't we represent this many body system just as the tensor product of Hilbert space itself? In short what is the difference between a Fock space and a tensor product of Hilbert space $H_n$?

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    $\begingroup$ The tensor product represents specifically $n$ particles, while the Fock space can represent any number of particles. $\endgroup$
    – Slereah
    Jan 14, 2020 at 6:59
  • $\begingroup$ You can do that, but then you need to manually make sure you are working with wavefunctions that have the correct symmetry properties under particle permutations (symmetric/anti-symmetric for bosons/fermions). By using the Fock space constrution, particle statistics are automatically implemented. Also note that in a field theory your Hilbert space is not just the space of $N$ particles, but a direct sum over all possible $N$ as number of particles can dynamically fluctuate. $\endgroup$
    – Heidar
    Jan 14, 2020 at 7:09
  • $\begingroup$ Trying to make sure the wavefunctions are symmetric/anti-symmetric while particle number can change is complicated without the Fock space. $\endgroup$
    – Heidar
    Jan 14, 2020 at 7:09

1 Answer 1

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A Fock space is just one special construction of a Hilbert space. The basic idea is that the Fock space allows you to superpose tensor products of distinct degree. In other words, it allows you to make sense of expressions of the form $$|a\rangle+|b\rangle\otimes |c\rangle.$$

where $|a\rangle,|b\rangle,|c\rangle$ are one-particle states. From the quantum mechanical point of view, if ${\cal H}_0$ is the "one-particle Hilbert space", representing the states of a single particle, then the states of a collection of $N$ particles of this kind, with this $N$ fixed, form a subspace of the tensor product of ${\cal H}_0$ with itself $N$ times: ${\cal H}_0\otimes\cdots \otimes {\cal{H}}_0$.

The Fock space allows you to superpose such states and hence allows you to have a state on which for every $N\in \mathbb{N}$ you have probabilities for the number of particles being $N$: speak differently, you are allowed to describe states on which the very number of particles is uncertain and becomes an observable with probabilities and mean values as any other observable.

A very transparent example where this would be necessary is in relativity theory. The relation $E = mc^2$ implies that given enough energy particles can be created and that particles can be destroyed. This makes relativistic quantum mechanics with fixed number of particles problematic and the Fock space picture helps quite a lot.

So the construction is to simply form the direct sum of all symmetric or skew-symmetric tensor powers of ${\cal H}_0$. This yields either the bosonic or fermionic Fock space: $${\cal F}_\pm ({\cal H}_0)= \bigoplus_{n=0}^\infty (\cal H_0)^{\pm \otimes n}$$ where $({\cal H}_0)^{\pm \otimes n}$ means in my notation to take the tensor product of ${\cal H}_0$ with itself $n$ times and symmetrize for $+$ or anti-symmetrize for $-$.

To answer your question the distinction between the Fock space and a tensor product of Hilbert spaces is simply that the Fock space is a direct sum of infinitely many tensor products of one Hilbert space with itself.

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  • $\begingroup$ Thanks for your answer. If one wants to go from Fock space to Hilbert space, for eg, (assume a system with indistinguishable bosons) can I write |2,0> in Fock space as |LL> in Hilbert space? As a follow-up, how does one write |1,1>? As |LR> + |RL> in Hilbert space? $\endgroup$
    – ksr
    Feb 17 at 19:15
  • $\begingroup$ I'm not sure I understand your question. A Fock space is a kind of Hilbert space, so I'm not sure what "go from Fock space to Hilbert space" means. Could you clarify? $\endgroup$
    – Gold
    Feb 17 at 19:19
  • $\begingroup$ Thanks. I just mean from the occupation number representation to particle state representation... So for example |2,0> tells me that there are 2 particles in the left mode. So if I write the wavefunction of the 2 particles as a tensor product, I get |L>|L>. I just want to know if I am reasoning correctly being intuitive, and whether there is a rigorous way to show this transformation in terms of annihilation and creation operators. $\endgroup$
    – ksr
    Feb 17 at 20:50

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