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In a book, it says, Fock space is defined as the direct sum of all $n$-body Hilbert Space:

$$F=H^0\bigoplus H^1\bigoplus ... \bigoplus H^N$$

Does it mean that it is just "collecting"/"adding" all the states in each Hilbert space? I am learning 2nd quantization, that's why I put this in Physics instead of math.

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    $\begingroup$ Are you asking about what a "direct sum" is or are you asking what the physical motivation to take that direct sum is? $\endgroup$ – ACuriousMind Dec 3 '16 at 15:44
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    $\begingroup$ en.wikipedia.org/wiki/Direct_sum but you probably read this , and the wikipedia page looks a bit unsure of itself.... $\endgroup$ – user108787 Dec 3 '16 at 16:47
  • $\begingroup$ It means space did something to piss you off. $\endgroup$ – PiKindOfGuy Oct 23 at 7:21
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Suppose you have a system described by a Hilbert space $H$, for example a single particle. The Hilbert space of two non-interacting particles of the same type as that described by $H$ is simply the tensor product

$$H^2 := H \otimes H$$

More generally, for a system of $N$ particles as above, the Hilbert space is

$$H^N := \underbrace{H\otimes\cdots\otimes H}_{N\text{ times}},$$

with $H^0$ defined as $\mathbb C$ (i.e. the field underlying $H$).

In QFT there are operators that intertwine the different $H^N$s, that is , create and annihilate particles. Typical examples are the creation and annihilation operators $a^*$ and $a$. Instead of defining them in terms of their action on each pair of $H^N$ and $H^M$, one is allowed to give a "comprehensive" definition on the larger Hilbert space defined by taking the direct sum of all the multi-particle spaces, viz.

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

known as the Fock Hilbert space of $H$ and sometimes also denoted as $e^H$.

From a physical point of view, the general definition above of Fock space is immaterial. Identical particles are known to observe a definite (para)statistics that will reduce the actual Hilbert space (by symmetrisation/antisymmetrisation for the bosonic/fermionic case etc...).

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    $\begingroup$ Superb answer! I wish they write the QFT textbooks like this. $\endgroup$ – rainman Feb 17 '19 at 11:54
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Great answers, but just for completeness maybe it will be illustrative to have an example.

Suppose your $H^1$ contains some single-particle states $|a\rangle$, $|b\rangle$, etc. The Fock space removes the limitation on being a single particle, and is composed of $H^0$ (which is 1-dimensional), $H^1$, $H^2 = H \otimes H$, etc. This allows states like

  • the vacuum state, let's call it the empty ket $|\rangle$,
  • all single particle states, $|a\rangle, |b\rangle, \ldots$,
  • all two-particle states, $|aa\rangle, |ab\rangle, |ba\rangle, \ldots$ (NB that this construction deems them distinguishable),

but most importantly

  • any superposition of the above, like $\frac{e^{i\pi/4}}{\sqrt2}|\rangle + \frac12 |a\rangle - \frac12|aab\rangle\otimes\left(\frac1{\sqrt2}|a\rangle + \frac i{\sqrt2}|b\rangle\right)$.

This space is inherently infinite-dimensional even if you start with something small like a qubit. If you want to imagine the result with the help of a basis, simply concatenate the lists of the basis states of all the components:

$$\{|\rangle, |0\rangle, |1\rangle, |00\rangle, |01\rangle, |10\rangle, |11\rangle, |000\rangle, |001\rangle, \ldots\}$$


In the most trivial setting the single particle does not really have any distinct states, so $H^1$ is 1-dimensional. It still makes sense to pick a fiducial state $|{}\circ{}\rangle \in H^1$ and construct the Fock space with basis

$$\{|\rangle =: |0\rangle, |{}\circ{}\rangle =: |1\rangle, |{}\circ{}\circ{}\rangle =: |2\rangle, |{}\circ{}\circ{}\circ{}\rangle =: |3\rangle, \ldots\},$$

an example of a state might be, say, a coherent state

$$|\alpha\rangle = \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{e^{|\alpha|^2}n!}} |n\rangle$$

and you have a nice example of why people can speak of excitations as of "phonons" in a harmonic oscillator even though there's just a single particle oscillating!

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Yes, it does. You build a "large" Hilbert space from the "small" ones, if you like.

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