4
$\begingroup$

I have troubles to understand the concept of a Fock space.

We defined it as a direct sum of the 0-particle, single particle, two particle etc. Hilbert space.

Unfortunately, I am not sure if I understood this concept of the direct sum.

So this Fock space $F$ consists of $F = \{(x_1,x_2,....) ; \text{what?}\}$? I mean what are the conditions on each component in this sequence?

This seems to be related to the occupation number representation, but I don't quite see how:

We said that for bosonic states

$|n_1,n_2,... \rangle = S_{+} |i_1,i_2,...,i_N \rangle \frac{1}{\sqrt{n1!n2!...}}$

I interpreted this as $n_i$ tells me how many particles are in state $i_i$, but since we only consider finitely many cases, I would conclude from this that $n_{m}$ for $m>N$ are zero, is this true?

EDIT: Although I got interesting answers about the Fock space, nobody so far has actually talked about this occupation number representation. You can also find this representation here at page 6: click

$\endgroup$
17
  • $\begingroup$ Roughly speaking, the Fock space consist of the "infinite" vectors $(\psi_0,\psi_1,\dotsc,\psi_n,\dotsc)$, where $\psi_n$ belongs to the $n$-particle space for any $n\in\mathbb{N}$ (the space for $n=0$ is the vacuum); with the additional condition $\sum_{n=0}^\infty \lVert \psi_n\rVert^2 <+\infty$ (finite norm). So you can see that it is possible to have vectors with non-zero components in any $n$-particle sector, provided the norm converges. The space of vectors for which there exists $N\in\mathbb{N}$ such that $\psi_n =0$ for any $n> N$ is called the finite particle vectors... $\endgroup$
    – yuggib
    Nov 13 '14 at 20:20
  • $\begingroup$ ...space and is dense in the Fock space. $\endgroup$
    – yuggib
    Nov 13 '14 at 20:23
  • $\begingroup$ but in QM you always deal with finite particle systems, right? By the way: I like your way of writing this down, could you please comment on $|n_1,n_2,...\rangle \in $ Fock space, I currently don't see the relationship $\endgroup$ Nov 13 '14 at 20:23
  • $\begingroup$ Yes in QM you usually have only finite particle systems, that for convenience may be written as vectors of the Fock space. Anyways, in that case the Hamiltonian would preserve the number of particles, i.e. it would leave each $n$-particle subspace invariant. Concerning your last notation, I am not familiar with it; if it means that you have $n_j$ particles in the state $i_j$, then the total number of particles would be $\sum_j n_j$, provided it is finite. Anyways, it does not seem (to me) the usual way of representing a state of the Fock space. $\endgroup$
    – yuggib
    Nov 13 '14 at 20:30
  • $\begingroup$ @yuggib: I have a "stupid" question: why at each $\psi_n$ there is no the energy exponential $\rm{e}^{-\rm{i}E_n t/\hbar}$? $\endgroup$ Nov 13 '14 at 21:15
7
$\begingroup$

First, the direct sum $S=V\oplus U$, of two vector spaces $V$ and $U$ is just the vector space constituted of "sums of vectors" $s=v+u:=(v,u)$ of each original vector space. The multiplication by scalar is viewed as obbeying the distributive property: $\alpha s=\alpha(v+u)=(\alpha v,\alpha u)$. The sum of two vectors of $S$ is just $s_1+s_2=(v_1+v_2,u_1+u_2).$

In the case of the Fock space, the idea is to have just one space with all possible combinations of states with different number of particles, including superposition of different numbers. So the direct sum is used to add up all the spaces $\mathcal{H_n}=\mathcal{H}\otimes\mathcal{H}\otimes\cdots\otimes\mathcal{H}$ of $n$ particles that is the $n$-fold tensor product of $\mathcal{H}$, the hilbert space of just one particle, with itself. We then define the Fock Space as $$\mathcal{F}(\mathcal{H})=\bigoplus_{n=0}^\infty\mathcal{H_n}$$ where, $\mathcal{H}_0=\Bbb{C}$ corresponds to no particles. So, a general element of the $\mathcal{F}$ will be a sequence $$\psi=(\psi_0,\psi_1,\psi_2,\dots),$$ where $\psi_n\in\mathcal{H}_n$ is an $n$ particle state. So a general state is the superposition of different states of different particle numbers. This discussion has not concerned the case of bosons and fermions statistics, that requires the space to be symmmetric or anti-symmetric. For this we just put positive or negatives permutations over each $\mathcal{H}_n$: $$\mathcal{H}_n^\pm=\Pi^\pm\mathcal{H}_n$$

$\endgroup$
7
  • 4
    $\begingroup$ Strictly speaking, the direct sum is not a usual "sum of vectors", but an ordered pair that satisfies additional properties, depending on which structure it acts upon (e.g. groups, rings, vector spaces, etc). The notation as an usual sum may be, in my opinion, quite misleading. $\endgroup$
    – yuggib
    Nov 13 '14 at 21:40
  • $\begingroup$ It can be viewd as a sum if you keep in mind that the operations in the new space are induced in the old ones via this sum, but I agree that this can be sometimes misleading. I'll change the explanation. $\endgroup$ Nov 13 '14 at 21:47
  • $\begingroup$ unfortunately, you did not comment on my operator with the occupation numbers. i just don't see how this one fits into this pattern. $\endgroup$ Nov 13 '14 at 22:58
  • $\begingroup$ If I got your state in the right way $|\psi\rangle=|n_1,n_2,\dots\rangle$ is just an element of the $N$ particle space $\mathcal{H}^+_N$, with the restriction that $\sum_i n_i=N$. $\endgroup$ Nov 13 '14 at 23:10
  • $\begingroup$ but then this would not be an element of the Fock space, right? I also found this state on page 6 in this pdf nt.ntnu.no/users/johnof/second-quantization-2013.pdf $\endgroup$ Nov 13 '14 at 23:22
2
$\begingroup$

Maybe I could clarify the concept of a direct sum with a matrix example:

Assuming you have three matrices $A_0$, $A_1$, $A_2$, the direct sum of them is $$A_0\oplus A_1\oplus A_2 = \left( \begin{array}{ccc} A_0 & 0 & 0 \\ 0 & A_1 & 0\\ 0 & 0 & A_2 \end{array} \right)$$ You can see the resulting matrix (~ the Fock space) is a block-diagonal combination of the constituent matrices (~ Hilbert spaces).

Now, you can imagine that $A_0$ is acting on the 0-particle Hilbert space, $A_1$ acts on the 1-particle Hilbert space, and so on. In this toy model, a 0-particle state is represented by, say $$\left| \psi_0 \right\rangle = \left( \begin{array}{c} 1 \\ 0\\ 0 \end{array} \right) $$ and a 1-particle state is, say, $$\left| \psi_1 \right\rangle = \left( \begin{array}{c} 0 \\ 1 \\ 0 \end{array} \right) $$ A creation operator must move from $\left| \psi_0 \right\rangle$ to $\left| \psi_1 \right\rangle$, therefore it has to have off-diagonal elements, e.g.: $$a^{\dagger} = \left( \begin{array}{ccc} 0 & 0 & 0 \\ \sqrt{1} & 0 & 0\\ 0 & \sqrt{2} & 0 \end{array} \right)$$ Likewise the annihilation operator is $$a = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0 \end{array} \right)$$ And the number operator $$n = a^{\dagger}a = \left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{array} \right)$$ You can generalize this model to larger Fock spaces (to accommodate more than 2 particles) and build on it (ultimately leading to infinite-dimensional matrices, if you will), but it should give you a sense of what it means when people say Fock space is a direct sum of Hilbert spaces.

$\endgroup$
0
$\begingroup$

Considering direct sums, plainly, if you have one coordinate axes and you put another one on it you made a 2-D space, and that is it...another way of looking at it is this: if you have some space V and two sub spaces s1 and s2, in this space V, AND if they dont cross, meaning no elements of one set are simultaneously in the other set, then V is direct sum of s1 and s2...in the same way, you combine hilbert spaces to form all possible combinations that can occur...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.