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I'm reading about the second quantization formalism. I can see the advantages of using number states to represent multiparticle states. Here's my question:

Let's say we're given a single-particle basis of plane waves with spins (so each particle can be described by some wavevector $k$ and some spin $\sigma$). Extending this formalism to $N$ such particles and writing this in occupation number representation, one could write states like $|n_1, n_2, \ldots, n_i,\ldots\rangle$. In doing so, what does each number $n_i$ mean? Does it mean, for example, that there are $n_i$ particles in some state $\{k_i, \sigma_i\}$? If that's the case, then supposedly we have enumerated as many $k$ and $\sigma$ combinations as could possibly exist, and only put non-zero values for those that describe any of the $N$ particles. Or is there another interpretation that I'm missing?

Any references that clarify this material are also welcome. Most books quickly skip from single-particle wavefunctions to those of many particles by saying something along the lines of "each $n_i$ means there are $n_i$ particles in state $i$," but I'm not sure what exactly this state describes.

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You are right. The $i$ indices of the occupation numbers $n_i$ can be thought of as multi-indices meaning that each $i$ encodes both the wavevector $k$ (which is quantized once boundary conditions are imposed) and the spin sign $\sigma$. In general, $i$ will enumerate all the allowed combinations of quantum numbers that characterize the single particle states.

It is also true that only the entries $n_i$ of those states which are populated will be non-zero. If the total particle number $N$ is fixed, the constraint $\sum_in_i=N$ has to hold.


I added this in response to your comment:

The space the occupation number states naturally live in is referred to as Fock space. It can be written as

$\mathcal F^\pm = \bigoplus_{N=0}^\infty\mathcal H_N^\pm = \mathcal H_0 \oplus \mathcal H_1 \oplus \mathcal H_2^\pm \oplus \dots$

where $\mathcal H_N^\pm$ refers to the usually $N$ particle Hilbert spaces and the $\pm$ indicates whether we restict ourselves to symmetric or antisymmetric states, i.e. bosons and fermions respectively. It is convenient to work in Fock space because annihilation and creation operators will take you from one Hilbert space with fixed particle number to another.

Now, assuming that you mean 'dimension' when you talk about the 'size' of a Hilbert space, keep in mind that you have to obey the (anti-)symmetry constaint. This reduces the dimension of each of the $N$ particle subspaces.

As an example, consider a two state fermionic system, e.g. electrons with spin either $\downarrow$ ($i=1$) or $\uparrow$ ($i=2$). Due to the Pauli principle (i.e. as a consequence of the antisymmetry requirement) the only occupation numbers allowed are 0 and 1. $\mathcal H_0$ is one-dimensional as always, it contains only the vacuum state $|00\rangle$ (and multiples thereof, but we are interested in normalized linear independent states). The one particle Hilbert space is two-dimensional as the electron can be in either one of the states (and symmetry still doesn't matter since there are no two particles to swap). The two particle Hilbert space $\mathcal H_2^-$ has again only one linear independent state, $|11\rangle$, because Mr. Pauli doesn't allow us to populate any of the two states twice. Thus, the two-state fermionic Fock space is four-dimenional.

More generally, the $M$-state fermionic Fock space will be $\sum_{N=0}^M\binom{M}{N}=2^M$ dimensional (which follows from simple combinatorics) and bosonic Fock spaces are infinite dimensional because bosons can occupy the same state as many times as they want.

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  • $\begingroup$ Excellent. So, following this argument, are you saying that each occupation number state $| \ldots, n_i, \ldots\rangle$ is actually of size $\Pi_i(number~of~possible~values~of~quantum~number~i)$ and resides in a Hilbert space of that many dimensions? $\endgroup$ – eqb Sep 27 '13 at 19:17
  • $\begingroup$ Ah, of course. It would make no sense to talk about dimensions without antisymmetrization constraints. Thanks for that quick calculation of an M-state fermionic Fock space. $\endgroup$ – eqb Sep 27 '13 at 21:41
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Hmm, $n_i$ is simply the number of times that the $i^{th}$ quantum state is occupied. Jonas, your explanation is confusing. However, your statement that the sum of all $n_j$ must be $N$ for a system of $N$ particles is correct though. Also note that he used $i$ as an index of a specific particle in the system, you need to choose a different index to sum over for clarity.

eqg: Consider the case of no spin (for simplicity) where momentum is a good quantum number (plane wave basis) this $n_i$ will be the number of particles which have momentum $k_i$. For fermions, there can only be one particle in each state and for bosons there may be up to $N$ per state for a system of $N$ particles (all which make up the $0^{th}$ state are said to be a part of the BEC fraction). You can add spin to this trivially.

This being said, your second question about the size of the Hilbert space can only be answered if you supplement the question with the basis with which you choose to work and the type of particles in the system. And I am not sure what you mean by the "size" of a quantum state, I think this needs to be re-worded. If by size you mean the degeneracy of the state (number of ways to construct it) this will also depend upon the symmetry of the particles involved as well as the distinguishable nature (is the system all one type of particle? etc..).

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  • $\begingroup$ Ah, you see this is the statement I was trying to clarify -- you said "the number of times that the $i^{th}$ quantum state is occupied" without explaining how you're making up these states. What exactly is, for example, the zeroth state? I think, based on Jonas' description, it is some combination of the allowed values for the quantum numbers, and each index refers to a possible combination of these numbers in some defined order. Can you please point me to a good reference, and possibly explain what you mean by zeroth state? Is it the zero momentum state? (Again, index vs meaning) Thanks! $\endgroup$ – eqb Sep 27 '13 at 21:37
  • $\begingroup$ Yes, in my example it was the zero momentum (ground) state. These are made up differently for different systems and you always will start by defining what your good quantum numbers (your first example was of spin and momentum). I learned about second quantization in a graduate solid state course (I am a M.S. student working in a condensed matter theory group) using various textbooks but it was originally developed (if I am not mistaken) in the context of nuclear physics. You can simply Google 'second quantization pdf' and will find numerous write-ups that people have done. $\endgroup$ – codeAndStuff Sep 27 '13 at 21:51
  • $\begingroup$ You can read about constructing basis sets and when quantum numbers are good (or not) from a textbook such as Griffiths, Bransden and Joachain (my favorite basic QM text) or slightly more advanced texts such as Sakurai or Shankar. $\endgroup$ – codeAndStuff Sep 27 '13 at 22:01

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