What is the physical interpretation of second quantization?

One way that second quantization is motivated in an introductory text (QFT, Schwartz) is:

1. The general solution to a Lorentz-invariant field equation is an integral over plane waves (Fourier decomposition of the field).
2. Each term of the plane wave satisfies the harmonic oscillator equation.
3. Therefore, each Fourier component is interpreted as a harmonic oscillator in ordinary QM
4. The $n$'th energy level of each Fourier component is now interpreted as $n$ particles.

Everything in 1-3 looks like a sensible application of ordinary QM to a field. But how did 4 come about? What is the justification?

$\renewcommand{ket}[1]{|#1\rangle}$ Item #4 in your list is best thought of as the definition of the word "particle".

Consider a classical vibrating string. Suppose it has a set of normal modes denoted $\{A, B, C, \ldots\}$. To specify the state of the string, you write it as a Fourier series

$$f(x) = \sum_{\text{mode } n=\in \{A,B,C,\ldots \}} c_n [\text{shape of mode }n](x) \, .$$

In the typical case, $[\text{shape of mode }n](x)$ is something like $\sin(n\pi x / L)$ where $L$ is the length of the string. Anyway, the point is that you describe the string by enumerating its possible modes and specifying the amount by which each mode is excited by giving the $c_n$ values.

Suppose mode $A$ has one unit of energy, mode $C$ has two units of energy, and all the other modes have zero units of energy. There are two ways you could describe this situation.

Enumerate the modes (good)

The first option is like the Fourier series: you enumerate the modes and give each one's excitation level: $$|1\rangle_A, |2\rangle_C \, .$$ This is like second quantization; we describe the system by saying how many units of excitation are in each mode. In quantum mechanics, we use the word "particle" instead of the phrase "unit of excitation". This is mostly because historically we first understood "units of excitations" as things we could detect with a cloud chamber or Geiger counter. To be honest, I think "particle" is a pretty awful word given how we now understand things.

Label the units of excitation (bad)

The second way is to give each unit of excitation a label, and then say which mode each excitation is in. Let's call the excitations $x$, $y$, and $z$. Then in this notation the state of the system would be $$\ket{A}_x, \ket{C}_y, \ket{C}_z \, .$$ This is like first quantization. We've now labelled the "particles" and described the system by saying which state each particle is in. This is a terrible notation though, because the state we wrote is equivalent to this one $$\ket{A}_y, \ket{C}_x, \ket{C}_z \, .$$ In fact, any permutation of $x,y,z$ gives the same state of the string. This is why first quantization is terrible: particles are units of excitation so it is completely meaningless to give them labels.

Traditionally, this terribleness of notation was fixed by symmetrizing or anti-symmetrizing the first-quantized wave functions. This has the effect of removing the information we injected by labeling the particles, but you're way better off just not labeling them at all and using second quantization.

Meaning of 2$^{\text{nd}}$ quantization

Going back to the second quantization notation, our string was written $$\ket{1}_A, \ket{2}_C$$ meaning one excitation (particle) in $A$ and two excitations (particles) in $C$. Another way to write this could be to write a single ket and just list all the excitation numbers for each mode: $$\ket{\underbrace{1}_A \underbrace{0}_B \underbrace{2}_C \ldots}$$ which is how second quantization is actually written (without the underbraces). Then you can realize that $$\ket{000\ldots \underbrace{N}_{\text{mode }n} \ldots000} = \frac{(a_n^\dagger)^N}{\sqrt{N!}} \ket{0}$$ and just write all states as strings of creation operators acting on the vacuum state.

Anyway, the interpretation of second quantization is just that it's telling you how many excitation units ("quanta" or "particles") are in each mode in exactly the same way you would do it in classical physics.

In introductory quantum we learn about systems with a single particle, say, in a 1D box. That particle can be excited to a variety of different energy levels denoted $\ket{0}, \ket{1},\ldots$. We refer to this system as having "a single particle" regardless of which state the system is in. This may seem to run contrary to the statements made above in this answer in which we said that the various levels of excitation are referred to as zero, one, two particles. However, it's actually perfectly consistent as we now discuss.

Let's write the equivalent first and second quantized notations for the the single particle being in each state: $$\begin{array}{lllll} \text{second quantization:} & \ket{1,0,0,\ldots}, & \ket{0,1,0,\ldots}, & \ket{0,0,1,\ldots} & \ldots \\ \text{first quantization:} &\ket{0}, &\ket{1}, &\ket{2}, & \ldots \end{array}$$ Although it's not at all obvious in the first quantized notation, the second quantized notation makes clear that the various first quantized states involve the particle occupying different modes of the system. This is actually pretty obvious if we think about the wave functions associated to the various states, e.g. using first quantized notation for a box of length $L$ \begin{align} \langle x | 0 \rangle & \propto \sin(\pi x / L) \\ \langle x | 1 \rangle & \propto \sin(2\pi x / L) \, . \end{align} These are just like the various modes of the vibrating string. Anyway, calling the first quantized states $\ket{0}$, $\ket{1}$ etc. "single particle states" is consistent with the idea that a particle is a unit of excitation of a mode because each of these states has one total excitation when you sum over all the modes. This is really obvious in second quantized notation.

• I should have explained why I think #4 needs a justification. In ordinary QM, let's say the single-particle harmonic oscillator, each possible energy level is associated with that single particle. We don't say that exciting the particle to the $n$-th energy level turns it into $n$ particles. What accounts for the appearance of this interpretation in QFT? – yjc Jun 1 '15 at 7:55
• @yjc: Together with my answer given, your comment explains why one says that in QFT, elementary particles are the elementary excitations of the quantum field. Indeed, this is a much better notion of particles than the semiclassical picture generally considered. – Arnold Neumaier Jun 1 '15 at 10:18
• @yjc For what it's worth, we do actually identify the nth energy level of a harmonic oscillator (i.e. a massive particle in a harmonic trap) with a state containing n quanta of vibrational energy aka n "phonons". – Mark Mitchison Jun 1 '15 at 17:11

In the statistical mechanics of the grand canonical ensemble, one needs to allow for superpositions and mixtures of of states with different particle number. Thus one is naturally led to considering the tensor product of the $N$-particle spaces with arbitrary $N$. It turns out (and is very relevant for nonequilibrium statistical mechanics) that one can reinterpret the resulting any-number-of-particles quantum mechanics as a nonrelativistic field theory, in which the number operator is defined to have the eigenvalue $N$ on $N$-particle space. (If one considers a single Fourier mode, this explains your 4.)

The resulting field formalism is called the second quantization (of the first quantized 1-particle space). You can read about this e.g., in the appendix of Reichl's statistical physics book.

If one replaces the 1-particle Schroedinger equation by the Klein-Gordon or Dirac equation one gets (after normal ordering) the relativistic version.

• I don't understand this. Second quantization is useful whenever you have more than one particle. You don't need an ensemble for it to be useful. – DanielSank Jun 1 '15 at 14:52
• My statement was just that it is necessary when particle number is indefinite. In ordinary QM, the only relevant situation of indefinite particle number that I know of is the multicanonical ensemble, and one relevant case is enough to motivate the concept (which was the question of the OP). But I agree that it may be useful in other circumstances as well, though it is overkill in case $N$ is fixed. – Arnold Neumaier Jun 1 '15 at 18:40
• I definitely think second quantization is important even with fixed $N$. See my answer. – DanielSank Jun 1 '15 at 19:27
• @DanielSank: To write a general $3$-particle antisymmetric wave function $\psi(x_1,x_2,x_3)$ for the analysis of a 3-atom molecule in terms of creation and annihilation operators is already awkward. People use creation and annihilation operators for fixed particle number, e.g., in coupled cluster quantum chemistry computations, but there the meaning is quite different, describing excitations of a few quasiparticles although the total number of particles is fixed. – Arnold Neumaier Jun 2 '15 at 12:30

The term is practical, but in reality, there is only one quantization. This was to stretch it a little, as I then consider the Klein-Gordon equation as classical. That is motivated by that I consider Maxwell's equations as classical. Both equations are on the same level and so I consider Klein-Gordon to be classical. The classical normal mode vibrations of Maxwell or Klein-Gordon fields are standing waves. The individual vibrations are quantized as one-dimensional harmonic oscillators.

The second quantization of fermions is only a way to manage the antisymmetry condition. Here the first quantization (as given by Heisenberg, Schrödinger, and Dirac) is the real quantization.