# What is the physical interpretation of second quantization?

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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?

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user yjc
retagged Jun 8, 2015
Please read this other Physics SE answer. I think it will explain a lot of what you want to know.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user DanielSank
Second quantization is just a way to construct, given a one particle quantum space, a space that can describe all at once states with an arbitrary probability of having $n$ of those identical particles, for any $n\in\mathbb{N}$. And that is the correct definition. The "heuristic" explanation you refer to may help to understand the structure of the so-called creation/annihilation operators associated to second quantization. However, it makes the straightforward interpretation in terms of number of particles more obscure in my opinion.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user yuggib

I take it to be less confusing to think of "2nd quantization" as quantization of a classical field theory, not as quantization of a quantization of a classical particle theory. Taking that view, interpretation of a quantized field theory is not so much different from quantization of a classical theory of some other kind, in contrast to trying to bend one's head around 2nd quantization, where we already have quite a lot of difficulty interpreting 1st quantization. One has to worry about dealing with an infinite number of DoFs and choosing a basis, for which the (improper) 4-momentum basis diagonalizes the commutation relations (at least for the free field, because of translation invariance), but then raising, lowering, and number operators provide a discrete structure. Not an answer, and not true to the historical development, more a suggested redirection, so I place it as a comment. 2nd quantization is less talked about now than it used to be, right?

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$\renewcommand{ket}{|#1\rangle}$ 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 work "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.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user DanielSank
answered Jun 1, 2015 by (60 points)
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?

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user yjc
@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.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user Arnold Neumaier
+1 Fantastic. Probably the best explanation of the second quantization that I've read.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user WetSavannaAnimal aka Rod Vance
@WetSavannaAnimalakaRodVance thanks much. If you like this one also see this post which is sort of the same thing but worded differently.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user DanielSank
@yjc It's because those excitations are massless that we don't initially think of them as particles. However, photons (which are also excitations of a harmonic oscillator mode) are massless and still we think of them as particles.

This post imported from StackExchange Physics at 2015-06-08 13:22 (UTC), posted by SE-user DanielSank
"We actually refer to the different excitation levels in our microwave resonators as having one, two... "photons" I believe this is because it is experimentally verifiable that the system emits a single photon when it transitions to a lower adjacent energy level. In that case, it seems like there is nothing radically new in the quantization of fields, compared to ordinary QM. The text says that these interpretations are unique to second quantization, and that may have misled me.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user yjc
@yjc: Yeah, quantum mechanics of continuous modes is not really all that different from quantum mechanics of localized ones. I'd agree with that.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user DanielSank
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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.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user Arnold Neumaier
answered Jun 1, 2015 by (14,437 points)
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.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user DanielSank
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.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user Arnold Neumaier
I definitely think second quantization is important even with fixed $N$. See my answer.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user DanielSank
@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.

This post imported from StackExchange Physics at 2015-06-08 13:23 (UTC), posted by SE-user Arnold Neumaier

@ArnoldNeumaier One always quantizes quasi-particles - collective motions of particles: normal modes of compound systems. Even for SHO one can have uncertain number of excitations (quasi-particles) if the energy of a given state ψ is uncertain. It has nothing to do with two constituent particles in SHO.

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I do not like the old-fashioned textbook approach which says: Abrakadabra! - and the Fourier coefficients are creation and annihilation operators of relativistic particles! (aka step 4 in OP) It usually fails to properly argue why the requirements and relations we imposed on the "classical field" $\phi(x^\mu)$ are necessary or even make sense in a many-particle context.

I am convinced that the only truly consistent approach is the "from-below" approach, building the quantum fields by many-particle considerations as presented e.g. by Weinberg - Quantum theory of fields, vol. 1. The basic outline is the following.

We do not care about relativistic fields at all, because we know that in fact we are faced with particles. For starters, we will only be interested in single particles with a definite momentum $p$ because the respective quantum states transform as  $e^{-ipa}$ under an $a$ translation and we can characterize them purely by the homogeneous Lorentz group representations. We find that the possible unitary representations break up naturally into several classes of particles including massive particles, massless particles, tachyons, and the trivial vacuum ($p=0$).

After properly documenting the properties of single-particle definite-momentum states and the extension to Fock many-particle formalism, we investigate asymptotic scattering theory via the properties of the $S$-matrix.

The $S$-matrix yields Lorentz-invariant results and satisfies the cluster decomposition principle (faraway experiments yield independent results) only if (the "only" if has not been proven rigorously, I believe) the interaction of the particles can be written as

$$V(t) = \int d^3 x \mathcal{H} (x,t)$$

where the Hamiltonian $\mathcal{H} (x,t)$ is a Lorentz-invariant scalar built from single-particle creation and annihilation operators which conserve momentum (i.e. the sum of annihilated momenta in each term must be equal to the sum of created momenta).

Imposing other causality and unitarity conditions, we are gradually able to construct the Hamiltonian as a function of the so-called causal fields $\phi$ and their formal derivatives $\partial_\mu \phi$. The causal fields are sums (integrals) of creation and annihilation operators which look exactly as if the operators  $a_p, a^\dagger_p, b_p, b^\dagger_p$ were Fourier coefficients of a classical field $\phi$.

This way, one never sees the "classical fields" as more as an interesting structure arising through many-particle relativistic quantum mechanics. Another consistent way to construct QFT is to see the causal field $\phi(x^\mu)$ for what they are from the very beginning: as a single-particle definite-position ($x^\mu$) creation operator. However, it is then harder to explain what $-i \partial_\nu \phi (x^\mu)$ is and what's the fuss about the wave equation. As far as I can imagine, one has to pass to the definite-momentum operators $a_p$ and show the wave equation is just the $p^2=-m^2$ relation. But then you are basically back at Weinberg.

answered Jun 9, 2015 by (1,635 points)
edited Jun 9, 2015

If you mention tachyons, you should also mention the unitary irreps with  (in the $+ - - -$ metric) $p^2=m^2, p_0<0$. Of course both are ruled out by causality requirements.

@ArnoldNeumaier : But $p^2=\mu^2$ are exactly tachyons and $p_0<0$ are just variants of the mentioned travelling back in time.

I do not like the old-fashioned textbook approach which says: Abrakadabra! - and the Fourier coefficients are creation and annihilation operators of relativistic particles!

Instead you propose an approach that says : Abrakadabra! - and posits the creation and annihilation operators of relativistic particles ex nihilo. Weinberg introduces them in (5.1.4/5+15/16) as something that reduces to Fourier transforms in the case of scalar fields.

We do not care about relativistic fields at all, because we know that in fact we are faced with particles.

But Weinberg cares, as he doesn't assume that in fact we are faced with particles - which is a very dubious assumption in view of the fact that it is very unclear what defines a quantum particle.

The fundamental theory of matter is not without reason called quantum field theory, and not quantum particle theory. There are many books titled the former, and not a single one titled the latter. Thus we do not care about quantum particles at all, because we know that in fact we are faced with quantum fields

The existence of particles is an experimental fact; a physical theory indeed needs some elements given "from outside", by experiment. We have never seen massive fields but we know an abundance of massive particles. We know that these particles behave in a quantum way, their dispersion relations are relativistic and they can be destroyed or created in reactions, hence we build a relativistic quantum many-particle theory. The fact is that it is actually believed these particles can have internal structure and be in no way fundamental, but starting from the phenomenological postulate of a particle makes our conclusion as robust as the phenomenological fact itself.

As we are already discussing Weinberg, I recommend the first chapter Historical Introduction, it clarifies why QFT is best built from below and it also becomes clear what historical developments lead to the conventional name "Quantum field theory". (As for creation and annihilation operators, they are in fact defined in section 4.2 Creation and annihilation operators.)

The only facts about existence in the subatomic domain are that of clicks or marks in detectors and tracks in bubble chambers (or other records of collision experiments).

Whether these are interpreted as the results of interactions of the detectors with a quantum field (suggested by quantum field theory) or quantum particles (suggested by the historical development) is a matter of interpretation, not a fact. Most of what are called quantum particles are just indirectly inferred as resonances based upon massive data analysis.

We see massless (electromagnetic) fields daily in everything we see. And we experience massive quantum fields daily in everything we touch; classical field theory (hydrodynamics or elasticity theory) is nothing else than the theory of expectation values of quantum fields in the classical (macroscopic) limit.

I know Weinberg's book quite well. In the first chapter that you recommend, he says that creation and annihilation operators were introduced in 1926 - long before Wigner came along and classified the irreducible unitary representations of the Poincare group. On p. 15 (1995 edition) he mentions that the photon was known as a field long before it was interpreted as a particle, and on p.16 he introduces creation and annihilation operators through standard second quantization of the electromagnetic field. - Nothing like the story you want to make us believe.

Well, yes, the parts you cite clarify why the theory is called "field" for historical reasons - because it was historically developed that way. I am referring to p.2 (also 1995)

The reason that our field theories work so well is not that they are fundamental truths, but that any relativistic quantum theory will look like a field theory when applied to particles at sufficiently low energy. On this basis, if we want to know why quantum field theories are the way they are, we have to start with particles.

That the history had it's own story to tell does not mean that following it is the truly consistent conceptual structure of the theory. On the contrary, I believe that in this case sticking to the historical development is wrong pedagogically and often even fundamentally.

And I believe that making particles fundamental and fields secondary is wrong both pedagogically and fundamentally.

Quantum particles are very weird objects with properties far away from what deserves to be called particle, while quantum fields behave in every respect like fields - even when they show particle (i.e., high local density) effects.

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