Is there something wrong with interpreting second quantization as "quantizing twice"?

Question

Second quantization is sometimes considered to be a bad name, because a single quantization is enough. For electrons, we can either start from a many body viewpoint and introduce field operators or we can start from an unphysical "classical" spin-1/2 field and quantize this. In both cases, there is only a single quantization.

My question is, if there's something wrong with the viewpoint that we are actually quantizing two times. First, we introduce the canonical commutation relations and then we promote the coefficients of the wavefunction to operators.

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user Fritz

Comments

"First quantization is a mystery, but second quantization is a functor"--Edward Nelson.

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib

---

Related: Baez' take on nth quantization.

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user ACuriousMind

---

There is nothing wrong with second quantization, except perhaps the name since second quantization works without any first quantization. For example, the (first) quantization of the classical Maxwell equations is identical to the second quantization of the (less classical) phase space of a single photon.

---

Don't say "quantizing two times", as this leads one to think you are referring to the multiple-time theory of Dirac, or other theories with two time variables. Edited to "twice".

Answer 1

There is a lot of people that do not like the name second quantization, mostly because the second quantization is often introduced in a not so clear way, at least in my opinion.

Second quantization has nothing wrong, if you see it as a precisely defined mathematical object: it is a functor between Hilbert spaces, that associates to the original one-particle space the suitable (anti)symmetric Fock space, to self adjoint operators $H$ their second quantization $d\Gamma(H)$ and to unitary operators $e^{−itH}$ the second quantization $\Gamma(e^{−itH})=e^{−itd\Gamma(H)}$.

The physical interpretation is quite simple; to a single particle Hilbert space $\mathscr{H}$, it is associated an Hilbert space that contains information about all the spaces with an arbitrary number of particles (the Fock space); to an operator $H$ that acts on a single particle it can be associated an operator [$d\Gamma(H)$] that acts on any space with $n$ particles as a sum of the action of H on each particle, or an operator [$\Gamma(H)$] that acts as the $n$-product of $H$, each one acting on a different particle.

With such second quantization functor however, you do not give meaning to important operators of the theory, e.g. the creation/annihilation operators; for they are not the second quantization of anything (the second quantization functor gives, roughly speaking, only operators on the Fock space that preserve the number of particles).

To introduce annihilation/creation operators one has to use "standard" quantization. The meaning is analogous to the quantum mechanical one, simply starting from a classical phase space that is infinite dimensional. So to functionals on that classical phase space are associated operators on a bigger space (the Fock space) almost exactly as in the usual quantization of finite dimensional phase spaces (with problems of ordering, domains of definition and so on).

But if second quantization can be defined mathematically as a functor (without entering too much into details), there is no satisfactory functorialization of the first quantization procedure, even with finite dimensional phase spaces. This is because it is not possible, roughly speaking, to define a quantization procedure that is coherent, for each phase space function, both with the symplectic structure of the classical space (Poisson brackets) and the irreducibility requirements of the Weyl commutation relations. This explains the quote of Edward Nelson that you find in the comments (this does not mean, however, that the quantization procedure is not well-defined mathematically, simply it is not an elegant categorical notion such as a functor).

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib

Comments

Actually, there are two distinct second quantization functors, as one can map a Hilbert space to a bosonic or a fermionic Fock space. 

Also, the definition of c/a operators does not need a phase space. The creation operator $a^*(f)$ with $f$ in the 1-particle space simply maps $\psi$ to the symmetrization of $\psi\otimes f$, and the annihilation operators are their adjoints. 

---

@ArnoldNeumaier Historically (see e.g. the papers of Segal and Cook in the 50s, from where the functorial picture actually originates), the functor $\Gamma$ from the category of Hilbert spaces with unitary maps as morphisms into itself is defined before symmetrization. In particular $\Gamma (\mathscr{H})=\bigoplus_{n=0}^\infty\mathscr{H}_n$, where $\mathscr{H}$ is a Hilbert space, $\mathscr{H}_0=\mathbb{C}$ and $\mathscr{H}_n=(\mathscr{H})^{\otimes n}$; $\Gamma(U)$, where $U$ is unitary on $\mathscr{H}$, is the operator that acts as the identity on $\mathbb{C}$ and as $(U)^{\otimes n}$ on $\mathscr{H}_n$. The symmetrization is then introduced a posteriori for obvious physical reasons (and also the associated c/a operators).

I agree that the c/a operators can be defined given the structure of the (anti)-symmetric Fock space. Nevertheless, they are also seen as the "canonical variables" of those spaces, and therefore the point of view of quantizing a classical infinite dimensional phase space is quite natural, and actually useful for many considerations. In addition, from an algebraic point of view, it is slightly more general to start with a one-particle space that is a real Hilbert space with a non-degenerate symplectic structure than with a complex Hilbert space (but it is not so important).

Finally, just to be pedantic, in your definition of the creation operator a $\sqrt{n}$ factor is missing (and it is not irrelevant).

---

in your definition of the creation operator a n‾√ factor is missing (and it is not irrelevant)

Ah, I don't like these factors, and put instead a $1/n!$ into the inner product. This is equivalent to the standard usage but makes most formulas much nicer.

---

@ArnoldNeumaier Ah ok, you use the Friedrichs notation. It is a nice one also for me, but unfortunately it has not been adopted by many in the literature, at least the mathematical physics literature.

---

Actually, I dind't remember correctly - it is the annihilation operator that then has no factor. My preferred conventions are those of Chapter 20.5 of my online book.

---

Well, not exactly in this way; in the sense that the functorial setting has not been much exploited (probably because it may not give additional information). Anyways, a precise mathematical description of second quantization (meaning broadly theory of operators in Fock spaces, and the definition of $d\Gamma$ and $\Gamma$ morphisms) can be found in the following references, in increasing level of difficulty (at least for me): Reed-Simon vol.2 chap X.7; Cook classical article of the 50s...

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib

---

Derezinski and Gérard "Mathematics of quantization and quantum fields"; Baez, Segal, Zhou "Introduction to algebraic and constructive quantum field theory". These are all mathematical references, because the approach I described is preferred by mathematicians, while physicists prefer a more hands-on and direct approach generalizing the concept of harmonic oscillators and so on.

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib

Answer 2

My question is, if there's something wrong with the viewpoint that we are actually quantizing two times. First, we introduce the canonical commutation relations and then we promote the coefficients of the wavefunction to operators.

Introducing commutation relations is already quantization - promoting something "classical" to operators. Either one promotes $x$ and $p$ to operators, or one promotes the amplitudes $a$ and $a^*$. These quantizations are related with a variable change, nothing else. So it is better to call "the second quantization" "a second way of quantization". In the second way of quantization the wave functions appear naturally at $a$ and $a^+$. The wave functions are not operators there!

P.S. Downvotes on my answer mean that there are indeed two quantizations - one after another.

Comments

When you Fourier decompose for example a scaler field $\phi(x,t)$ and promote the coefficients to creation and annhihilation operators $\hat{a}^{\dagger}$ and $\hat{a}$, the r.h.s contains operators, so the l.h.s becomes a (field) operator $\hat{\phi}(x,t)$ too and it is not a wave function. The last sentence in this answer makes no sense to me.

---

I meant the wave functions at $a$ and $a^+$, the plane waves (harmonics), for example. The whole construction $\hat{\phi}$ is an operator, of course, due to $a$ and $a^+$.

Instead of promoting $a$ and $a^*$, you can promote $\phi$ and $\dot{\phi}$ to the operators and make the Fourier decomposition of your operator $\hat{\phi}$. It will contain $a$ and $a^+$ automatically.