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

+ 2 like - 0 dislike
671 views

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

asked Mar 24, 2015 in Theoretical Physics by Fritz (10 points) [ revision history ]
edited May 17, 2015 by Ron Maimon
"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".

2 Answers

+ 5 like - 0 dislike

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
answered Mar 24, 2015 by yuggib (360 points) [ no revision ]
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Thanks! I have a bit trouble with the abstractness, though. Just checking if I get this right: the functor takes a Hilbert space and maps it to another one. So in this case, the functor is an endomorphism on the "set" of Hilbert spaces. For some reason, we shouldn't call the "set" of Hilbert spaces a set but we call such a thing a category. And maps between categories are called functors.

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user Fritz
Yes, you almost got it right. The set of all Hilbert spaces cannot exist as a set (such as it cannot exist the set of all sets), so these "generalized" collections are called classes. A category is a very general mathematical construct that has objects (that are members of a class) and morphisms (that are maps between objects) that behaves in a suitable manner. A functor is a map between categories, that again satisfies suitable assumptions. In the case at hand, the category is that of Hilbert spaces as objects, and two type of morphisms: the self-adjoint operators and the unitary operators...

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib
The second quantization functor takes the category of Hilbert spaces with s.-a. and unitary operators into itself, associating to a given Hilbert space (the object) the corresponding (anti)symmetric Fock space; to a given self-adjoint operator (morphism) $H$ its second quantization $d\Gamma(H)$ (that is again self-adjoint) and finally to a given unitary operator $U$ its second quantization $\Gamma(U)$ (again unitary). So you see it makes sense as a functor of the category into itself (and actually satisfies all the suitable axioms)

This post imported from StackExchange Physics at 2015-05-13 18:54 (UTC), posted by SE-user yuggib
Thanks for your comments! I think I basically get it now. Maybe one more thing: do you know a paper or a textbook where second quantization is explained this way?

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

+ 1 like - 1 dislike

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.

answered May 17, 2015 by Vladimir Kalitvianski (22 points) [ revision history ]
edited May 22, 2015 by Vladimir Kalitvianski

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.

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