The Story of Nth Quantization

March 14, 2016

This is a story that leads up to a puzzle. I wrote it with some help from Toby Bartels.

Chapter 0: Zeroth Quantization

In the beginning there is the empty set. But since we're doing quantum theory, we'll hit the empty set with the functor $$F: \mathrm{Set} \to \mathrm{Hilb}$$ that takes any set and spits out a Hilbert space having that set as an orthonormal basis. This functor carries us out of the world of set theory, into the world of quantum theory...

[insert spooky sound effects here]

... so now we have a Hilbert space: the 0-dimensional Hilbert space. Let's call this Hilbert space "0" for short. The only vector in this Hilbert space is the zero vector. This is the Hilbert space of the system with no states at all. The empty set forms a basis of states.

So far, life is dull. But now comes the fun....

Chapter 1: First Quantization

As Edward Nelson said, "first quantization is a mystery, but second quantization is a functor."

This story is not a mystery, but rather a creation myth! So let us apply the second quantization functor $$K: \mathrm{Hilb} \to \mathrm{Hilb}$$ to our Hilbert space 0, and see what happens. This functor takes any Hilbert space $$H$$ and spits out a new Hilbert space $$K(H)$$, called the 'Fock space' of $$H$$. If $$H$$ has an orthonormal basis of states of the form $$|e_i \rangle$$ where $$i$$ ranges over some set $$S$$, then by definition $$K(H)$$ has an orthonormal basis of states of the form $$|e_{i_1} \cdots e_{i_n} \rangle$$ where $$n$$ is any natural number, and the order of the $$e_{i_j}$$ does not matter.

If we apply this functor to 0, we get $$K(0) = \mathbb{C}$$, the complex numbers. This the Hilbert space of a system with no degrees of freedom, and thus only one state (modulo phase). Michael Weiss calls this system a 'noton', since it's a nice watered-down version of a photon, suitable for learning about quantum field theory. However, we could also call it a 'quantum', for reasons soon to be clear.

$$K(0)$$ has a basis of states consisting of just $$|\;\rangle$$ that is, the 'empty ket'—the ket consisting of only the empty set! We also abbreviate the empty ket as "1", just as in set theory one uses "1" to stand for the set whose only element is the empty set.

One state is better than none, but life is still not very interesting.

Chapter 2: Second Quantization

Let us apply the functor $$K$$ to the Hilbert space $$K(0) = \mathbb{C}$$.

$$K(K(0)) = K(\mathbb{C})$$ is the Hilbert space of an arbitrary finite collection of indistinguishable quanta. A basis of states is given by $$\begin{array}{l} | \; \rangle \\ || \; \rangle\rangle \\ || \; \rangle,\; | \; \rangle\rangle \\ || \; \rangle,\; | \; \rangle, \; | \; \rangle \rangle \\ || \; \rangle,\; | \; \rangle, \; | \; \rangle, \; | \; \rangle \rangle \\ \end{array}$$ and so on, where the $$n$$th of these states contains $$n$$ quanta. If this seems weird, use the abbreviation $$1$$ for the empty ket; then these states look like this: $$\begin{array}{l} | \; \rangle \\ |1 \rangle \\ |1, 1 \rangle \\ |1, 1, 1 \rangle \\ |1, 1, 1, 1 \rangle \\ \end{array}$$ and so on. If this still looks weird, use '$$n$$' to stand for the state consisting of $$n$$ quanta; then these states look like this: $$\begin{array}{l} |0 \; \rangle \\ |1 \rangle \\ |2 \rangle \\ |3 \rangle \\ |4 \rangle \\ \end{array}$$ and so on, one for each natural number. It should now be clear that we are dealing with the Hilbert space of the quantized harmonic oscillator! The $$(|n\rangle$$ are the eigenstates of an operator known as the 'number operator', defined by $$N|n\rangle = n|n\rangle$$ This operator is actually $$dK(1_\mathbb{C})$$, where $$1_\mathbb{C}: \mathbb{C} \to \mathbb{C}$$ is the identity operator on the complex numbers. Let me explain! The functor $$K$$ sends unitary operators on any Hilbert space $$H$$ to unitary operators on $$K(H)$$; in fact, it is a homomorphism of Lie groups. Differentiating, we get a Lie algebra homomorphism $$dK$$, which sends self-adjoint operators on $$H$$ to self-adjoint operators on $$K(H)$$. (Mathematicians prefer skew-adjoint operators here, but we're being physicists here, so we'll throw in a factor of $$i$$ and work with self-adjoint operators instead!)

The operator $$N$$ also goes by the name of the 'harmonic oscillator Hamiltonian'. So, some interesting physics is beginning to emerge!

Chapter 3: Third Quantization

Here's a basic rule of thumb: when something was fun, try it again.

$$K(K(K(0))) = K(K(\mathbb{C}))$$ is the Hilbert space of the right-moving modes of a string! A basis of states is given by finite multisets of natural numbers, e.g. $$| n_1, \dots, n_k \rangle$$ where, as per the usual definition of 'multiset', the the order doesn't matter, and a number can appear more than once. In fact this is how the whole story works: the basis vectors for the Hilbert space in each chapter consists of finite multisets of basis vectors in the previous chapter.

So, here's an example of an orthonormal basis vector for $$K(K(K(0)))$$: $$| 3, 1, 2 \rangle$$ This is really short for $$|| 1, 1 , 1\rangle, \; |1\rangle, \; |1 , 1\rangle \rangle$$ And this, in turn, is really short for $$|||\; \rangle,\; |\; \rangle , \; |\; \rangle\rangle, \quad ||\; \rangle\rangle, \quad ||\; \rangle ,\; |\; \rangle \rangle \rangle$$ We can think of this state as the state of a massless scalar field theory on a cylindrical spacetime in which there is one particle of momentum 3, one of momentum 1 and one of momentum 2. Even better, the Hamiltonian for this theory is $$K(N)$$, where $$N$$ was the number operator of the previous chapter.

Yes, it's true! If we take the harmonic oscillator Hamiltonian $$N = dK(1)$$ on $$K(K(0))$$ and second quantize it, we get a Hamiltonian $$dK(N)$$ on $$K(K(K(0)))$$ which is precisely the usual Hamiltonian for a massless scalar quantum field on a cylindrical spacetime! The only odd part is that we're getting only states containing particles of nonnegative momentum. But this is not so bad: physicists study this sometime. In physics lingo, we're getting a field theory with only "right-moving" particles. We'd have to tensor $$K(K(K(0)))$$ with another copy of itself to get left-moving particles as well.

Now, a massless scalar field theory on a cylindrical spacetime is basically the same as a string in 1-dimensional spacetime in covariant gauge, but here we are considering only the right-moving modes. (Or, if we use lightcone gauge, I guess we can alternatively think of this system as a string in (2+1)-dimensional spacetime!)

So, starting with the system with no states and repeatedly applying the second quantization functor, we have gotten to string theory. It would be crazy to stop now....

Chapter 4: Fourth Quantization

$$K(K(K(K(0))))$$ is... well, this is the puzzle. One easy answer is that it's the Hilbert space for a second quantized string, i.e. the system consisting of an arbitrary collection of indistinguishable strings. That's actually quite nice. But you should try to find an even better answer....

Appendix

In response to my puzzle, Toby ventured this answer:

"The Hilbert space for a 2-brane?"

This sounds interetsing. But as you can see, I'm a little bit confused about the dimensions even in Chapter 3. I hope I got it right, but there are two choices, and I don't know which interpretation is best. A harmonic oscillator is a particle in 1+1-dimensional spacetime... so maybe Chapter 3 should be about a string in 2+1-dimensional spacetime. And then....

Anyway, I urge everyone to figure out if there's a cool pattern going on here. I had a different, less exciting answer in mind — but there should be something more exciting, too. Personally I do not see the connection to 2-branes, but that doesn't mean it doesn't exist!

A different sort of idea, not exactly an answer to my mathematical puzzle, is here:

He uses third quantization to describe the creation and annihilation of universes, and fourth quantization to describe the creation and annihilation of multiverses.