## Where does the Hilbert space come from?

Continuing the discussion from last time, today we can put some of the pieces of the puzzle together. It is helpful to switch the discussion direction from classical to quantum and start for a moment from the quantum mechanics side to see where we want to arrive.

Why do we use a Hilbert space in quantum mechanics? This is a big topic and we cannot cover it in one (or even several posts). Right away we will restrict ourselves to the finite dimensional degrees of freedom case, thus excluding the field theory considerations and avoiding the issues raised by Haag's theorem, or the breakdown of the Stone-von Neumann uniqueness theorem. We will also skip the treatment of unbounded operators which require the theory of rigged Hilbert spaces and we will stick with boring but well behaved bounded operators.

For a bounded operator T on a Hilbert space it is easy to prove that $$||T^{\dagger} T|| = {||T||}^2$$ as follows:

$${||T \Phi||}^2 = \langle T \Phi, T \Phi\rangle = \langle T^{\dagger} T \Phi, \Phi\rangle \leq ||T^{\dagger} T\Phi || ||\Phi|| \leq ||T^{\dagger} T || {||\Phi||}^2$$
therefore
$${||T||}^2 \leq ||T^{\dagger} T ||$$
and since
$$||T^{\dagger} T || \leq ||T^{\dagger}|| ||T|| = {||T||}^2$$
we have:
$${||T||}^2 \leq ||T^{\dagger} T || \leq {||T||}^2$$

An algebra of bounded operators on a Hilbert space is the prototypical example of a C* algebra. A remarkable fact is the correspondence between states and representations of C* algebra given by the GNS construction. Here a representation is a linear map from the elements of the C* algebra to bounded operators on some Hilbert space.

From categorical considerations one can obtain a C* algebra without the norm axioms. To distinguish math from physics one needs to be able to make experimental predictions and this is where the states enter the picture. A state on a C* algebra gives rise to a representation of the algebra as bounded linear operators on some Hilbert space and this is how Hilbert spaces are introduced. The key ingredient for this to work is the C* norm condition: $$||T^{\dagger} T|| = {||T||}^2$$. However, this norm is unique and is given by the spectral radius - an algebraic concept! So there is hope we can arrive at quantum mechanics using only algebraic methods. Now we will show how.

Coming back to the quantization discussion from the prior post, what we need to achieve is a prescription which constructs operators on a Hilbert space from functions on the phase space (also known as the cotangent bundle). Even better we should be able to start from either a Kahler, symplectic, or Poisson manifold.

We can start with the simplest case where we replace the position $$q$$ and the momenta $$p$$ with the operators: $$x$$ and $$\frac{h}{i}\frac{\partial}{\partial x}$$ in any observable $$f(p, q)$$ provided $$f$$ contains no products $$pq$$ because the position and momenta operators in the Hilbert space do not commute and the order of the operators is ambiguous.

The next level of sophistication is Weyl quantization procedure and the details can be found here. (Please excuse me for skipping typesetting it in LaTeX.) Weyl quantization tends to preserve well symmetry properties, but a better quantization prescription is Berezin quantization which work on all Kahler manifolds when positivity is guaranteed by the Kodaira embedding theorem.

 Erich Kahler

A Kahler manifold is a truly outstanding mathematical object where three concepts meet:

• a metric structure
• a symplectic structure
• a complex structure
and any two define the third one. The main example is the complex projective space (endowed with the Fubini-Study metric) which is essential for quantum mechanics. It is very enlightening to see how it all works out in quantum mechanics and I'll attempt to show it below.

In classical and quantum mechanics there are two products, one symmetric $$\sigma$$ and one anti-symmetric $$\alpha$$ corresponding to observables and generators as follows:

Observables: $$\sigma$$ = regular function multiplication on phase space OR Jordan product
Generators: $$\alpha$$ = Poisson bracket OR commutator

There is also a 1-to-1 map $$J$$ between observable and generators called dynamic correspondence where $$J^2 = 0$$ for classical mechanics and $$J^2 = -1$$ for quantum mechanics. This map corresponds to Noether's theorem.

Composing two physical systems 1 and 2 gives rise to the following fundamental composition relationship:

$$\sigma_{12} = \sigma_1 \otimes \sigma_2 + J^2 \frac{\hbar^2}{4}\alpha_1 \otimes \alpha_2$$
$$\alpha_{12} = \sigma_1 \otimes \alpha_2 + \alpha_1 \otimes \sigma_2$$

and so the symmetry and anti-symmetry of the products is preserved.

Now we want to deform the Poisson bracket and regular function multiplication of classical mechanics which respects the composition with $$J^2 = 0$$ into two products which respect $$J^2 = -1$$. We can do this term by term in powers of $$\hbar$$ preserving associativity at each step. This is the essence of deformation quantization.

Without ado, here is the solution given by Moyal sine and cosine brackets in terms of the Poisson bracket $$\{ , \}$$ in the simplest case of a flat space:

$$\alpha = \frac{2}{\hbar} sin (\frac{\hbar}{2} \{ , \})$$
$$\sigma = cos (\frac{\hbar}{2} \{ , \})$$

The star product is then $$\star= \sigma+ J\frac{\hbar}{2} \alpha$$ and we arrived at quantum mechanics in phase space.

First a note: I demanded earlier to preserve associativity at each power of $$\hbar$$. This is a physical requirement to be able to compose experiments sequentially and not care where we draw the boundaries between them. But this has a very interesting consequence: we have freedom of pick how we carry out the quantization at each power of $$\hbar$$ step and this makes the subject of quantization non-trivial. In particular it turns out that the equivalence classes of star products on symplectic manifolds are in 1-to-1 correspondence with the second de Rham cohomology $$H^2_{dR} (M)$$!

Second, we can see where the inner product is coming from. From the Moyal sine bracket we extract a symplectic form $$\omega^{IJ}$$ and construct it's inverse $$\Omega_{IJ}$$. So we have one of the three structures of a Kahler manifold: the symplectic structure. But we also have the complex structure as well because we have $$J^2 = -1$$. It can be shown that $$J$$ is actually a tensor or rank (1,1): $$J = J^{I }_{ J}$$ and from this we get a metric tensor $$g_{IJ}$$:

$$g_{IJ} = \Omega_{IK} J^{K}_{ J}$$

The complex inner product is defined now by: $$g+ \sqrt{-1}\Omega$$:

$$\langle X, Y \rangle = X^{T} g Y + i X^T \Omega Y$$

where X and Y are column vectors : $$q_1, q_2,...q_n, p_1, p_2, ..., p_n$$

Time evolution preserves $$J$$ and $$\Omega$$, meaning they preserve the metric structure by preserving a normalization constraint:

$$\langle g \rangle - 1= X^I g_{IJ} X^J - 1 = 0$$

The constraint Hamiltonian motion which preserves the metric structure is nothing but the Schrodinger equation is disguise!

I do not want to create the impression that this is all as simple as this. I only discussed the flat $$R^{2n}$$ case above. There are many subtle and hard problems, as well as open questions. As an example, there are Poisson manifolds which do not admit a Kahler structure, but all Poisson manifolds are quantizable. How would the quantization of such a system look like? Perhaps there are no bounded operators in this case, I don't know.

Next time I'll present a concrete calculation of a standard problem in the phase space formalism of quantum mechanics. This will challenge the ontic interpretation of quantum mechanics.