Monday, December 30, 2013

Solving Hilbert’s sixth problem (part three of many)

A tale of two products

Thinking how to proceed on deriving quantum mechanics I came to the conclusion to reverse a bit the order of presentation and review first the goal of the derivation: a symmetric and a skew-symmetric product.

Let us start with classical mechanics and good old fashion Newton’s second law: F = ma. Let’s consider the simplest case of a one dimensional motion of a point particle in a potential V(x):

F = -dV/dx
a = dv/dt with “v” the velocity.
Introducing the Hamiltonian as the sum of kinetic and potential energy: H(p,x)= p2/2m + V(x) we have:

dp/dt = - ∂V/∂x = -∂/∂x (p2/2m + V) = -∂H/∂x
dx/dt=v=∂/∂p(p2/2m + V) = ∂H/∂p

In general, one talks not of point particles, but it is customary to introduce generalized coordinates q and generalized momenta p to take advantage of various symmetries of the problem (q = q1, q2,…,qn p = p1, p2,…,pn).

dp/dt = -∂H/∂q
dq/dt = +∂H/∂p

with H = H(q,p,t)

We observe two very important things right away: the equations are linear and q’s and p’s are in one-to-one correspondence.

Now we can introduce the Poisson bracket of two functions f(p,q) and g(p,q) as follows:

{f,g} = ∂f/∂q ∂g/∂p - ∂f/∂p ∂g/∂q

as a convenience to express the equations of motion like this:

dp/dt = {p, H}
dq/dt = {q, H}

The Poisson bracket defines a skew-symmetric product between any two functions f,g:

{f,g} = f o g = -g o f = -{g, f} and more importantly this product obeys the so-called Jacoby identity:

{f,{g,h}} + {h,{f,g}} + {g,{h,f}} = 0

This identity follows identically from the definition of the Poisson bracket and expanding and canceling the partial derivatives.

The Jacoby identity and the skew-symmetry property define a Lie algebra.

So in classical mechanics in phase space one defines two products: the regular function multiplication: f(q,p).g(q,p) which is a symmetric product, and the Poisson bracket {f(q,p),g(q,p)} which is a skew-symmetric product.

Now onto quantum mechanics. In quantum mechanics one replaces the Poisson bracket with the commutator [A, B] = AB-BA which can be understood as a skew-symmetric product between operators on a Hilbert space. There is also a symmetric product: the Jordan product defined as the anti-commutator: {A,B}=1/2 (AB+BA)

The commutator also obeys the Jacoby identity:

[A,[B,C]] + [C,[A,B]] + [B,[C,A]] =

[A, BC-CB] + [C, AB-BA] + [B, CA-AC]=


and the commutator also defines a Lie algebra, just like in classical mechanics.

How can we understand the Jordan product? In quantum mechanics operators do not commute and we cannot simply take the function multiplication.

To generate real spectra and positive probability predictions, observable operators must be self-adjoint: O=O meaning that in matrix form they are the same as the transposed and complex conjugate. However, because of transposition, the product of two self-adjoint operators is not self-adjoint:

(AB) = BA=BA != AB

However, the Jordan product preserves self-adjoint-ness:

{A,B} = ½ ( (AB) + (BA) ) = ½ (BA + AB) = {A,B}

if A=A and B=B

In quantum mechanics the Jordan product is a symmetric product.

Both classical and quantum mechanics have a symmetric and a skew-symmetric product:

                                    CM                              QM

Symm                           f.g                                Jordan product
Skew-Symm                Poisson bracket            Commutator

Both classical and quantum mechanics have dualities:

CM: duality between qs and ps: q <---> p
QM: duality between observables and generators:q <---> -i ħ ∂/∂q = p

So in this post we solved the simple direct problem: extract a symmetric and a skew symmetric product.

In subsequent posts we will show two important things:
1)      we will derive the symmetric and skew-symmetric products of classical and quantum mechanics from composability

2)      we will solve the inverse problem: derive classical and quantum mechanics from the two products.

In the meantime: HAPPY NEW YEAR!

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