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
and
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).
The Hamilton’s equations of motion are:
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]=
ABC-ACB –BCA+CBA + CAB-CBA -ABC+BAC
+BCA-BAC -CAB+ACB =0
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)† = B†A†=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
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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In the meantime: HAPPY NEW YEAR!
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