Derivator said:
[...]ahh, I see. But don't you define this way a ''force-operator'' \hat{\frac{dH}{dx^i}}=\hat{F}_i and thus would calculate the expectation value of the force as <\Psi|\hat{\frac{dH}{dx^i}}|\Psi> anyway?
Getting very technical, the force is in this context a
super-operator i.e. it is an operator acting on the operator algebra, not on the Hilbert space itself.
You can let
x be an operator instead of a parameter (so that [p,x]=1 makes sense.) In that context the Hamiltonian is a bit more abstract than an operator, it is an operator valued function of operators (so we can take derivatives) and we should use a different notation for its derivative w.r.t.
x.
\frac{\partial H}{\partial x^i} \to \frac{\Delta H}{\Delta x^i}
Simply think of H as a polynomial over a non-commutative ring. Then for example
H(x,y) = xyx has x-derivative \frac{\Delta H}{\Delta x} which is the super-operator mapping an arbitrary operator via: z\mapsto zyx + xyz (replacing one
x with
z everywhere it occurs and summing the result).
Note that once you allow the variables to commute this is just multiplication by 2xy the "super-operator'' is just multiplication by the the "operator" which we usually think of as the derivative. The problem in the non-commutative case is that we can't always write that super-operator as simply left (or right) multiplication by an operator but rather a combination of both. Fundamentally these derivatives reside in a bigger space than the operator algebra. Note that we can write Heisenberg's equation as:
\frac{d}{dt} = \frac{i}{\hbar}\Delta H
where \Delta H is the commutator
super-operator mapping operators via:\Delta H: G\mapsto [H,G].
Anyway, these operator derivatives for polynomials can then be extended to convergent limits of polynomials i.e. analytic functions. You can then define a "force" as an operator derivative in the case where coordinates are operators. This however is a bad way of doing it IMNSHO. Especially in relativistic mechanics coordinates become parameters rather than observables.
In the context of this thread you work with parameters instead of operators then the trouble is with the "commutators" [x,p] in my "derivation". I was using that as a shortcut which glosses over a great deal of mathematical theory.
Let's see...
How I would say it is that you expand the Hamiltonian in terms of some basis of of the operator algebra:
H = \omega_k A^k
where the A's are the operators and the omegas are the parametric velocities both of which may depend on the parameters \theta_k.
One then extends from the operator algebra of the quantum theory into the operator algebra plus the differential algebra of the parameter calculus and define the extended Hamiltonian:
\eta \doteq i\hbar \frac{d}{dt} + \Delta H
(\Delta H: A\mapsto [H,A] is again a super-operator specifically "take the comutator with H of".)
The Heisenberg equation becomes the constraint:\eta = 0
(Note general covariance typically leads to a zero extended Hamiltonian, the dotted equality reads "is constrained to be equal to")
All this technical business then allows us to write for arbitrary operator B (depending on the parameters):
\eta B(\theta)= i\hbar \frac{d}{dt}B +[H,B]\doteq 0
or\frac{d}{dt}{B} \doteq \frac{i}{\hbar} \omega_k [A^k,B]
but since \frac{dB}{dt} = \dot{\theta}_k \frac{\partial B}{\theta_k}= \omega_k \frac{\partial B}{\theta_k}
one has:\frac{\partial}{\partial \theta_k}B \doteq \frac{i}{\hbar}[A^k,B] = \frac{i}{\hbar}\Delta A^k
for any operator B!
(again \Delta A: B\mapsto [A,B] is the commutator super-operator.)
So
\frac{\partial}{\partial \theta_k}\doteq \frac{i}{\hbar}\Delta A^k
We view the A^k as generalized momenta canonically dual to the parameters \theta_k.
Then the "force" equations become...wait for it...here is the real punch line...
\frac{d}{dt}A^k = \frac{i}{\hbar}[H,A^k] = -\frac{i}{\hbar}[A^k,H] \doteq -\frac{\partial}{\theta^k} H
And applying the Hellmann-Feynman theorem we have:
\frac{d}{dt}\langle A^k\rangle \doteq -\frac{\partial}{\theta^k}\langle H\rangle = -\frac{\partial}{\theta_k} E
In summary, when we constrain our parameters to evolve over time according to the dynamics of the system, and assuming the Hamiltonian is the generator of time evolution, then the forces = the rate of change of the generators of parameter evolution (momenta) must equal the negative of the parameter derivatives of the Hamiltonian.
The above is very general. In practice the Hamiltonian is some function of the usual momenta p^\mu plus some other operators but you can choose a basis of the operator algebra which includes the usual momenta \{A^k\} = \{p^\mu\}\cup\{C^1,C^2,\cdots\}. We of course use the usual coordinates x_\mu as parameters dual to the usual momenta.
Thus if your Hamiltonian is some function of the momentum operators, it will --upon expansion in the basis-- take the form:
H(p)=\dot{x}_\mu p^\mu + o.t.
where o.t. means other terms linearly independent from the momenta but which still depend on the x's. Those terms may arise from quadratic and higher order terms of the momenta as well as other components of the Hamiltonian. Especially note that in relativistic mechanics the mass operator M=g_{\mu\nu}p^\mu p^\nu which for a single particle is must it's mass times the identity operator is only proportional to the identity when we impose the dynamic constraints:
M\doteq m\mathbf{1}
We can still take parametric derivatives of this operator "off shell" and get non-trivial results, then impose the dynamic constraint.
What is happening is that we are introducing a gauge extension via the parameter algebra and then imposing the dynamics as a gauge constraint (and dynamic evolution is the unfolding of a gauge transformation). This allows us to talk about the classical c-numbers and quantum system variables in the same context. This is done implicitly in all the standard QM where we "choose a picture" and introduce parameters.
