A
Archaea
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I was originally going to post this in the thread: An alternative derivation of special relativity. But this isn't really relativity and that thread's a mess, so I thought I'd start this one.
This isn't really an alternative to the whole of quantum theory. What I've been looking for is a way to rigorously derive the Hamiltonian used in Schrödinger's equation. However, because the Schrödinger equation is so fundamental to much of quantum theory, changes to it mean there many changes to the theory.
The first thing to do is to take out the minus sign in the momentum operator in the position representation and put it into the position operator in the momentum representation:
P = iћ(d/dx) and x = -iћ(d/dp)
This has the effect of changing the signs of the commutators, so: [x, P] = -iћ and [P, x] = iћ. This needs to be done in order to make the commutators of the Hamiltonian work. This also means that the momentum eigenstates take the form e-ikx and the position eigenstates are eirp which shows up in the Fourier transforms.
The second thing we need to do is define:
F = dP(t)/dt and v = dx(t)/dt
These are supposed to be relativistic variables and not operators, so they are in need of a precise definition and the statement that they're variables needs to be justified... I think: The momentum eigenstates ψk(x) and ψj(x) will have a definite momenta pk = ћk and pj = ћj respectively. In the presence of the same "force" these eigenstates will have their momenta changing at a constant rate, so that:
dk/dt = dj/dt
And since a wave function can be decomposed into eigenstates, F will not depend on the wave function itself and so doesn't need to be an operator. We can do the same thing with v using position eigenstates.
The next thing to do is define the Hamiltonian as:
Ht = iћ(d/dt)
And look at the commutators in the position representation:
[Ht, P] = -ћ2[d/dt, d/dx] = 0
[Ht, x] = iћ(dx/dt) = iћv
And the commutators in the momentum representation:
[Ht, x] = -ћ2[d/dt, d/dp] = 0
[Ht, P] = iћ(dP/dt) = iћF
It seems to me that in quantum mechanics if something commutes with the right hand side then it commutes with the left hand side (except in the mainstream Schrödinger equation). With this in mind we can create two separate Hamiltonians for the position representation and the momentum representation:
Hx = vP
HP = -Fx
This is a bit funny, but after a bit of thought I decided it was okay, my thinking is that potential energy doesn't have any effect in the position representation and kinetic energy doesn't have any effect in the momentum representation. Also it makes sense not to add x and P together in the same equation because they don't commute.
Both the equations above are trivially correct due to the chain rule. The Hamiltonian equations of motion are satisfied as well:
F = -(dHP/dx)
v = (dHx/dP)
Finally, if we turn these into differential equations and cancel out the i's and the ћ's we get:
And combine these with the Fourier transforms (which I hope I got right):
I don't know if there's a way to get these into single differential equations for ψ(x) and φ(p) as I haven't tried yet. These equations are relativistic equations, which is good. The big problem I'm having now is that trying to find solutions gives weird answers, so I think it's a nice theory, but it might not give nice predictions. :)
This isn't really an alternative to the whole of quantum theory. What I've been looking for is a way to rigorously derive the Hamiltonian used in Schrödinger's equation. However, because the Schrödinger equation is so fundamental to much of quantum theory, changes to it mean there many changes to the theory.
The first thing to do is to take out the minus sign in the momentum operator in the position representation and put it into the position operator in the momentum representation:
P = iћ(d/dx) and x = -iћ(d/dp)
This has the effect of changing the signs of the commutators, so: [x, P] = -iћ and [P, x] = iћ. This needs to be done in order to make the commutators of the Hamiltonian work. This also means that the momentum eigenstates take the form e-ikx and the position eigenstates are eirp which shows up in the Fourier transforms.
The second thing we need to do is define:
F = dP(t)/dt and v = dx(t)/dt
These are supposed to be relativistic variables and not operators, so they are in need of a precise definition and the statement that they're variables needs to be justified... I think: The momentum eigenstates ψk(x) and ψj(x) will have a definite momenta pk = ћk and pj = ћj respectively. In the presence of the same "force" these eigenstates will have their momenta changing at a constant rate, so that:
dk/dt = dj/dt
And since a wave function can be decomposed into eigenstates, F will not depend on the wave function itself and so doesn't need to be an operator. We can do the same thing with v using position eigenstates.
The next thing to do is define the Hamiltonian as:
Ht = iћ(d/dt)
And look at the commutators in the position representation:
[Ht, P] = -ћ2[d/dt, d/dx] = 0
[Ht, x] = iћ(dx/dt) = iћv
And the commutators in the momentum representation:
[Ht, x] = -ћ2[d/dt, d/dp] = 0
[Ht, P] = iћ(dP/dt) = iћF
It seems to me that in quantum mechanics if something commutes with the right hand side then it commutes with the left hand side (except in the mainstream Schrödinger equation). With this in mind we can create two separate Hamiltonians for the position representation and the momentum representation:
Hx = vP
HP = -Fx
This is a bit funny, but after a bit of thought I decided it was okay, my thinking is that potential energy doesn't have any effect in the position representation and kinetic energy doesn't have any effect in the momentum representation. Also it makes sense not to add x and P together in the same equation because they don't commute.
Both the equations above are trivially correct due to the chain rule. The Hamiltonian equations of motion are satisfied as well:
F = -(dHP/dx)
v = (dHx/dP)
Finally, if we turn these into differential equations and cancel out the i's and the ћ's we get:
And combine these with the Fourier transforms (which I hope I got right):
I don't know if there's a way to get these into single differential equations for ψ(x) and φ(p) as I haven't tried yet. These equations are relativistic equations, which is good. The big problem I'm having now is that trying to find solutions gives weird answers, so I think it's a nice theory, but it might not give nice predictions. :)