Force as a Time Derivative of ihk

[Bob Dylan]

If energy is ihw and p is ihk, can force be written as derivatives of these? Might the fundamental forces just be some patterned change in the change of the wave functions of Dirac's equation?

Edit: the title should be "Time derivative of ihk" but I can't edit the title.

 

[PeterDonis]

If energy is ihw and p is ihk

First, where did the i come from?

Second, these equations are not general; they refer to a particular kind of quantum state of a particular kind of system (a plane wave).

can force be written as derivatives of these?

Derivatives with respect to what?

Might the fundamental forces just be some patterned change in the change of the wave functions of Dirac's equation?

I'm not sure what you mean by this, but it looks to me like a personal theory. Please review the PF rules on personal theories.

 

[hilbert2]

You can, in principle, take the time derivative of the momentum expectation value $<\mathbf{p}>$ of a particle (or of the CMS of a system of particles) to obtain something similar to a "force", but this is not a concept that has much use in QM - for instance the "force" defined this way stays at a constant value of zero for the electron in the ground state of a hydrogen atom (which isn't what happens in classical central-force motion).

 

[Bob Dylan]

Well, I may have worded it wrong, but it's an honest question. If force changes momentum, isn't it necessarily the case that fundamental forces are going to alter the momentum of particles. How then could the particle not experience a change in its momentum operator over time?

 

[PeterDonis]

I may have worded it wrong, but it's an honest question.

I'm not sure what honest question you think you are asking in your OP (the original post in this thread), but your OP is not asking the question you are asking in your post #4:

If force changes momentum, isn't it necessarily the case that fundamental forces are going to alter the momentum of particles. How then could the particle not experience a change in its momentum operator over time?

This is an honest question, yes, and the answer to it is that "force changes momentum" is classical thinking, not quantum thinking. In quantum mechanics, as @hilbert2 has pointed out, "force" is not really a useful concept, and you cannot reason about things like the Coulomb interaction between an electron and the nucleus in an atom by thinking of that interaction as a "force" that changes the momentum of the electron.

Also, @hilbert2 did not say the "momentum operator" of an electron in a atom doesn't change; he said the expectation value of the electron's momentum doesn't change. These are not the same thing. (It is also true that the momentum operator doesn't change, but that's because of the definition of the momentum operator, not because of anything to do with the interaction between the electron and the nucleus.)

 

[jtbell]

The closest one can get to what I think the OP is after is this version of Ehrenfest's theorem: $$\frac{d}{dt} \langle p \rangle = - \left\langle \frac {\partial V} {\partial x} \right\rangle$$ In classical mechanics, $- \partial V / \partial x$ is the conservative force associated with the potential $V(x)$, or more generally the x-component of the conservative force associated with $V(x,y,z)$.

 

[hilbert2]

If you set a displaced Gaussian initial state

$\Psi (x,t_0 ) = Ae^{-b(x+\Delta x)^2}$

in motion in a harmonic oscillator potential $V(x) = \frac{1}{2}kx^2$, the expectation values $\left<x\right>$ and $\left<p\right>$ will evolve with cosine and sine time dependence just like the classical harmonic oscillator set in motion from rest at a point with distance $\Delta x$ from equilibrium. However, if the function $V(x)$ has any powers of $x$ higher than $x^2$, this will not work exactly like that.

Some ions can have so called "Kepler wavepackets" orbiting around them, if I can remember correctly. This means that there's a relatively localized electron wavepacket at a quite large distance from the nucleus, and it moves somewhat similarly to a planet orbiting a star, except that it probably comes down gradually because of photon emission. But I don't think that the $m\frac{d\left<x\right>}{dt} = \left<p\right>$ correspondence is true in that case.