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Say electron A is nearby another electron (B), so that they may repel each other. Electron B is in a position eigenstate (so it has a definite position). But electron A is not. How does electron A affect the acceleration of electron B? Does it "divide up" its electromagnetic force as if it were a charged object spanning the space that the wave function occupies, whose charge density is proportional to the value of the probability density function? Otherwise, how can electron B decide where to move?

Simply: if an electron can be in "multiple places at once", and the force it produces depends on its location, which location is "chosen" for that force?

...I know that $\exists$ a whole theory on this, Quantum Electrodynamics (thanks Feynman!!!), but I have not studied it. I have only ever taken an intro QM class as an undergraduate.

Edit: If the position eigenstate causes problems, let B be in an arbitrary eigenstate as well. The question is rephrased: if the positions are indeterminate, how is the force, which depends on them, calculated?

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  • $\begingroup$ Does it "divide up"..... Yes, it does just that. But if we draw a gaussian surface that encompasses electron A (mostly) and we compute the electric field due to that, we find that it acts like a point charge centered at the average position of A (weighted by the probability cloud of course) with a charge equaling the total charge encompassed by the gaussian surface. What I'm saying is "You can use the average position of the electron usually" $\endgroup$ – Jim Jan 7 '15 at 20:13
  • $\begingroup$ @Jim, you are told that A is fix. $\endgroup$ – Sofia Jan 7 '15 at 20:46
  • $\begingroup$ @Jim : I apologize, I don't know how it came that I read oppositely. $\endgroup$ – Sofia Jan 7 '15 at 20:58
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Note that the problem you pose is non-realistic. If at a certain moment B is in a position eigenstate, $\delta (\vec r)$, at an extremely short time after , B can be everywhere is the universe with equal probability. You will see the effect of this, below.

But let's first calculate the force $<\vec F>$. In QM, the influence of between A and B goes as follows: let $\psi_A(\vec r)$ be the wave-function of the electron A, where the vector $\vec r$ connects A, wherever A is, with B.

Then the force of interaction is

$\vec F(\vec r) = -\frac {e^2 \vec r}{4 \pi \epsilon_0 |r|^2}$.

The average force between the two electrons is

$<\vec F> = \int d\vec r \int d\vec r' \psi_A^* (\vec r)\delta (\vec r') \frac {e^2 (\vec r - \vec r')}{4 \pi \epsilon_0 |\vec r - \vec r'|^3} \psi_A (\vec r) \delta (\vec r')$.

$=\int d\vec r d\vec r' \delta (0) |\psi_A (\vec r)|^2 \frac {e^2 \vec r}{4 \pi \epsilon_0 |\vec r|^2} = \delta (0)\int d\vec r |\psi_A^* (\vec r)|^2 \frac {e^2 \vec r}{4 \pi \epsilon_0 |\vec r|^2}$

So, we have a problem because the function $\delta (\vec r')$ has infinite norm. On the other side, if $\psi_A$ is spherically symmetrical, one gets $<\vec F> = 0$. For the case that $\psi_A$ is not spherically symmetrical we have to replace the wave-function of B by another function, let's name it $\psi_B (r')$, highly localized around the point $\vec r' = 0$, but normalized. In that case

$<\vec F>=\int d\vec r d\vec r' |\psi^* _B (\vec r')|^2 |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r - \vec r')}{4 \pi \epsilon_0 |\vec r - \vec r'|^3}$,

and since $\psi_B (r')$ is highly localized around $\vec r' = 0$ we can approximate,

$<\vec F>=\int d\vec r d\vec r' |\psi^* _B (\vec r')|^2 |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r)}{4 \pi \epsilon_0 |\vec r|^3} = \int d\vec r |\psi_A^* (\vec r)|^2 \frac {e^2 (\vec r)}{4 \pi \epsilon_0 |\vec r|^3}$.

Now I return to the next moment after localization. The function $\psi_B (\vec r')$ will be practically zero, everywhere. So, in the before last equation we will get $<\vec F> = 0$.

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  • $\begingroup$ interesting. So my question wasn't well posed, because the position eigenstate thing messed it up. But the approach shows that it IS the average of all of the forces, weighted by the probability density, that acts on another particle. I'm assuming that if the other particle also has a PDF, you'll include its PDF in the average, essentially averaging over all possibilities of each position. Am I right in this summary? And also, is this an EXACT answer, quantum electrodynamically (ignoring relativity), or does humanity have a more precise way of doing this? $\endgroup$ – doublefelix Jan 8 '15 at 1:20
  • $\begingroup$ @user31415926535897932384626433 , wow, couldn't you choose a shorter name? It's difficult. Now, what is PDF? I know that PDF is "Portable Data File", which for sure it's not what you mean. The position eigenstate is a problem, its norm is infinite (not physical) and a particle which is in such a state, after an extremely short time is everywhere in the space. So, the calculus of the force with the localized B, isn't worth much, it can last for an extremely short time. It's not a stable situation. $\endgroup$ – Sofia Jan 8 '15 at 1:32
  • $\begingroup$ sorry, "Probability Density Function". I'm gonna change my name soon, too... $\endgroup$ – doublefelix Jan 8 '15 at 7:13
  • $\begingroup$ How is a position eigenstate not physical? Isn't the idea that it occurs after a position measurement? And by definition, doesn't the dirac delta integrate to 1? $\endgroup$ – doublefelix Jan 8 '15 at 7:19
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    $\begingroup$ Yes, $\delta (r)$ integrates to 1, but not $\delta^2(r)$. $\endgroup$ – Sofia Jan 8 '15 at 7:47
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If your charged particle is not in a position eigenstate you can always write the position as a superposition of positional eigenstates. Therefore you would have a quantum superposition of forces on a test particle (weighted by the probability amplitudes).

For instance suppose you have a negative particle which is initially has the wavefunction $$\psi^- = \delta(0),$$ i.e. is in the definite position state located at the origin. Now suppose you have another positive charge in the superposition state $$\psi^+ = \left(\delta(-x) + \delta(+x)\right)/\sqrt{2}.$$ Now a short time later both particles will spread out a little, but the negative particle will be in a superposition of moving to the left and right, i.e. $$\psi^-\approx\left(\delta(-\Delta x) + \delta(+\Delta x)\right)/\sqrt{2}.$$ Of course the positive charge will also change position (and spread out).

If we plotted this, the wavefunctions might look like the picture below. enter image description here

If you tried to measure the position (of either or both particles), then you would get a wavefunction collapse and particles would both be on either the right or left (with 50% probability for each case).

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  • $\begingroup$ The other answer says that the particle in a position eigenstate could, at the next instant, be anywhere in the universe with equal probability. This sounds somewhat reasonable to me because the uncertainty in its momentum (velocity) would be infinite, so who knows how many meters it would move in the next second. How can I reconcile this? $\endgroup$ – doublefelix Jan 8 '15 at 7:16
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    $\begingroup$ @user31415926535897932384626433 This is correct, a delta wavefunction has infinite velocity components. However, basic quantum mechanics is not a relativistic theory and so is only an approximation. In a fully relativistic quantum field theory you will find that you can't perfectly localize a particle (without creating particle-antiparticle pairs), and can only localize a particle to roughly it's Compton wavelength, which is determined by the particle's mass (en.wikipedia.org/wiki/Compton_wavelength) $\endgroup$ – Punk_Physicist Jan 21 '15 at 21:39

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