Does force change with wave function?

Question

Assuming an electron is moving along a path with constant velocity. According to the square modulus of it’s wave function, it has lower probability to be present at points A and B and higher probability at point C. Then should we expect the average electric force induced by another charged particle be less effective in the areas that electron is less present than the peaks? In other words, in double slit experiment with a charged particle, should we expect stronger force interactions in the peaks compared to the valleys- where particle is less likely to be present?

PS: to avoid the second charged particle’s effect on the wave function, let’s assume the second particle appears instantaneously from vacuum or just shows up to the proximity from distance. Just to eliminate its effect on the test particle’s wave function.

Answer 1

Electrons are elementary particles and as such, they are modeled with quantum mechanics, which the wave function obeys. To calculate the wavefunction one introduces the potentials that the electron is subject to, to the wave equation and solves the problem with the boundary conditions . The wavefunction $Ψ$ , leads to the probability of location of the electron given by $Ψ^*Ψ$.

The functional form of the potentials is what gives the probability of finding the electron at a given (x,y,z,t), There are no classical force interactions of the electron and the slits.

Answer 2

I'll consider the question of the OP as it's formulated in a comment to annav's answer:

Even though the wave function is dictated by the potential, but it’s related probability still follows a sine shape over the length, with higher probability in some areas and lower in others. My main confusion is if that electron feels same average force/ potential over all those areas ?

For an electron with the simple Hamiltonian

$$\hat H=\frac{\hat p^2}{2m}+\hat U$$

we can define the acceleration operator as

$$\hat{\vec{a}}=-\frac{\hat{\nabla U}}m,$$

which is compatible with the Newtonian notions of force and acceleration (see a derivation here). Now we can try to find the average acceleration using the usual method of computation of the expectation value of an observable:

$$\langle \vec{a}\rangle=\langle\psi|\hat{\vec{a}}|\psi\rangle.$$

If you compute this in position representation, you'll find that it's just an integral of the acceleration at different points in space weighted by probability density:

$$\langle \vec{a}\rangle=\int_{\mathbb{R}^3}\vec a(\vec x)|\psi(\vec x)|^2\,\mathrm{d}V.$$

So yes, the wavefunction's magnitude does affect the contribution of acceleration at different points in space.

Answer 3

The wave function of an electron is a complex-valued function. The probability distribution for the electron as a function of position is given by the modulus square of this complex-valued wave function. Therefore, the force that would be experienced by the electron would not have the nulls that one would have with a real valued wave.

Answer 4

As said in the previous answer, the WF (of a free electron) is normally complex and cannot be represented by an oscilating function, only the real or the imaginairy part has such a representation. Therefore it doesn't make sense to speak about peaks and valleys. Moreover the WF of two free particles can be orthogonal, therefore the interaction might be less efficient.

In constrast, under an attractive potential the WF tends to be a real-valued function, this argument about peaks and valleys is correct in this case : in localized WFs there is a strong Coulomb repulsion at the peaks and less at the valleys.