Winigefx

2022-06-21

How can an electron be fired at a target when uncertainty principle says it will spread out around axis of motion?

Consider an electron fired at a target. Taking the axis of motion to be $x$, and position to be $(x,y,z)$ then

$\mathrm{\Delta}y=\mathrm{\Delta}z=0$

Therefore by the uncertainty principle

$\mathrm{\Delta}{p}_{y}=\mathrm{\Delta}{p}_{z}=\mathrm{\infty}$

So the electron will spread out in $y$ and $z$ and never hit its target (unless that target was very close i.e. on the same scale as the initial effective size of the electron). How can it hit a target in reality?

I realise there is a possible duplicate to this question: How can particles travel in a straight line? The accepted answer for the duplicate is "because $\hslash $ is really small so the particles don't spread out much before hitting the target". I'm not sure this stands though. The time-dependent Schrodinger equation is:

$i\hslash \frac{\delta}{\delta t}\psi (r,t)=[-\frac{{\hslash}^{2}}{2\mu}{\mathrm{\nabla}}^{2}+V(r,t)]\psi (r,t)$

In the absence of external potential $V$ and dividing both sides by $\hslash $

$i\frac{\delta}{\delta t}\psi (r,t)=-\frac{\hslash}{2\mu}{\mathrm{\nabla}}^{2}\psi (r,t)$

So if we change $\hslash $ (or $\mu $), sure the particle will spread out less as a function of time, but won't it also move slower in exactly the same proportion to its lack of spreading out, so it still won't hit the target?

Consider an electron fired at a target. Taking the axis of motion to be $x$, and position to be $(x,y,z)$ then

$\mathrm{\Delta}y=\mathrm{\Delta}z=0$

Therefore by the uncertainty principle

$\mathrm{\Delta}{p}_{y}=\mathrm{\Delta}{p}_{z}=\mathrm{\infty}$

So the electron will spread out in $y$ and $z$ and never hit its target (unless that target was very close i.e. on the same scale as the initial effective size of the electron). How can it hit a target in reality?

I realise there is a possible duplicate to this question: How can particles travel in a straight line? The accepted answer for the duplicate is "because $\hslash $ is really small so the particles don't spread out much before hitting the target". I'm not sure this stands though. The time-dependent Schrodinger equation is:

$i\hslash \frac{\delta}{\delta t}\psi (r,t)=[-\frac{{\hslash}^{2}}{2\mu}{\mathrm{\nabla}}^{2}+V(r,t)]\psi (r,t)$

In the absence of external potential $V$ and dividing both sides by $\hslash $

$i\frac{\delta}{\delta t}\psi (r,t)=-\frac{\hslash}{2\mu}{\mathrm{\nabla}}^{2}\psi (r,t)$

So if we change $\hslash $ (or $\mu $), sure the particle will spread out less as a function of time, but won't it also move slower in exactly the same proportion to its lack of spreading out, so it still won't hit the target?

Nia Molina

Beginner2022-06-22Added 21 answers

In essence you are mixing the notion of electron as a point particle, and quantum physics.

While classical mech have us believe that electron are point-wise particles, Quantum mech actually forbids it, by the same Uncertainty principle you are invoking.

Thus the electron has some dimensionality in the z axis as well as the y axis, I'll go ahead ad generalize by calling this the transverse direction.

One can consider an electron as basically a wave packet, one naive representation of which is a Gaussian wave packet that, while has a locale in the classical sense, still is smeared over an effective region.

While it is true that the Lorentz transformation changes the x-axis, and thus the wave-form in the x direction, the transverse coordinates are left unchanged.

What IS true though is that the wave numbers for the transverse component are affected, so the shape of the wave PACKET as such, and the dispersion relations might change.

While the electron's wave packet smears a bit, its location in the quantum sense is still around x=vt, y=o, z=0 and will hit the target.

Another way to view it might be, that the electron doesn't actually hit the target, only gets close enough to the target for the interaction between the target potential and the electron charge to be very significant.

While classical mech have us believe that electron are point-wise particles, Quantum mech actually forbids it, by the same Uncertainty principle you are invoking.

Thus the electron has some dimensionality in the z axis as well as the y axis, I'll go ahead ad generalize by calling this the transverse direction.

One can consider an electron as basically a wave packet, one naive representation of which is a Gaussian wave packet that, while has a locale in the classical sense, still is smeared over an effective region.

While it is true that the Lorentz transformation changes the x-axis, and thus the wave-form in the x direction, the transverse coordinates are left unchanged.

What IS true though is that the wave numbers for the transverse component are affected, so the shape of the wave PACKET as such, and the dispersion relations might change.

While the electron's wave packet smears a bit, its location in the quantum sense is still around x=vt, y=o, z=0 and will hit the target.

Another way to view it might be, that the electron doesn't actually hit the target, only gets close enough to the target for the interaction between the target potential and the electron charge to be very significant.

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