I If you shoot an electron at a wall...

1. Apr 24, 2017

mike1000

Lets pretend there is a gun of some type that can fire electrons one at at time at a wall. The plane of the wall is oriented perpendicular to the x axis, which is to say that the wall is in the y-z plane some distance down the xaxis.

There is no intervening screen with slits in it. We are just firing an electron at a wall.

The electron is fired at the wall, at some well defined momentum in the x-direction, call it $p_x$. I am going to assume that the momentums of the electron in the transverse directions (y,z) are also well defined but I really do not know if that is true. I think the momentum in the y and z directions, $p_y$ and $p_z$ is such that the electron actually follows a helical path to the wall, but I am not sure of that. (I will deal with that later, because if that is true, isn't the motion of the electon in the plane transverse to its direction of motion equivalent to an angular momentum about the x-axis?)

For the moment, my question is this...

Where on the wall will the electron hit? Will it hit the point where the gun is aiming? Or will it hit all over the place? If I understand correctly, if $p_y$ and $p_z$ are well defined then the position of the electron in the y and z directions is completely undefined making it equally probable that the electron can hit anywhere on that wall because the wave functions in the y and z directions should be plane sine waves.

2. Apr 24, 2017

Staff: Mentor

It is perfectly possible in principle to have well-defined linear momentum in all three directions, since the three linear momentum operators commute. Whether this happens in practice will depend on the electron source. (And in any case having perfectly well-defined linear momentum is an idealization that can't be realized in practice--see below.)

In this highly idealized case, yes. But as noted above, this highly idealized case can't be realized in practice, because it is impossible to have infinite uncertainty in position in a real case. In a real case, if you run the numbers using the electron's mass, you will see that you can achieve, in the y-z plane, a very narrow uncertainty in position while still also having a very narrow uncertainty in velocity, so that the electron can be aimed very accurately at a particular point on the wall. This is how actual electron guns (for example, the ones that used to be used in TV picture tubes) work.

3. Apr 24, 2017

Staff: Mentor

If you have an electron gun, then you can say something about the position of the electron as it comes out the gun, so it can't have a definite momentum.

4. Apr 24, 2017

vanhees71

You can't answer this question, because you are not specific enough, how your gun is constructed! In more technical terms you need to specify in which state $\hat{\rho}$ the gun prepares the electron.

As a model you may think about pure states of free electrons, described by wave packets,
$$\psi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} A(\vec{p}) \exp \left (-\mathrm{i} t \frac{\vec{p}^2}{2m} \right) \frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}).$$
If you choose $A(\vec{p})$ as a Gaussian, you can evaluate the integral exactly and then you get the probability distribution of finding the electron at a given place by $|\psi(t,\vec{x})|^2$.

A single electron, hitting your screen will give just a dot, not some clould-like something. The conclusion is still the old one by Born [1]: the wave function (modulus squared) describes probabilities and nothing than probabilities. You'll find the Gaussian distribution only firing a lot of such prepared electrons on your screen.

[1] M. Born, Zur Quantenmechanik der Stoßvorgänge, Z. Phys. 37, 863 (1926)

5. Apr 24, 2017

mikeyork

As far as I can see, there is nothing about this question that needs any consideration of QM uncertainty. If you have an idealized gun firing electrons with a precise momentum, then the direction of that momentum will precisely determine the spot where they hit the wall. And if, in vanhees71's example, you have a narrow Gaussian distribution of momentum then you'll get an appropriately narrow distribution of hits on the wall.

6. Apr 24, 2017

mike1000

I thought that if the momentum is precise, ie $A(\hat{p})$ in vanhees equation is a $\delta()$ function, then the corresponding position wave function is completely undetermined. The electron could hit anywhere with equal probability. ( I realize that $\hat{p}$ in vanhees equation is a vector for $p_x,p_y$ and $p_z$ components of momentum. Shouldn't only $p_y$ and $p_z$ be of importance for this? Those are the momenum's that can cause the electron to deviate from a straight path in the x-direction)

7. Apr 24, 2017

Staff: Mentor

But you can't have that, because a "gun" also determines the position of the electron, within some finite uncertainty, when it leaves the gun, as DrClaude pointed out. But there is no such thing as a state with precise momentum and anything less than infinite uncertainty in position. So there is no such thing as a "gun" that fires electrons with a precise momentum, not even an idealized gun. (Unless you want to admit that your idealized "gun" must have an infinitely wide muzzle, which IMO makes the term "gun" meaningless.)

No, it won't, because a precise momentum means infinite uncertainty in position, therefore infinite uncertainty in the starting point of the electron. And it's a precise direction plus a precise starting point that would be required to precisely determine the spot where the electron hits the wall.

In an actual gun, as has been pointed out in previous posts, the state of the electron when it leaves the gun has a narrow spread in position (based on the finite size of the gun's opening), and a narrow enough spread in velocity (because of the electron's mass) to make it possible to aim the electron to a good accuracy. But such a state does not have a precise momentum.

8. Apr 24, 2017

mikeyork

Regarding your claim about precise momentum implying infinite uncertainty in the position of the gun, I think that is true only if you know the precise time of emission (because precise momentum means precise velocity) -- which is something that has not been specified. The position of the gun is not the same as the position of the electron -- particularly if it has a precise momentum. So I see no problem with an idealized gun if the gun has a barrel of non-vanishing length and one does not care about the time the electron passes a specific location in the barrel. But this discussion is getting into esoteric regions far from reality as all idealized discussions do and arguing about what an idealized gun would look like will not be a useful way to occupy our time, since we all agree that in practice there will be a narrow distribution in both momentum and location at the wall..

In particular, a gun with a wide barrel will give a wide distribution in momentum. So the width of the distribution on the wall is not inverse to the width of the barrel as one might expect for a QM effect. This is why I say we are dealing with an essentially classical question.

And how will we measure the momentum in practice? A good place would be at the wall, the momentum determined by its impact on a very narrow spot on the wall. Then one really doesn't care much about the gun except that it gives us a clue where to place a detector if we remove the wall. Would QM effects be observable? I very much doubt it.

What about protons shooting around the LHC? They have a very sharply determined momentum and they collide in a very sharply determined location. Yes, there is an implied "cross-section", but that is not what is directly measured, it is constructed mathematically from the detected results of the collision. My point is that I don't think this is a place to look for QM effects except in terms of the results of the collisions.

Another example would be momentum measurement in a bubble chamber. I'm no experimentalist, but if I recall correctly, the momentum is determined in a purely classical way by the track it leaves in space. Sure, there is QM uncertainty, but no one cares about it.

Last edited: Apr 24, 2017
9. Apr 24, 2017

Staff: Mentor

No, it has nothing to do with the time of emission. The question is, what state is the electron in when it is "emitted"? If it is in a state of precise momentum, then that state is also a state of infinite uncertainty in position. That's all there is to it. The time of emission is irrelevant.

Irrelevant. If the gun has a barrel of finite width, then it imposes a finite limit on the possible positions of the electron when it exits the gun. Which means the state of the electron at the time of emission can only have a finite uncertainty in position, if it is emitted from a gun with a barrel of finite width.

Who said it should be?

If the state at emission is a state of precise momentum, then the width of the position distribution is constant in time (infinite at emission, infinite at detection). This is because a state of precise momentum is also a state of precise energy, which means it's an eigenstate of the Hamiltonian, which means it doesn't change with time.

If the state at emission is not a state of precise momentum, then the width of the position distribution will change with time.

Why?

And this claim of yours is incorrect. This is an essentially quantum question.

In practice the electron's state exiting the gun will not be a state of precise momentum. It will be a state with a narrow spread in position and a narrow spread in velocity. That is what allows us to predict where on the wall we should place our momentum detector.

You appear to be trying to apply intuitions from that case to the idealized case where the electron at emission does have a state of exactly precise momentum. That won't work.

"Very sharply" is not the same as "exactly". Instead of making intuitive guesses, you really should step back and do the math. Assume that the protons in the LHC have wave functions that are Gaussians in both position and momentum, with the right spreads in both to exactly satisfy the uncertainty principle (i.e., $\Delta x \Delta p =h$, with an exact equals sign). Then calculate the spread in position and the spread in velocity (the latter by dividing the spread in momentum by the proton's mass). What do you get?

No, the direction of travel is determined by the track left in space. You can't derive the magnitude of momentum (or velocity) from this data because there is no way to measure the time at which each bubble in the track is formed. All you can derive from the bubble chamber data is the direction of momentum (or velocity). You have to use other data to determine the particle's energy, which then tells you the magnitude of its momentum.

Also, the tracks in bubble chambers are explained using Gaussian wave functions of the form I described above. They are most certainly not explained by plane wave momentum eigenstates.

10. Apr 25, 2017

mikeyork

Momentum of an electron in a bubble chamber is determined from the track curvature in an electromagnetic field.

As for the rest of what you write, I see no further point in arguing with you, since we both understand QM but you have clearly do not want to understood why I say it is not the important issue in the situation posed, but would rather pretend there is an argument about QM.

11. Apr 25, 2017

vanhees71

$A(p)$ must be square integrable, otherwise your wave packet doesn't represent a state. That's what's behind the uncertainty relation. A distribution is a distribution (generalized function)!

12. Apr 25, 2017

vanhees71

It is very well known from the early days of QT, why an electron leaves a "track" in a bubble or cloud chamber. It's due to the interaction of the electron with the material. It's well worth thinking about the famous treatment by Mott to understand QT (in the real sense without philosophical gibberish about "Interpretation"):

https://en.wikipedia.org/wiki/Mott_problem

It's clearly demonstrating the correctness of the Born rule rather than socalled "realistic interpretations" of the "wave function".

13. Apr 25, 2017

ftr

14. Apr 25, 2017

mike1000

Whether it is an argument about QM or not I will leave to the experts, but for me this has been an informative thread.

It is clear now, that as diameter of the barrel of the gun increases, the momentum of the electron in the transverse direction, the y-z plane, becomes more precise at the expense of uncertainty in the starting position of the electron when the the gun is fired. This was the piece to the puzzle that I was missing. I was not thinking in terms of the starting position of the electron. I was thinking that the electron was taking some erratic path to the target, but that is not the case (when the transverse momentum is well defined) ... the uncertainty is in the starting position not the path. It follows, I think, that as the diameter of the gun decreases, the starting position becomes precise but the momentum in the transverse plane become more uncertain. This is the case where the path of the particle becomes more erratic to the target. (A helical path? As if, the smaller diameter imparts angular momentum to the electron. If this logic is correct, then one can conclude that a completely free electron is in a state which does not have any angular momentum. ( I am not talking about spin angular momentum, I think it is $L$ type angular momentum.)

Last edited: Apr 25, 2017
15. Apr 25, 2017

ftr

I think in terms of QM the path is undefined, we only know starting position probability and the end position probability precisely because of the UP that thread has been analyzing. Each electron is part of an statistical ensemble so to speak.

That is why if you measure it you will ruin the outcome of the experiment.

Last edited: Apr 25, 2017
16. Apr 25, 2017

Staff: Mentor

Yes.

Yes.

No. As ftr points out, in the absence of a measurement of the path, the particle does not have a precise path.

It isn't, because your logic assumes that the particle has a definite path, when it doesn't.

17. Apr 25, 2017

Staff: Mentor

Yes, you're correct. I was thinking of a neutral particle, but of course we are talking about a charged particle in this thread.

I understand why you think QM is not important. I just disagree with you on that point. The relationship between the width of the gun barrel and the uncertainty in momentum, which is the key to answering the OP's question (as shown by his post #14 and my post #15 in response), is a quantum relationship.

To put it another way, here is the key claim you made in post #5:

This statement would be correct if classical physics applied to electrons being fired by a gun. But it is wrong if quantum physics applies: in QM, the more precise the momentum is, the wider the gun barrel has to be (so that the electron's wave function when it exits the gun is closer and closer to an exact plane wave), and therefore the more uncertainty there is in the spot where the electron hits the wall.

So either you are claiming that QM is wrong when applied to electrons being fired out of a gun (which is a very strong claim that you would need to back up with a lot of experimental evidence), or you are simply mistaken when you say this is not an essentially QM problem. I'm going for option #2.

18. Apr 25, 2017

mike1000

You misinterpret what I said. I never said the particle has a definite path. I said just the opposite. I said the path is erratic, I suspect it is helical because of the non-zero momentums in the tranverse plane.

19. Apr 25, 2017

Staff: Mentor

An erratic path is still a definite path. If there is no definite path, then you can't say anything about it, because it doesn't exist. You can't say it's erratic or not erratic. All you can say is that there is no definite path.

"Helical" is a definite shape. How can a path that does not exist (because there is no definite path) have a shape? Helical or anything else?

20. Apr 25, 2017

mike1000

Helical, as I am using the term is not a definite shape. Maybe I should use the word "corkscrew". I am suggesting that a path DOES exist, but it is not well defined, and I think it is reasonable to predict that if will be "helical" or "corkscrew" in character because the momentums in the transverse plane are not zero.

Obviously, the electron takes some path to get to the wall.