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I'm reading about uncertainty principle, and something has been bothering me for quite a while. There is the formula:

$$\sigma_x \sigma_p \ge \frac{\hbar}{2}$$

I know what this means: the more you know about the position, the less you know about the momentum, and vice-versa.

As an implication of this principle, I see this kind of image:


And thats my point: as we narrow the hole, we know more about the position, so we are uncertain about... about what? The trajectory? But the state is that we know more about the position, we know less about the momentum. So that means:

Momentum = Trajectory ?

I don't think so, once momentum is mass times velocity. The mass must be constant - I assume - so the velocity must be changing, and that will make the bean spread out after passing through the hole? How is that?

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Maybe this site can help you – anna v Sep 3 '13 at 2:57
hmmm... really interesting scource! – user2489690 Sep 3 '13 at 3:03
Recall that momentum is a vector. When you limit it with the hole you only limit the transverse directions. – dmckee Sep 3 '13 at 3:03
And you only know that the photon passed through a certain position (the hole) you don't know when - so it's position at any given moment is still uncertain – Martin Beckett Sep 3 '13 at 3:17
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I am not quite sure what is your doubt so I will try to describe this as precise as possible what I think the doubt you might have.

Q. Momentum == trajectory.

A. Yes... Kind-of! Momentum does describe the trajectory of the free-particle if you knew the starting position in transverse plane (say $(x_o, y_o)$). As a matter of fact (in technical words) momentum is generator of position.

If you are okay with this then no need to read forward you could just let this go, but if you are new to quantum then I would suggest that you read further as it might give a clear pathway to understanding of the problem itself which get blurred by weird orthodox interpretation still taught in classrooms.

In simple words when you try to find the position of photon's position in xy-plane (assuming that light is traveling along z-axis), then it's momentum becomes uncertain "in XY-Direction ONLY" (well from energy conservation you might argue that z-momentum as well becomes uncertain, well yes but the effect is more pronounced in x-/y-direction as prior to slit the momentum was exactly zero). The momentum (hence the trajectory in space) was perfectly known in before photon encounters slit. Later the photon get dispersed after it has passed through slit purely due to the quantum effects.

Following is just mathematical details of "what happens" from "Before" to "After".

Lets say you start with plane wave, $|\psi_{Before}> = |\vec{p}>$. Now when photon are passing through the slit measurement is being carried by the atoms making the slit and they are making observation in position basis so (orthodox interpretation suggest) we better express the state-ket in terms of position basis $|x>$. Since the slit is only making the measurement on x and y position of photons we expand the state in $|x>$ and $|y>$.

$$ |\psi_{Before}> = \frac{1}{2\pi \hbar^2} \int dp_x dp_y e^{i(p_x x + p_y y)/\hbar} |x,y> $$ It basically means that we have a superposition state i.e. prior to measurement by slit the transverse position-wise incoming photon is dispersed though-out the measuring plane.

Say the state of photon (after it has passed through slit) is $|\psi_{After}> = |x_\circ, y_\circ> $. Now we have to evolve it to screen in unitary fashion. With known Hamiltonian, $ H = \frac{p^2}{2m} $ it can be seen easily that $ <x,y|U|x_\circ, y_\circ> \neq 0 \Rightarrow $ The beam spreads.

Also trajectory is just a colloquial term it sometime helps connecting the quantum mechanics to classical notions, but you might have realised that it is quite confusing to use. That is why you should proceed with more concrete way of expressing the quantum-ideas.

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