B Trajectory of photon or electron in double slit

learis
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Hello, first post. I recently became interested in quantum physics and its mysteries. To my understanding, In the double slit experiment, the photon or electron will scatter and behave like a wave when both slits are open and neither slit is measured. Over time their cumulative scattering mimics what the interference pattern for what a wave would look like upon hitting the wall.

Now, it is my guess that each individual photon or electron upon passing through the slit (... or both slits?) ends up having a particular trajectory that aims it towards one of the locations on the final wall where its wave would register as not canceling itself out.

My question:
Do we have any knowledge or formulas to show what would cause that photon or electron to take that particular trajectory as opposed to another acceptable trajectory for where its wave would register and not cancel out? Or... is it instead believed that no further causes can exist at this point and its trajectory is truly a random probablity? And that the best we can have is a probability function for its possible trajectories?
 
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learis said:
it is my guess that each individual photon or electron upon passing through the slit (... or both slits?) ends up having a particular trajectory

This is not correct. Quantum mechanics does not work this way.
 
PeterDonis said:
This is not correct. Quantum mechanics does not work this way.

So it doesn't have a particular trajectory after passing through the slit(s)? I'm not sure what the alternative is. Does it have something to do with superposition and having multiple trajectories at once?
 
learis said:
it doesn't have a particular trajectory after passing through the slit(s)?

Or before. It doesn't have a particular trajectory, period.

learis said:
I'm not sure what the alternative is.

There isn't one that is intuitively obvious. That's why it's difficult to deal with quantum mechanics when you first learn about it; it's highly counterintuitive.

learis said:
Does it have something to do with superposition

Yes.

learis said:
and having multiple trajectories at once?

No, because that's not what superposition means (although many pop science sources misleadingly imply that it is). Superposition means that there is an amplitude for all of the different possible trajectories, and to get the probability for arriving at a particular point on the detector, we add up the amplitudes for all the possible trajectories that arrive at that point, then square the result (more precisely, we take the squared modulus of the result, since amplitudes are complex numbers). But that's all it means; talk about "having multiple trajectories at once" is not required or implied.
 
learis said:
Hello, first post. I recently became interested in quantum physics and its mysteries. To my understanding, In the double slit experiment, the photon or electron will scatter and behave like a wave when both slits are open and neither slit is measured. Over time their cumulative scattering mimics what the interference pattern for what a wave would look like upon hitting the wall.

Now, it is my guess that each individual photon or electron upon passing through the slit (... or both slits?) ends up having a particular trajectory that aims it towards one of the locations on the final wall where its wave would register as not canceling itself out.

My question:
Do we have any knowledge or formulas to show what would cause that photon or electron to take that particular trajectory as opposed to another acceptable trajectory for where its wave would register and not cancel out? Or... is it instead believed that no further causes can exist at this point and its trajectory is truly a random probablity? And that the best we can have is a probability function for its possible trajectories?

I suggest you watch this:

 
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PeroK said:
I suggest you watch this:



Thank you! That looks right up my alley. I was really interested in what situations quantum mechanics asserts as random probability being the final cause. I'll give it a watch!
 

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