Basic QM: Probability Density w/ 3 Slits Open

  • Thread starter Thread starter phosgene
  • Start date Start date
  • Tags Tags
    Qm
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 2K views
phosgene
Messages
145
Reaction score
1

Homework Statement



Suppose that we have a source of particles (e.g. photons) S, then three slits labelled 1,2 and 3, followed by a screen. For a particle that has passed through slit i, where i=1,2,3, let ψi(x) be the amplitude for the particle arriving at a position x units along the screen.

(a) Write down the probability density function for detecting a particle at a position x on the screen when:

1. all three slits are open,
2. slits 1 and 3 are open,

[Note: you don't need to determine explicit expressions for the amplitudes ψi(x)]

(b) If we applied purely classical physics, how would your answers to the above differ?

(c) Suppose we know the probability density functions for detecting the particle at a position x when only one particular slit is open. That is, we know P1(x), P2(x) and P3(x). Are we able to express the probability density function for the case of all three slits open in terms of P1(x), P2(x) and P3(x)? Can we do this if we apply purely classical physics?

Homework Equations





The Attempt at a Solution



(a) |ψslits 1,2,3|2= (ψ*slit 1 + ψ*slit 2 + ψ*slit 3)(ψslit 1 + ψslit 2 + ψslit 3)

= |ψslit 1|2 + |ψslit 2|2 + |ψslit 3|2 + ψ*slit 1ψslit 2 + ψ*slit 1ψslit 3 + ψ*slit 2ψslit 1 + ψ*slit 2ψslit 3 + ψ*slit 3ψslit 1 + ψ*slit 3ψslit 2

(b) The probability density functions would add linearly in classical physics, giving:

slits 1,2,3|2= |ψslits 1|2 + |ψslits 2|2 + |ψslits 3|2

(c) Well I would just plug the values for P(x) into the above equations for QM and classical physics, respectively, right?
 
Physics news on Phys.org
Maybe not, because those probabilities don't take into account that the electron could interfere with itself. But it is possible in classical physics because the electron is just a particle.
 
phosgene said:
Maybe not, ...

You should be able to give a definite answer by looking at your mathematical expression for the answer to (a). How would you write P1(x) in terms of ψslit 1, etc.?
 
P1(x) could be written as ψ*slit 1ψslit 1, but my expression of the probability in a) includes terms like ψ*slit 1ψslit 3 which cannot be expressed in terms of P1(x), P2(x) or P3(x). Therefore I cannot express the probability in terms of just P1(x), P2(x) and P3(x). But in classical physics, the probability density functions add linearly, so I could express the probability in terms of P1(x), P2(x) and P3(x).