Double Slit Question

  • #1
There is something I never understood in the double slit. Here is a picture.

http://www.physics.colostate.edu/users/pavol/anims/physics/double_slit.gif [Broken]

When the particle is emitted from a source, if it truly acts like a wave, then wouldn't it sometimes hit the wall in-between the slits? And, wouldn't this cause the wave function to collapse? Please help me understand. Thanks in advance.
 
Last edited by a moderator:

Answers and Replies

  • #2
dx
Homework Helper
Gold Member
2,011
18
Yes, the electron can hit the wall.
 
  • #3
So, would the wave function collapse when it does this?
 
  • #4
13
0
The wave function does not collapse from a superposition of eigenstates until a measurement is performed. If the quantity being measured has a discrete spectrum, the wave-function will collapse to a single eigenfunction. If it has a continuous spectrum (like position in your case), the theory predicts that the wave-function will collapse to a localized wavepacket (still a superposition of eigenfunctions). This wave-packet is of course to satisfy the normalization condition on the wavefunction.

The little applet you are showing is not actually observed. There is no wave observed crashing into the double slits and producing the nice video you showed. The wavefunction by itself has no physical meaning and cannot be observed crashing into the double slit aparatus. The Psi*Psi however has a physical meaning and that is the probability of the particle at position x, x+dx (for one dimension).

What is observed is the interference pattern that is produced on the back screen.

So to answer your question, "then wouldn't it sometimes hit the wall in-between the slits?" I would say the answer is sometimes. There is a finite probability that the particle will be at the the wall in-between the slits. So it is possible to look at the system (make a position measurement) and see the particle at the wall between the slits. But to ask "And, wouldn't this cause the wave function to collapse? " I don't really understand this. What causes the wavefunction to collapse is YOU making the position measurement to see if the particle is at the wall in between the slits. Before making the measurement there is no way to know exactly where the particle is. The only thing that can be predicted is the probability of the particle being somewhere. So the particle being at the wall doesn't collapse the wavefunction, you observing the particle at the wall does.
 
  • #5
104
0
There is something I never understood in the double slit. Here is a picture.



When the particle is emitted from a source, if it truly acts like a wave, then wouldn't it sometimes hit the wall in-between the slits? And, wouldn't this cause the wave function to collapse? Please help me understand. Thanks in advance.
The wave associated to a particle is localized inside its compton wave length.

To have a significative interference the distance between the two slits must be smaller than the Compton wave length of that particle (this is the case represented in your picture), otherwise the wave pass in one of the two slits without self interference.
 
  • #6
jtbell
Mentor
15,613
3,635
Suppose that the barrier with the slits, itself has the same kind of photodetectors that the final screen on the right has, and the source is faint enough that you can "watch" the arrival of individual photons at the detectors. Then you would "see" that most of the photons represented by the incoming wave hit the barrier at uniformly random locations, and stop there. A few of them pass throuigh the openings and hit the final screen, at non-uniformly random locations, with a probability distribution given by the usual double-slit interference equation.
 
  • #7
I guess what I'm trying to understand is the difference between observation and collision with the barrier. The barrier is made of particles, and we observe with photons. No matter what, it hits something, or is hit by something, so why does it matter what particle hits it? Why doesn't the wave function collapse when it hits the wall?
 
  • #8
1,838
7
I guess what I'm trying to understand is the difference between observation and collision with the barrier. The barrier is made of particles, and we observe with photons. No matter what, it hits something, or is hit by something, so why does it matter what particle hits it? Why doesn't the wave function collapse when it hits the wall?
What matters is the state of the sytem the photon collides with. When a photon hits a detector, the state of the detector changes in such a way that you can tell in an unmabiguous way that the detector has interacted with the photon.

If a photon bounces off a wall, you cannot (even in principle) examine the wall and tell that a photon has bounced off it. If you have a floating mirror in space and do interference experiments, you could argue that due to conservation of momentum you could still tell that a photon had reflected off the mirror. But this is not true either. The reason is that for a interference pattern to exist, the mirror's position must be determined to within a wavelength of the photon. It then follows from the uncertainty relation that the mirror's momentum has an uncertainty larger than the photon's momentum.

This then means that the wavefunction of the mirror as a function of momentum is broader than the momentum it absorbs if the photon bounces off. The wavefunction thus shifts by an amount that is less than its own width. Therefore a momentum measurement of the mirror cannot tell you with certainty if a photon has bounced off it.
 
  • #9
13
0
If you want to know where the particle is, you must make some sort of position measurement. If you want to know if the particle collides with the barrier, you must make some sort or measurement that will detect this collision. You must see that these two measurements are related. If you are trying to measure a particle collision with a barrier you are limiting the particle to exist in some position range where the barrier is. This is a position measurement! Therefore, the wavefunction collapses. In otherwords, you are measuring a collision of a particle with a wall at a particular position in space. For the particle to be detected at this position in space it must be at that position in space. Therefore measuring the collision is just another way of measuring the position of the particle.

So the wavefunction will collapses when you measure a collision. You ask "Why doesn't the wave function collapse when it hits the wall?" Like i just stated, the only way to know if the particle collides with the wall, is to observe or measure, which in itself collapses the particle's wave function. You speak of the wave function collapsing as IT hits the wall. The wavefunction is not real it doesn't exist as something you can visualize as hitting a barrier. It by itself has no meaning, no interpretation.
 
  • #10
I'm getting it now. Thanks guys.
 
  • #11
198
0
Here's a page with an applet where you see that there's reflection of part of the plane wave approaching the wall with the slits.
 
  • #12
So, just for quick clarification, the wavefunction collapses only if it changes the state of the barrier it hits. That is how detectors work. Is that correct?
 
  • #13
1,838
7
So, just for quick clarification, the wavefunction collapses only if it changes the state of the barrier it hits. That is how detectors work. Is that correct?
Yes. Newguy was wrong to say:

...the only way to know if the particle collides with the wall, is to observe or measure, which in itself collapses the particle's wave function.
What matters is that after the particle reflects off the wall, the wall's wavefunction is unaltered. You can come up with cases where this is not the case. Then the particle's wavefunction would be entangled with the new wavefuncton of the thing it bounces off with. This is then what we would effectively manisfest itself as a "collapse of the wavefunction" and that doesn't involve any measurement.

Measurement leading to a "collapse of the wavefunction" is ultimately nothing more than the observer's wavefunction getting entangled with the wavefunction of the measured system.

There can be no real "collapse of the wavefunction", because ultimately everything consists of particles that exactly obey the Schrödinger equation.
 

Related Threads on Double Slit Question

  • Last Post
Replies
6
Views
747
  • Last Post
Replies
5
Views
659
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
6
Views
3K
  • Last Post
6
Replies
143
Views
6K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
19
Views
781
  • Last Post
Replies
5
Views
2K
  • Last Post
2
Replies
35
Views
5K
Top