Question on double slit experiment

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I have read that if you set up a double slit experiment and let photons go through one at a time they build up an interference pattern. I have also read that this interference pattern is similar to having a pool with two wave generators on one side of the double slit and looking at the interference pattern on the other side. This "makes sense" to me in some sense as the photon travells as a wave, interferres with itself and the interference pattern is the probality that a photon will be detected. What I don't understand is the fact that you need multiple wave crests to make the interference pattern, how many wave crests does a photon have?
 

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I believe there is only one wave generator on one side of the double slit. As the one wave passes through both slits it creates two waves on the back side. These two waves interact to form the interference pattern.
 
It depends on its energy. Any photon detected is essential confined between the emitter and the detector. The number of nodes in the probability wave depends on the energy of the photon. The lowest energy level of a confined particle has one node. The next has two, the next three, then four, etc. In the double slit experiment, monochromatic light (I think) is used, so the number of 'crests' will be the distance travelled divided by the wavelength of the photon.

You're talking about 'photons', so I assume you're referring to the quantum theory, not classical EM theory, of light. Note, then, that the photon doesn't interfere with itself (unlike in EM). It is the wavefunction, i.e. the probability waves of the two trajectories, that interfere with themselves. At any position between the emitter and the detector you have a probability of finding the photon for each path (left slit or right slit). This probability can be zero, since the the probability is given by the square of the sums of the wavefunctions, and the wavefunctions may be negative (so 0.5 + -0.5 squared is zero). It's not a case of "needing" multiple crests - multiple crests are a consequence of having a photon that may be anywhere between the emitter and detector.
 
jtbell
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BigMacnFries said:
how many wave crests does a photon have?
You're apparently thinking of a photon as a little bundle or pulse of waves, a "wave packet". Don't do that! The wave associated with a photon extends over a large region of space, and is related to the probability of finding the photon at any particular location.

I think the size of the region that the wave function covers has little or nothing to do with the size of the photon itself, but has more to do with the conditions under which the photon is produced. If your setup contains a light source that is "on" continuously, but whose intensity is reduced enough so that it produces photons at random, say once every second on the average, then the wave function fills the entire volume of space that is illuminated by the light source (regardless of intensity). If your light source has a shutter that opens for a very short time, then the wave function covers a smaller volume of space, corresponding to the length of time that the shutter is open.
 
BigMacnFries
Can you tell me more about how a photon is emitted. I have read that it is when an electron jumps down an energy level. I have also read a free electron cannot emit a photon because if your were in the same frame of reference as it this would violate energy conservation.
 
BigMacnFries said:
Can you tell me more about how a photon is emitted. I have read that it is when an electron jumps down an energy level. I have also read a free electron cannot emit a photon because if your were in the same frame of reference as it this would violate energy conservation.
Yes, a photon is emitted when an electron makes a transition from a higher energy level to a lower one. Free electrons are known to radiate when they deccelerate. Since this requires a change in kinetic energy, there is no inertial frame in which photon emission would violate energy conservation.
 
BigMacnFries
When an electron makes a bigger jump down it emits a photon of a higher frequency. Apart from the equation e=hf is there an explantion as to why a bigger jump ultimately leads to a narrower interference pattern?
 
jtbell
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For any type of wave, a higher frequency gives you a shorter wavelength. A shorter wavelength gives you a narrower spacing between interference maxima, note for example [itex]n \lambda = d \sin \theta[/itex]
 
BigMacnFries
In classical physics waves are created by something oscillating. Is the frequency of the photon related to anything that oscillates?
 

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