What is the relation between wave function on a photon

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SUMMARY

The discussion centers on the relationship between the wave function of a photon and its classical wave equation, particularly in the context of the double-slit experiment. The final wave equation derived is y=2A*cos(u/2)*sin(kx-wt+u/2), where 'u' represents the phase difference and 'q' the path difference. The intensity of the wave is proportional to the square of this wave equation, linking it to the probability of a photon hitting a certain point on the screen. The conversation also touches on the implications of superposition in quantum mechanics and its connection to classical wave behavior.

PREREQUISITES
  • Understanding of wave equations, specifically y=Asin(kx-wt)
  • Familiarity with the double-slit experiment and its implications in quantum mechanics
  • Knowledge of phase difference and path difference in wave interference
  • Basic grasp of quantum mechanics, particularly the wave function and superposition principle
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  • Explore the derivation of the Schrödinger equation and its relation to classical wave equations
  • Study the Helmholtz Equation and its applications in quantum mechanics
  • Investigate the concept of superposition in quantum systems and its experimental validations
  • Learn about the mathematical formulation of wave functions in quantum mechanics
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Physicists, quantum mechanics students, and anyone interested in the foundational principles of wave-particle duality and interference phenomena.

Tigersharkmks
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... and its classical wave equation?
Suppose in our double sit experimental setup with the usual notion of d,D we have a light of known frequency (v) and wavelength (L)- so its y=Asin(kx-wt). It passes through the two hole and move ahead doing the usual interference stuff, so final wave equation will be y =y1 + y2 = 2A*cos(u/2)*sin(kx-wt+u/2) where u is the phase difference. We can convert phase difference 'u' to path difference 'q'. Now we choose the point of interest on screen (s) which essentially depend upon q. So the final y at s is y=2A*cos(a*s)sin(kx-wt+a*s) where a is constant. Now this y implies a set of wave (because different values s corresponds to interference of different wave with different phase difference) and is function of variable s,x,t. Now since I have put screen at some x (=D), therefore the y reduces to function of two variable s,t. Now rewriting y=2A*cos(u/2)sin(u/2 -wt +kD) which is also a wave equation (but with different meaning). Now the screen acts as our x-axis (or to be precise u-axis). Now intensity is proportional to to sq of wave eqn written above which in turns depends upon u. And what we know is intensity is proportional to number of photon, so we postulate that probability of a photon to hit certain u is proportional to intensity = wave eq ^2. Now is the function y I have written above is psi(wave function) of quantum mechanics with u acting as x(in psi). If not, then what is their relation between them. I have other things to ask depending upon your response. Thank you!
 
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Forget this wave particle analysis of the double slit:
http://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

What's really going on is this.

Consider just one slit for a moment and we will use electrons for definiteness. Just behind the slit it has a definite position along the screen. This means its momentum along the screen is totally unknown so it will scatter in an unknown direction and hit the screen at an unknown location. Now imagine you have two slits - its not in a definite position behing the screen because a detector placed just behing the screen will give either one or the other slit. This means its state is a superposition of the states of exact position - and that why you get interference.

If you insist on that kind of analysis see the following:
http://arxiv.org/abs/1009.2408

But its a more difficult approach showing the Schroedinger's equation reduces to the Helmholtz Equation.

Thanks
Bill
 

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