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Tomsk
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Homework Statement
The question starts with a diagram of a source S emitting particles towards two slits 1 and 2 a distance d apart. They are a distance L from the screen. Let the amplitude for a particle from source S to reach slit 1 be <1|S>, to get from slit 1 to point x on the screen <x|1> etc. Assume each slit is infinitely narrow but that if a particle hits the slit it goes through with amplitude 1.
i)Write down expressions for the amplitude for a particle to leave S and reach x via slit 1, via slit 2, and via both slits.
ii)Assume that the source is infinitely far away from the slits (so that the probability amplitudes for the particles at slits 1 and 2 are equal and in phase) Given that the amplitude for a particle of momentum p starting at y to end at x is
\langle x|y\rangle=\frac{1}{|x-y|}e^{ip.(x-y)/\hbar}
Compute exactly the probability distribution P(x) for particles arriving at the screen when both slits are open.
Homework Equations
P(x)=|\langle\phi|\psi\rangle|^2
The Attempt at a Solution
i)S -> 1 -> x = <x|1><1|S>
S -> 2 -> x = <x|2><2|S>
S -> 1 and 2 -> x = <x|1><1|S>+<x|2><2|S>
ii)P(x) = |\langle x|1\rangle\langle 1|S\rangle+\langle x|2\rangle\langle2|S\rangle|^2
=|\langle x|1\rangle\langle 1|S\rangle|^2 + |\langle x|2\rangle\langle 2|S\rangle|^2 + \langle x|1\rangle\langle 1|S\rangle\langle 2|x\rangle\langle S|2\rangle+\langle 1|x\rangle\langle S|1\rangle\langle x|2\rangle\langle 2|S\rangle
=\left(\frac{1}{|x-(1)||(1)-S|}\right)^2 +\left(\frac{1}{|x-(2)||(2)-S|}\right)^2+\frac{2}{|x-(1)||(1)-S||x-(2)||(2)-S|}
Then I said:
|x-(1)| = |x\hat\vec{i}-d/2\hat\vec{i}-L\hat\vec{j}|=\sqrt{(x-d/2)^{2}+L^{2}}
|x-(2)| = \sqrt{(x+d/2)^{2}+L^{2}}
|(1)-S| = |(2)-S| = \sqrt{(L-S)^{2}+(d/2)^{2}}
(Taking the origin at the screen at x=0.) I think that's right anyway. I can't seem to simplify it much further- if S is infinitely far away don't all those terms go to zero?! And I was expecting a sinc^2 term somewhere, but I can't see where it would come from. Any help much appreciated.
