Double Slit Experiment Mathematics

  • #1
James Brady
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4
Electrons are shot thru two slits separated by a distance s at a screen a distance ##z_0## away. The wave function for the particles is proportional to ## e^{ik \sqrt{(x-s/2)^2+z_0^2}} +e^{ik \sqrt{(x+s/2)^2+z_0^2}}##

Taking the first one, we can manipulate the square root algebraically ##z_0\sqrt{1 + (x-s/2)^2/z_0^2}##, this is where I run into issues, according to the text we can use the fact that ##\sqrt{1+a} = 1 + a/2## in a power series expansion to get ##z_0 + (x-s/2)^2/2 z_0## for small values of x.

I don't know how they got that. When I do a Taylor series expansion on ##z_0\sqrt{1 + (x-s/2)^2/z_0^2}## I get ##z_0\sqrt{1 + (s/2)^2/z_0^2} - \frac{-s/2}{z_0*\sqrt{\frac{-2/2}{z_0^2}+1}} *x## ...

How did they come up with such a simple answer using ##\sum_{n=0}^1 \frac{f^n(0)}{n!} x^n## ?

Am I using the Taylor series wrong?

The book goes on to use ##2cos(\theta) = exp(i \theta) + exp(-i \theta)## to obtain a final wave equation (at the screen) of ##exp(i \theta) cos(ksx/2z_0)## where k is the wave number ##\lambda = 2 \pi/k##

How did they come up with that? I don't see the algebra at all.

Sincerely, confused.
 

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  • #2
PeroK
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Taking the first one, we can manipulate the square root algebraically ##z_0\sqrt{1 + (x-s/2)^2/z_0^2}##, this is where I run into issues, according to the text we can use the fact that ##\sqrt{1+a} = 1 + a/2##
That can be obtained from the binomial theorem, which is a special case of Taylor series.
The book goes on to use ##2cos(\theta) = exp(i \theta) + exp(-i \theta)## to obtain a final wave equation (at the screen) of ##exp(i \theta) cos(ksx/2z_0)## where k is the wave number ##\lambda = 2 \pi/k##

How did they come up with that? I don't see the algebra at all.
Do you know Eulers identity for the complex exponential?
 
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  • #3
James Brady
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When I apply the Taylor series about x = 0:

##\frac{f(0)}{0!}0^0:####z_0 \sqrt{1 + (0-s/2)^2/z_0^2} = z_0 \sqrt{1 + (-s/2)^2/z_0^2}##
##\frac{f^{(1)}(0)}{1!}0^1:##
##\frac{0-s/2}{z_0 \sqrt{1 + (0-s/2)^2/z_0^2}}## = ##\frac{-s/2}{z_0 \sqrt{1 + (-s/2)^2/z_0^2}}##
Adding these values gives a different answer than the book.

I'm not exactly sure what a fractional binomial theorem looks look, I understand the whole number BT, but for square roots and what not I can't find a formula online.

...

I know Euler's identity ##exp(i \pi) + 1 = 0## I just don't see how that applies here.
 
  • #4
PeroK
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You need to use the Taylor series around ##x = \frac s 2##.

You can make the calculation easier by the substitution ##a = \frac{(x -\frac s 2)^2}{z^2}##, as the book indicates. Then use the Taylor series about ##a = 0##.

Try Eulers formula!
 

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