I Double Slit Experiment Mathematics

[James Brady]

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.

confused.

 

[PeroK]

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?

 

[James Brady]

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.

 

[PeroK]

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!