Undergrad Double Slit Experiment Mathematics

Click For Summary
The discussion focuses on the mathematical interpretation of the double slit experiment involving electrons, particularly the manipulation of the wave function and Taylor series expansions. Participants express confusion over the application of the Taylor series and the simplification of the square root term in the wave function. The conversation highlights the use of the binomial theorem and Euler's identity to derive the final wave equation at the screen. Clarifications are provided on how to correctly apply the Taylor series and the significance of substituting variables for simplification. The thread concludes with a suggestion to utilize Euler's formula for further understanding.
James Brady
Messages
106
Reaction score
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.

confused.
 
Physics news on Phys.org
James Brady said:
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.
James Brady said:
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?
 
  • Like
Likes vanhees71 and 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.
 
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!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 7 ·
Replies
7
Views
625
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
877
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
8K