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Probability Density for Wavefunctions undergoing phase shifts.

  • #1
Sorry for not using template but you should find everything in the image provided:

Hey guys. All of the info for the problem is in a picture.

I've tried working on this for ours and I still can't seem to get the trig identities right :(

http://img208.imageshack.us/img208/1770/assignmentquestion2.jpg [Broken]

NOTE THAT THERE SHOULD BE ANOTHER BRACKET ON THE VERY END OF THE EQUATION FOR THE PROBABILITY DENSITY. IT SHOULD HAVE sin(delta)), NOT sin(delta) AS IT CURRENTLY HAS.

from that final step, i've done many things by both hand and scientific notebook and I just can't seem to get things to simplify down properly. There is no way I could possibly post all of the different things I've tried but don't worry, I'm not simply looking for a copy-paste answer into homework. I want to be able to understand the working.

Please clarify my initial working and steer me in the correct direction. I'm pretty sure that I understand the physics, it's just the maths....
 
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Answers and Replies

  • #2
dx
Homework Helper
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Hi Einstein2nd,

What you've done so far is correct. Now use Euler's formula exp(ix) = cos(x) + i sin(x). Once you've separated the real and imaginary parts using this formula, add their squares to get the answer.

The only trig identity you will need is cos²(x) + sin²(x) = 1.
 
Last edited:
  • #3
Thank you for your help so far. I'm not getting the right answer though. When doing your above procedure. I'm using Euler's formula on what I have inside the brackets in the last line of my opening post. This gives me (1 + icos(delta) - sin(delta)) because there was that existing i out the front.

I seperated the real and imaginary parts and squared each of them (not sure why this is legit) to get:

(1-sin(delta))^2 + (icos(delta))^2 = -cos2(delta) + sin2(delta) - 2sin(delta) + 1

this either isn't correct or I'm missing another step of simplification. I can't use that Pythag identity yet.
 
  • #4
dx
Homework Helper
Gold Member
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Hi,

It should be (-1 + i cos(δ) - sin(δ)), you forgot the minus before 1. And the absolute value squared of a complex number (x + iy) is x² + y², not x² + (iy)².
 
  • #5
so you're saying I should do:

|(-1-sin(delta))^2| + |(icos(delta)^2|

That does give me sin2(delta) + 2sin(delta) + 1 + cos(delta) if I do it as you say.

What I'd actually found before your most recent post is if I multiply what I have there by its complex conjugate. Please check if this is right:

(-1+ie^(i delta)) * (-1 - ie^(-i delta)

=ie^(-iδ) -ie^(iδ) + e^(-iδ)e^(iδ)+1

=2sinδ+(cosδ-isinδ)(cosδ+isinδ)+ 1

=2sinδ+2

That gives me 2sin(delta) + 2

I also tried multiplying with complex conjugate after taking into sin and cos first.

(icosδ - sinδ - 1)(icosδ - sinδ - 1)

=cos²δ + sin²δ + 2sinδ + 1

=2sinδ+2

Is what I've done alright? All I did was put a negative in front of i whereever it came up (twice when in exponential form, once when already converted to trig first)
 
  • #6
dx
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Yes it looks correct.
 
  • #7
Thank you very much for your time.
 

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