Bcs theroy, just an easy algebraic manipulation

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SUMMARY

The discussion centers on deriving formula 3.35 from Tinkham's book on superconductivity through algebraic manipulation. Participants suggest using the equations |u|^2 + |v|^2 = 1 and |v/u| = (E - xi)/|Delta| as starting points. A more efficient approach involves utilizing equation (3.29c) alongside the normalization condition, leading to simplified calculations for u^2 and v^2. This method streamlines the derivation process significantly.

PREREQUISITES
  • Understanding of algebraic manipulation techniques
  • Familiarity with Tinkham's book on superconductivity
  • Knowledge of the normalization condition in quantum mechanics
  • Basic concepts of energy (E), xi, and Delta in the context of superconductivity
NEXT STEPS
  • Review Tinkham's equations, particularly (3.29c) and its applications
  • Study normalization conditions in quantum mechanics
  • Explore algebraic manipulation methods in physics
  • Investigate the implications of |u|^2 + |v|^2 = 1 in quantum systems
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This discussion is beneficial for physics students, researchers in superconductivity, and anyone interested in advanced algebraic techniques in quantum mechanics.

Derivator
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bcs theroy, just an "easy" algebraic manipulation

Hi folks,

i'm sitting here for hous and I'm wondering, how one can find formula 3.35' on page 61 in Tinkham's book: http://books.google.de/books?id=k6A...superconductivity&pg=PA61#v=onepage&q&f=false

I think, it should be possible to find this formula just with some algebraic manipulation of

|u|^2 + |v|^2 = 1
and
|v/u| = (E - xi)/|Delta|

However, I can't see how... I'm getting nuts...

--
derivator
 
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Derivator said:
I think, it should be possible to find this formula just with some algebraic manipulation of

|u|^2 + |v|^2 = 1
and
|v/u| = (E - xi)/|Delta|

It is possible, but it's much easier to do as the author suggested and use (3.29c) together wuith the normalization condition:

[tex]u^2+v^2=1[/itex]<br /> [tex]v^2-u^2=-\frac{\xi}{E}[/tex]<br /> <br /> ...add the two equations together, divide by 2 and you get [itex]v^2[/itex]<br /> ...subtract the 2nd from the 1st and divide by 2 and you get [itex]u^2[/itex][/tex]
 


Thank you!
 

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