Cooper pairs and BCS

  • Thread starter Niles
  • Start date
  • #1
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Hi

In my book it says that a Cooper pair is formed by a single electron state and its time-reversed counterpart. The way I have understood it, a Cooper pair is formed by the states

[tex]
\left| {k \uparrow } \right\rangle \quad \text{and} \quad \left| { - k \downarrow } \right\rangle.
[/tex]

Where does the time-reversal come into play?

Best,
Niles.
 
Last edited:

Answers and Replies

  • #2
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Try to act with the time-reversal operator on the state [tex] \left| {k \uparrow } \right\rangle \quad [/tex] and see what you get. (You first need to look up how this operator works.)

Edit: Okay I might as well reveal that you should get [tex]
T\left| {k \uparrow } \right\rangle \quad = \quad \left| { - k \downarrow } \right\rangle.
[/tex]
 
  • #3
1,868
0
Try to act with the time-reversal operator on the state [tex] \left| {k \uparrow } \right\rangle \quad [/tex] and see what you get. (You first need to look up how this operator works.)

Edit: Okay I might as well reveal that you should get [tex]
T\left| {k \uparrow } \right\rangle \quad = \quad \left| { - k \downarrow } \right\rangle.
[/tex]
Ahh, I see. I'll take a closer look at the operator.

Thanks.
 

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