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Cooper pairs and BCS

  1. Jun 7, 2010 #1

    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

    \left| {k \uparrow } \right\rangle \quad \text{and} \quad \left| { - k \downarrow } \right\rangle.

    Where does the time-reversal come into play?

    Last edited: Jun 7, 2010
  2. jcsd
  3. Jun 7, 2010 #2
    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.
  4. Jun 7, 2010 #3
    Ahh, I see. I'll take a closer look at the operator.

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