Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Bcs equation (superconductor)

  1. Nov 23, 2005 #1

    i have superconductor question, need someone familiar with this field.

    does any one know how to prove normalization of bcs equation?


    given uk^2 + vk^2 =1

    i went through the heisenberg algebra but still can't solve it.

    any guide?
  2. jcsd
  3. Nov 23, 2005 #2

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    The BCS wavefunction can be written as a product of factors like [tex] u_k | 0 ,k\rangle + v_k | 1,k \rangle [/tex] where the states refer to an unoccupied or occupied Cooper pair (k spin up, -k spin down). The normalization condition boils down the normalization of these factor states. Using the fact the [tex] | 0,k \rangle [/tex] and [tex] |1,k \rangle [/tex] are orthogonal, can you prove that you must have [tex] u_k^2 + v_k^2 = 1[/tex] for normalization of the BCS state?

    Hope this helps, and if you still have trouble let me know.
    Last edited: Nov 23, 2005
  4. Nov 23, 2005 #3
    thanks. i am new here, please excuse my latex typing if there is error, i am not used to this:

    we have:

    [tex] |\psi>=\prod_k [u_k + v_k b*]|0> [/tex]

    i want to prove
    [tex] <\psi |\psi>=1 [/tex]

    i do the product but still cannot get rid of b and b* terms. is there an algerbraic trick there? thanks.

    btw does anyone know why whenever i open a thread the screen keeps on scrolling down to the bottom page until the download finishes? is something wrong with my browser? or is it normal for this website?
  5. Nov 24, 2005 #4

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Each of those factors adds (or removes, in the case of the adjoint) an even number of electrons: zero for the [tex] u_k [/tex] part and two for the [tex] v_k [/tex] part. Because all the states are different, any two of those factors can be exchanged since each such exchange involves an even number of fermion exchanges i.e. [tex] (-1)^2 = 1 [/tex]. The exception to this rule is when a factor meets its adjoint. If you call [tex] f_k = u_k + v_k b^+_k [/tex], then the normalization condition boils down to a bunch of terms like [tex] f^+_k f_k ,[/tex] and by exchanging such terms (see above) you should be able to evaluate them easily using the special properties of the vacuum state.
    Last edited: Nov 24, 2005
  6. Nov 24, 2005 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Alternatively, (well, actually it's much the same thing), you could show that the state [itex]|1 \rangle \equiv b^{\dagger} |0 \rangle [/itex] is orthonormal to the vacuum state, given that the sum of the squares adds to 1.
  7. Nov 25, 2005 #6
    is [b,b*]=1-ndown-nup ?
    then we have bb* - b*b = 1 -ndown - nup
    bb* = b*b + 1 - ndown - nup
    and b|0> = 0 so b*b terms dissapear?
    Last edited: Nov 25, 2005
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook