# Bcs equation (superconductor)

1. Nov 23, 2005

### michaeltorrent

hi

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

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

<psi|psi>=1

given uk^2 + vk^2 =1

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

any guide?

2. Nov 23, 2005

### Physics Monkey

The BCS wavefunction can be written as a product of factors like $$u_k | 0 ,k\rangle + v_k | 1,k \rangle$$ 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 $$| 0,k \rangle$$ and $$|1,k \rangle$$ are orthogonal, can you prove that you must have $$u_k^2 + v_k^2 = 1$$ for normalization of the BCS state?

Hope this helps, and if you still have trouble let me know.

Last edited: Nov 23, 2005
3. Nov 23, 2005

### michaeltorrent

thanks. i am new here, please excuse my latex typing if there is error, i am not used to this:

we have:

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

i want to prove
$$<\psi |\psi>=1$$

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?

4. Nov 24, 2005

### Physics Monkey

Each of those factors adds (or removes, in the case of the adjoint) an even number of electrons: zero for the $$u_k$$ part and two for the $$v_k$$ 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. $$(-1)^2 = 1$$. The exception to this rule is when a factor meets its adjoint. If you call $$f_k = u_k + v_k b^+_k$$, then the normalization condition boils down to a bunch of terms like $$f^+_k f_k ,$$ 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
5. Nov 24, 2005

### Gokul43201

Staff Emeritus
Alternatively, (well, actually it's much the same thing), you could show that the state $|1 \rangle \equiv b^{\dagger} |0 \rangle$ is orthonormal to the vacuum state, given that the sum of the squares adds to 1.

6. Nov 25, 2005

### michaeltorrent

is [b,b*]=1-ndown-nup ?
then we have bb* - b*b = 1 -ndown - nup
so
bb* = b*b + 1 - ndown - nup
and b|0> = 0 so b*b terms dissapear?

Last edited: Nov 25, 2005
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