Is There a Way to Prove Normalization of the BCS Equation?

[michaeltorrent]

hi

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

does anyone 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?

 

[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.

 

[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?

 

[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.

 

[Gokul43201]

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.

 

[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?