# Bogoliubov transformation

1. Jul 9, 2012

### arojo

Hello everybody,

I am having some trouble to understand the significance behind the Bogoliubov transformation in the case of the BCS theory in the mean field approximation (MFA). Without going into all the details of the calculation the final result is a Bogoliubov transformation like:

(1) $γ_{k, σ}=α_{k}^{*} c_{k, σ}+β_k c_{-k,-σ}^{\dag}$
(2) $γ_{-k, -σ}^{\dag}=-β_{k}^{*} c_{k, σ}+α_k c_{-k,-σ}^{\dag}$

where $γ_{k, σ}$ in the annihilation operator in the new basis, α and β are the Bogoliubov coeffs.

My question is the next, once we get the previous result why can not we get equation (2) from (1) by making the change of variables k→-k, σ→-σ and the applying $^{\dag}$ in both sides?
I know that the vector in (1) and (2) are eigenvectors of the BCS hamiltonian, therefore, linearly independent from each other. But I still do not get why we can not do the transformation I mentioned before.
Thanks

2. Jul 9, 2012

### nonequilibrium

Can't help you with the specifics, but I do know that you'll be more likely to get help if you make the latex work properly. The tags you'd want to use are
Code (Text):
$latex here$
.

3. Jul 10, 2012

### DrDu

I think you can do this and the second equation is given for convenience of the reader. Who sais you can't?

4. Jul 10, 2012

### arojo

Hey,

I received some comments asking for a more clear latex expression. So equations (1) and (2) become:

$\gamma_{k, σ}=\alpha_{k}^{*} c_{k, σ}+\beta_{k} c_{-k,-σ}^{\dagger}$ (1)

$\gamma_{-k, -σ}^{\dagger}=-\beta_{k}^{*} c_{k, σ}+\alpha_{k} c_{-k,-σ}^{\dagger}$ (2)

Where $c_{k, σ}$ and $\gamma_{k, σ}$ are the fermionic annihilation operators in the old and Bogoliubov basis, respectively. $k$ and $σ$ are the quasi-momentum and the spin projection in the z direction (which takes values of up and down), respectively.

$\alpha_{k} = \sqrt{ \frac{1}{2} \left( 1 + \frac{ε^{-}_{k}}{E_k} \right)}$ (3)

$\beta_{k} = \sqrt{ \frac{1}{2} \left( 1 - \frac{ε^{-}_{k}}{E_k} \right)}$ (4)

$ε^{-}_{k} = \frac{ε_{k} - ε_{-k}}{2}$ (5)

$ε^{+}_{k} = \frac{ε_{k} + ε_{-k}}{2}$ (6)

$E_{k} = \sqrt{(ε^{+}_{k})^2 + \Delta^2_{k}}$ (7)

Now from the set of equations (3)-(7) we can see that

$\alpha_{-k} = \beta_{k}$ (8)

So if to equation (1) we apply $^{\dagger}$ both sides we get:

$\gamma_{k, σ}^{\dagger}=\alpha_{k} c_{k, σ}^{\dagger}+\beta_{k}^{*} c_{-k,-σ}$ (9)

If now we change $k → -k$ and $\sigma → -\sigma$, we obtain:

$\gamma_{-k, -σ}^{\dagger}=\alpha_{-k} c_{-k, -σ}^{\dagger}+\beta_{-k}^{*} c_{k,σ}$ (10)

But because of (8) we can rewrite equation (10) as

$\gamma_{-k, -σ}^{\dagger}=\beta_{k} c_{-k, -σ}^{\dagger}+\alpha_{k}^{*} c_{k,σ}$ (11)

which is different from (2).

So I repeat my question, why is this procedure incorrect? Which is the wrong assumption or step in the calculation?
Thanks

5. Jul 10, 2012

### arojo

I am not sure of your comment, because if I follow the definitions for α and β, I can do the following

$\epsilon_{k}^- = \frac{\epsilon_{k} -\epsilon_{-k}}{2}$

So if we replace $k → -k$ we obtain:

$\epsilon_{-k}^-= - \epsilon_k^-$

For $\epsilon_{-k}^{+}= \epsilon_k^+$, therefore is we chose α and β to be real,

$\alpha_{-k}= \sqrt{\frac{1}{2} \left( 1 + \frac{\epsilon_{-k}^-}{E_{-k}}\right)} = \sqrt{\frac{1}{2} \left( 1 - \frac{\epsilon_{k}^-}{E_{k}}\right)} = \beta_k$

$\beta_{-k}= \sqrt{\frac{1}{2} \left( 1 - \frac{\epsilon_{-k}^-}{E_{-k}}\right)} = \sqrt{\frac{1}{2} \left( 1 + \frac{\epsilon_{k}^-}{E_{k}}\right)} = \alpha_k$

Then in this skim your comment turn to be wrong. In any case, I would thank you if you could point out an error in my calculation.

Maybe we do not have the same definition, I am using the definitions for α and β from my last post.

Thanks

6. Jul 10, 2012

### arojo

Maybe I was not very clear, my question is not related to the fact if it is allowed or not to follow the procedure I propose. My question is why we do not get equation (2) from (1) if we do the proposed change of variables?

7. Jul 10, 2012

### arojo


Actually the problem is little bit different. It will be more like:

$\gamma_{-k,-\sigma}^{\dagger} = \gamma_{k,\sigma}$

Which could have some physical meaning as:

Annihilating a particle with a momentum $k$ and an spin $\sigma$ is equivalent to create a particle with a momentum $-k$ and an spin $-\sigma$.

But still it does not explain the mathematical problem even if it could have some physical sense.

Thanks

8. Jul 10, 2012

### DrDu

There must be an error with your definitions, specifically in the dependence of alpha and beta on epsilon_k. Elektronic states with opposite value of k are degenerate due to time inversion symmetry, so
$\epsilon^-_k=(\epsilon_k−\epsilon_{-k})/2=0$
$\epsilon^-_k$ has to be replaced by $\epsilon_k=\epsilon^+_k$.
That's also what I found in the literature.

9. Jul 10, 2012

### arojo

I agree with you that there is a problem, but the definitions I am using are the most standard, actually for confirmation I took them from Bruus "Many body quantum theory in condensed matter". So in principle the definitions are right, besides error of my part recopying them.
Then if we change $\epsilon_{k}^{-}$ for $\epsilon_{k}^{+}$ gives the same problem even more dramatically.

10. Jul 11, 2012

### DrDu

11. Jul 13, 2012

### DrDu

I think I can now pinpoint the problem. I think it is not correct to use a general $\sigma$ as label for the anihilators/creators. Cooper uses explicitly either $\uparrow$ or $\downarrow$. That means that you cannot generate e.g. $\gamma_{k,\uparrow}$ from $\gamma^\dagger_{-k,\downarrow}$ by complex conjugation and permutation of labels, but you have to take the complex conjugate and rotate 180 deg. around an axis which is perpendicular to both k and the z-axis. It is clear that such a rotation will transform $|\uparrow\rangle$ into $|\downarrow\rangle$ and vice versa up to a phase factor. As repetition of such a transformation (i.e. rotation by 360 deg) transforms the spin wavefunction into minus itself, there is a relative phase of -1 in the transformation of $|\uparrow\rangle$ and $|\downarrow\rangle$. Alternatively, this can be understood in terms of the behaviour under time inversion (Kramers degeneracy).

12. Jul 13, 2012

Thanks