I Deriving Bogoliubov transformations correctly

[pines-demon]

Everytime that I deal with Bogoliubov transformations I get some sign wrong or a missing assumption. Let's take the case for fermions
$$c_{\mathbf p \sigma}=u_{\mathbf p \sigma} b_{\mathbf p \sigma} +v_{\mathbf p \sigma} b^\dagger_{-\mathbf p \bar{\sigma}}$$
where $\sigma$ is the spin variable, $\bar{\sigma}$ the opposite one and $b$ are the Bogoliubov operators.

Per commutation relations
$$|u_{\mathbf p \sigma}|^2+|v_{\mathbf p \sigma}|^2=1.$$

Now my deal is how to derive more properties.

Take the example of the BCS s-Hamiltonian:
$$H=\sum_{\mathbf p \sigma}\epsilon c^\dagger_{\mathbf p \sigma}c_{\mathbf p \sigma} -\sum_{\mathbf p} \Delta c_{\mathbf p \uparrow}^\dagger c^\dagger_{-\mathbf p\downarrow}+\Delta^* c_{-\mathbf p -\downarrow}^\dagger c^\dagger_{\mathbf p\uparrow}$$

No spin dependence so I drop the $\sigma$ dependence of the $u_{\mathbf p\sigma}=u_{\mathbf p}$, I guess that by particle-hole symmetry I can $u_{-\mathbf p}=u_{\mathbf p}$. So that simplifies things, but the Hamiltonian has imaginary term $\Delta$ so I guess I can't choose $u_{\mathbf p}$ real.

Then I would naively try
$$\begin{pmatrix}c_{\mathbf p\uparrow}^\dagger& c_{-\mathbf p\downarrow} \end{pmatrix}\begin{pmatrix}\epsilon& -\Delta\\-\Delta^* & -\epsilon \end{pmatrix}\begin{pmatrix}c_{\mathbf p\uparrow}\\ c^\dagger_{-\mathbf p\downarrow} \end{pmatrix}=\begin{pmatrix}b_{\mathbf p\uparrow}^\dagger& b_{-\mathbf p\downarrow} \end{pmatrix}\bar{U}\begin{pmatrix}\epsilon& -\Delta\\-\Delta^* & -\epsilon \end{pmatrix}U\begin{pmatrix}b_{\mathbf p\uparrow} \\ b^\dagger_{-\mathbf p\downarrow} \end{pmatrix}$$

and then find diagonal and off-diagonal terms to fix the values of $U$, where
$$U=\begin{pmatrix}u_{\mathbf p}&v_{\mathbf p}\\v^*_{\mathbf p} & u^*_{\mathbf p}\end{pmatrix}$$
now here is where more sources change signs. To me the matrix is ok as it is, if there is an extra sign in the elements, can be fixed by the complex conjugation.
Secondly, I understand that $U$ is not necessarily unitary, but then what is $\bar{U}$? Is it $U^{-1}$? Then I get:
$$U^{-1}=\frac{1}{|u_{\mathbf p}|^2-|v_{\mathbf p}|^2}\begin{pmatrix}u^*_{\mathbf p}&-v_{\mathbf p}\\-v^*_{\mathbf p} & u_{\mathbf p}\end{pmatrix}$$

What do I do with that determinant? I don't see it in most sources. Also how is this justified?
Clearly if
$$c_{\mathbf p \uparrow}=u_{\mathbf p} b_{\mathbf p \uparrow} +v_{\mathbf p} b^\dagger_{-\mathbf p \downarrow}$$
then taking complex conjugate
$$c_{\mathbf p \uparrow}^\dagger=u^*_{\mathbf p} b^\dagger_{\mathbf p \uparrow} +v^*_{\mathbf p} b_{-\mathbf p \downarrow}$$
but using $\bar{U}=U^{-1}$ as in the matrix equation above I get
$$c_{\mathbf p \uparrow}^\dagger=\frac{1}{|u_{\mathbf p}|^2-|v_{\mathbf p}|^2}(u^*_{\mathbf p} b^\dagger_{\mathbf p \uparrow} -v^*_{\mathbf p} b_{-\mathbf p \downarrow})$$

What is wrong here? How do I get Bogoliubov transformation right? What is a good source about general Bogoliubov transformations?

 

[div_grad]

If you set up two vectors $\gamma$ and $\beta$,
$$
\gamma = \begin{pmatrix} c_{\mathbf{p} \uparrow} \\ c^{\dagger}_{-\mathbf{p} \downarrow} \end{pmatrix} \quad{\rm{,}}\quad \beta = \begin{pmatrix} b_{\mathbf{p} \uparrow} \\ b^{\dagger}_{-\mathbf{p} \downarrow} \end{pmatrix} \rm{,}
$$
and relate them as $\gamma = U \beta$, then Hermitian conjugate gives $\gamma^{\dagger} = \beta^{\dagger} U^{\dagger}$. And if you compute $U^{\dagger}$ according to the matrix form of $U$ given in OP, then you readily obtain this result

(...) then taking complex conjugate
$$c_{\mathbf p \uparrow}^\dagger=u^*_{\mathbf p} b^\dagger_{\mathbf p \uparrow} +v^*_{\mathbf p} b_{-\mathbf p \downarrow}$$

I believe there is no need to compute $U^{-1}$ as $U$ is in general not unitary.

 

[div_grad]

What do I do with that determinant?

On Wikipedia (https://en.wikipedia.org/wiki/Bogoliubov_transformation), there result is that if ${\rm{det}}\, U = 1$ then the commutation relations between $c$'s and $b$'s are preserved (the Wiki notation is, of course, different from OP):

 

[pines-demon]

If you set up two vectors $\gamma$ and $\beta$,
$$
\gamma = \begin{pmatrix} c_{\mathbf{p} \uparrow} \\ c^{\dagger}_{-\mathbf{p} \downarrow} \end{pmatrix} \quad{\rm{,}}\quad \beta = \begin{pmatrix} b_{\mathbf{p} \uparrow} \\ b^{\dagger}_{-\mathbf{p} \downarrow} \end{pmatrix} \rm{,}
$$
and relate them as $\gamma = U \beta$, then Hermitian conjugate gives $\gamma^{\dagger} = \beta^{\dagger} U^{\dagger}$. And if you compute $U^{\dagger}$ according to the matrix form of $U$ given in OP, then you readily obtain this result

I believe there is no need to compute $U^{-1}$ as $U$ is in general not unitary.

Calculating just $U^\dagger$ does not give the correct diagonalization for the BCS Hamiltonian, you can check.

Edit: why the determinant needs to be 1 ?

 

[pines-demon]

On Wikipedia (https://en.wikipedia.org/wiki/Bogoliubov_transformation), there result is that if ${\rm{det}}\, U = 1$ then the commutation relations between $c$'s and $b$'s are preserved (the Wiki notation is, of course, different from OP):
View attachment 362831

The line you are citing is for bosons, obviously differs from that of fermions.

 

[div_grad]

The line you are citing is for bosons, obviously differs from that of fermions.

Right.

 

[Demystifier]

Bogoliubov transformation is unitary. For bosons, especially in quantum optics, this unitary transformation is known as squeezing. But I am talking about the transformation of states (on which the particle creation and destruction operators act), so I'm not sure if this note is relevant here.

 

[pines-demon]

Bogoliubov transformation is unitary. For bosons, especially in quantum optics, this unitary transformation is known as squeezing. But I am talking about the transformation of states (on which the particle creation and destruction operators act), so I'm not sure if this note is relevant here.

Actually according to various sources, Bogoliubov transformations of the boson operators (not of the states), are symplectic. See for example https://www.cambridge.org/core/book...ansformation/26C26F3B060A025B7550C8B766349F27

This note also says that Bogoliubov transformation for fermions should be unitary (more specifically orthogonal), meaning I have a wrong sign in my starting $U$, but I would like then to know how to show that.

 

[div_grad]

Edit: why the determinant needs to be 1 ?

In the case of bosons, apparently to ensure that the commutation relations are the same for both $c$'s and $b$'s. As you noted, for fermions and the associated anticommutation relations this would no longer seem to be the case, because when $U$ has the form given in OP then ${\rm{det}}\, U = |u_{\mathbf{p}}|^2 - |v_{\mathbf{p}}|^2$, while

Let's take the case for fermions (...) Per commutation relations
$$|u_{\mathbf p \sigma}|^2+|v_{\mathbf p \sigma}|^2=1.$$

with a "$+$" sign between the squared moduli of $u$ and $v$.

But it seems like the above condition for fermions can be imposed ad hoc on the determinant of $U$ by writing the fermionic matrix $U$ with one of its off-diagonal elements containing a minus sign, like in the source you provided. Then ${\rm{det}}\, U = |u_{\mathbf{p}}|^2 + |v_{\mathbf{p}}|^2$ and you can require that ${\rm{det}}\, U = 1$ in order to satisfy the above condition for fermions, i.e., in order to preserve the anticommutation relations between $c$'s and $b$'s.

What I'm getting at is that maybe this is one of these cases in which you need to impose some conditions ad hoc in order for the mathematics employed to be consistent with the fermion statistics. I mean, even the canonical quantization is "naturally" suited for boson fields, because you replace an anti-symmetric Poisson bracket with an anti-symmetric commutator. While in the case of fermion fields, the relevant symmetric anticommutators are either put into the theory "by hand" or by further introducing some Grassmann variables (same goes for choosing normal-ordering rules for fermion operators).

 

[pines-demon]

In the case of bosons, apparently to ensure that the commutation relations are the same for both c's and b's. As you noted, for fermions and the associated anticommutation relations this would no longer seem to be the case, because when U has the form given in OP then detU=|up|2−|vp|2, while

with a "+" sign between the squared moduli of u and v.

But it seems like the above condition for fermions can be imposed ad hoc on the determinant of U by writing the fermionic matrix U with one of its off-diagonal elements containing a minus sign, like in the source you provided. Then detU=|up|2+|vp|2 and you can require that detU=1 in order to satisfy the above condition for fermions, i.e., in order to preserve the anticommutation relations between c's and b's.

What I'm getting at is that maybe this is one of these cases in which you need to impose some conditions ad hoc in order for the mathematics employed to be consistent with the fermion statistics. I mean, even the canonical quantization is "naturally" suited for boson fields, because you replace an anti-symmetric Poisson bracket with an anti-symmetric commutator. While in the case of fermion fields, the relevant symmetric anticommutators are either put into the theory "by hand" or by further introducing some Grassmann variables (same goes for choosing normal-ordering rules for fermion operators).

Mmhm I see your point, however I still think there is something beyond the ad-hoc response. Note that once you fix $c_p$, $c^\dagger_{-p}$ gets fixed by complex conjugation and parity operations. I guess I need to review the particle-hole symmetry.

I guess then that there is some condition that fixes $u_{-p}^*\to u_p^*\,;\,v_{-p}^*\to- v_p^*$.

 

[Demystifier]

Check out the book Grosso, Pastori Parravicini, Solid State Physics, Sec. 18.4.1 The Bogoliubov Canonical Transformation. The calculations are quite detailed so maybe you will find something illuminating.

 

[pines-demon]

Check out the book Grosso, Pastori Parravicini, Solid State Physics, Sec. 18.4.1 The Bogoliubov Canonical Transformation. The calculations are quite detailed so maybe you will find something illuminating.

Thanks, I love how precise the derivation is in this book, it solved another doubt that I had. However it starts with the ad-hoc derivation, the postulate that the transformation is real and the minus sign that I am missing.

I am convinced now that, if
$$c_{\mathbf p \uparrow}=u_{\mathbf p} b_{\mathbf p \uparrow} +v_{\mathbf p} b^\dagger_{-\mathbf p \downarrow}$$
and
$$c^\dagger_{-\mathbf p \downarrow}=u^*_{-\mathbf p} b_{-\mathbf p \downarrow}^\dagger +v_{-\mathbf p}^* b_{\mathbf p \uparrow}$$
then for
$$U=\begin{pmatrix}u_{\mathbf p}&v_{\mathbf p}\\v^*_{-\mathbf p} & u^*_{-\mathbf p}\end{pmatrix}=
\begin{pmatrix}u_{\mathbf p}&v_{\mathbf p}\\-v^*_{\mathbf p} & u^*_{\mathbf p}\end{pmatrix}$$

to be true, I need some condition that makes $u_{-p}^*\to u_p^*\,;\,v_{-p}^*\to- v_p^*$. But this is most of the time not explained (note that I removed spin labels, but maybe it is a condition on that too).

 

[div_grad]

(...) I need some condition that makes $u_{-p}^*\to u_p^*\,;\,v_{-p}^*\to- v_p^*$. But this is most of the time not explained (note that I removed spin labels, but maybe it is a condition on that too).

I don't know if this may help, but under parity transformation $P$ the momentum changes sign (vector) but the spin does not (pseudovector). While under time-reversal $T$ both the momentum and the spin change signs. Maybe there is some interplay here that would justify this minus sign for fermions more convincingly?