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What is the Physical meaning of Bogoliubov transformation

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Yiping
#1
Sep3-08, 08:47 AM
P: 7
http://en.wikipedia.org/wiki/Bogoliubov_transformation

I try to know more about Bogoliubov transformation. I try to find some material to help me understand the transformation more. Most of the material I found are lack of physical explanation for me. Most of the many-body text book(Fetter and Flensberg) only try to convince me that the transformation can simplify the Hamiltonian and eliminate the cross terms. And we can diagonalize the many-body Hamiltonian as quadratic Hamiltonian. What I want to know is about the physical picture of this kind of transformation.
Also, I am confused about the phase term in the wikipedia
[tex]u=e^{i\theta}\cosh r[/tex]
I feel the phase term should mean something, which I do not find any reference about the phase factor.

Can anybody explain/(discuss about) what is Bogoliubov transformation and what is the meaning of the phase factor?Thanks very much. :)
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malawi_glenn
#2
Sep3-08, 12:00 PM
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It depends on what physical system you apply it to I should say, I have done it on Nuclear Structure physics.
Minich
#3
Sep3-08, 06:16 PM
P: 91
Bogoliubov transformation is a very simple thing.
You may consider it as a filter.

Imagine a chorus that sings. And You don't want to hear all singers at once. You are the conductor and you do want to know how well every singer sings during the concert.

Conductor's ear filters the chorus sounding and obtains sounding for every singer.

Physisists prefer name it as "reduction to normal coordinates'. Bogoliubov transformation is one of many such reductions (transformations), mainly used for reduction special hamiltonians in superfluid theory of interacting particles (helium, electron gas) to almost noninteracting quasiparticles (normal coordinates).

Yiping
#4
Sep4-08, 11:15 PM
P: 7
What is the Physical meaning of Bogoliubov transformation

Quote Quote by malawi_glenn View Post
It depends on what physical system you apply it to I should say, I have done it on Nuclear Structure physics.
I am reading the paper of one dimensional fermion system(Voits,1994,One dimensional fermi liquid, on p.998). In the paper, he use it to diagonalize the Hamiltonian. I know how Bogoliubov transformation canceled the cross term, but I cannot imagine the meaning of it. In quantum mechanics for one particle, the unitary transformation usually means transformation to another complete basis, and every state in this basis is for this particle. When it is in many body system, the basis has different number of particles, and the transformation is hard for me to understand.
malawi_glenn
#5
Sep9-08, 12:38 AM
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Well basically it leads yo to easier clalculations since H is diagonalized, easier to get eigenenergies and eigenstates.

In the case of nuclear structure physics, the Bogoliubov transformation leads you to interprent the states as Quasiparticles, where you have one probability to have a hole, and another probability to have a particle. So basically that is what Minich said to you in the last post, you reduce the problem into "normal coordinates" and you'll get an easier way to obtain a solution, but the solution must be interprented (quasiparticles).
Yiping
#6
Sep9-08, 03:31 AM
P: 7
If I write down the transformation
[tex]a_p^{\dagger}=u_p\alpha_p^{\dagger}-v_p\alpha_{-p}[/tex]
[tex]a^{\dagger}[/tex] is the field operator after the transformation. Assume the ground state is
[tex]|g\rangle[/tex]
[tex]a_p^{\dagger}|g\rangle=u_p\alpha_p^{\dagger}|g\rangle-v_p\alpha_{-p}|g\rangle[/tex]
p is momentum. Do you mean the first quasiparticle state is actually the linear composition of a particle with momentum p and a hole with momentum -p with specific weight? And Bogoliubov transformation is actually tuning the weighting such that the whole Hamiltonian can be imagined as noninteracting quasiparticles. I am not sure whether I understand it correctly.
malawi_glenn
#7
Sep9-08, 04:24 AM
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Quote Quote by Yiping View Post
If I write down the transformation
[tex]a_p^{\dagger}=u_p\alpha_p^{\dagger}-v_p\alpha_{-p}[/tex]
[tex]a^{\dagger}[/tex] is the field operator after the transformation. Assume the ground state is
[tex]|g\rangle[/tex]
[tex]a_p^{\dagger}|g\rangle=u_p\alpha_p^{\dagger}|g\rangle-v_p\alpha_{-p}|g\rangle[/tex]
p is momentum. Do you mean the first quasiparticle state is actually the linear composition of a particle with momentum p and a hole with momentum -p with specific weight? And Bogoliubov transformation is actually tuning the weighting such that the whole Hamiltonian can be imagined as noninteracting quasiparticles. I am not sure whether I understand it correctly.

I don't know your context, I know nuclear structure physics.
Minich
#8
Sep15-08, 08:15 AM
P: 91
Don't take Bogoliubov transformation too seriously for fermions.

It is very serious for boson gases because of very useful C-number substitution justification made by Lieb more than 30 years ago:
see recent more methodical paper at
Justification of c–Number Substitutions in Bosonic Hamiltonians
Method analogous to Lieb's (Agarwal and Wolf technics) helped me remove an error in my diploma in 1974.

But this method (C-number substitution, made by Bogoliubov for fermionic electron BCS hamiltonian) is obviously incorrect, because of number of electrons is going to infinity in thermodynamic limit and Lieb's theorem can't be applied. It may by correct occasionally.

Let me quote Gorkov (Ginsburg-Landau-Abricosov-Gorkov theory) paper at BCS 50 years annivesary confference:
http://www.conferences.uiuc.edu/bcs50/PDF/Gorkov.pdf

Bogolyubov announces his SC theory and is invited to talk at the Landau Seminar (October 1957)
Landau's refuses to understand the ad hoc “principle of compensation of the most dangerous
diagrams”, insists on the physics behind it
Exhausted N. N. finally gives up and produces the Leon Cooper's paper
N.N: Bogoliubov.

At the same time Gorkov's method of obtaining GL equations is almost equivalent Bogoliubov's C-number substitution (some operator expressions are substituted by their average over greate canonical ensуmble.

For formular references see attached images.
Bogoliubov suggested replace BCSint hamiltonian (see bcsint image) with effective mean field hamiltonian containing Hartree-Fock part (see hf image) and Heff (see Heff), where GAP function (see gap image) is C-number function.

The resulting hamiltonian is bilinear in creation and annihilation operators and can be diagonalizie by canonical transformation.

By the way in a book about Migdal (in russian, Migdal and Bogoliubov are mentioned in Bardeen's Nobel prize lecture) Migdal said about his talk with Bogoliubov:
АБ Мигдал передавал свой разговор с Боголюбовым:
— Как жаль, что я не догадался сделать Ваше каноническое преобразование, когда понял, что фононы вызывают притяжение между электронами.
На это Боголюбов ответил:
— А я сделал это каноническое преобразование в 1947 году, но не знал про фононы, а с кулоновским отталкиванием получил нулевой результат.
I try to translate:
- It is a pity, that I have not guessed to make your canonical transformation when i has understood, that phonons cause an attraction between electrons.

Bogoliubov has answered:
— I have made this canonical transformation in 1947, but did not know about phonons, and with coulomb repulsion has received zero result.

As for physical meaning of Bogoliubov transformation see Landau's words in Gorkov paper:
Landau... insists on the physics behind it.
Exhausted N. N.(Bogoliubov) finally gives up and produces the Leon Cooper's paper
Attached Thumbnails
bcsint.jpg   hf.jpg   heff.jpg  
lukasch
#9
Sep17-08, 08:51 AM
P: 3
Quote Quote by Yiping View Post
http://en.wikipedia.org/wiki/Bogoliubov_transformation

I try to know more about Bogoliubov transformation.
I would explain it in this way - for a system of noninteracting fermions with hamiltonian

ham = sum_k c^+_k_up c_k_up + c^+-_k_down c^-_k_down

you can apply the fermi-dirac statistic, which says that a one-electron state |i> is occupied with probability
< c^+_i c_i > = f(Ei) = 1/(1+exp( (E_i-mu)/T) ) where mu is chemical potential.

This is however no longer true for a system with interaction terms. Then all the statistical computations would be more complicated.
If the hamiltonian can be written as quadratic form (see bellow) with vectors containing the
electron creation and anihhilation operators, like in the mean-field approximation for BCS hamiltonian, you can transform these vectors into different base, in which the matrix of q-form is diagonal. Then you will have hamiltonian which looks like hamiltonian for
non-interacting fermions.
And you can continue making profit of fermi-dirac statistics.

Assure yourself that

hamiltonian = sum_k psi^T H psi

where
psi^T = ( c^+_k_up, c_-k_down )

H = (
e_k -Delta_k \\
-Delta_k -e_k )

Assume transformation U

psi = U * Gamma

Gamma is vector containing new operators
Gamma^T = ( gamma^+_k0, gamma_k1 )

It is not important which (or if) of gammas is creation or anihillation operator, but this is in accord with textbooks, and it will be convenient in the end. Neither does it matter if you write (k0) (k1) or (k up) (-k down).

The form of U ... search for genaral form of a general unitary 2x2 matrix.
The unitarity, and condition det(U)=+1, ensures that new operators fulfill fermionic commutation relations => the new "quasi-particles" are fermions, too.

U = (u v \\ -v u), try to work out coefficients u, v such that they diagonalize
the meanfield BCS hamiltonian above:

hamiltonian = sum_k Gamma^+ U^+ H U * Gamma
= sum_k Gamma^+ H_diag * Gamma

For case that U is just real... dont you see some analogy with rotations ? :)
Rotation matrix is an orthogonal matrix with det = 1.

Look for lecture notes by prof. Eschrieg from dressden or other on the net
or Stat. phys. by Reichel, or Landau + lifschitz ...
And let me know how you would diagonalize hamiltonian

H = sum_k c^+_k c_k - G_k c^+_(k+Q) c_k ... k+Q = k-Q

Have a good time with B-V transformation :-D
Lukas
KFC
#10
Apr8-09, 12:55 AM
P: 369
I am trying to simplify the single-mode bosonic hamiltonian H = A a^\dagger a + B a^2 + B^* {a^\dagger}^2. I use the transformation shown in http://en.wikipedia.org/wiki/Bogoliubov_transformation.

For making the hamilton diagonal, I have to make the off-diagonal element vanish, but it seems quite difficult to solve the equation. Any example?
peteratcam
#11
Apr8-09, 03:51 AM
P: 171
Work out a general Bogoliubov transformation once in your life, write it down on a piece of paper and keep it safe.

I have done that, and my piece of paper for bosons says:

[tex]
\begin{pmatrix}
a^{\dagger} & a
\end{pmatrix}
\begin{pmatrix}
1 & \gamma \\
\gamma & 1
\end{pmatrix}
\begin{pmatrix}
a \\
a^{\dagger}
\end{pmatrix}
=\begin{pmatrix}
\alpha^{\dagger} & \alpha
\end{pmatrix}
\begin{pmatrix}
\sqrt{1-\gamma^2} & 0\\
0 & \sqrt{1-\gamma^2}
\end{pmatrix}
\begin{pmatrix}
\alpha \\
\alpha^{\dagger}
\end{pmatrix}
[/tex]
when
[tex]
\begin{pmatrix}
a \\
a^{\dagger}
\end{pmatrix}
=\begin{pmatrix}
\cosh\theta & \sinh\theta \\
\sinh\theta & \cosh\theta \\
\end{pmatrix}
\begin{pmatrix}
\alpha \\
\alpha^{\dagger}
\end{pmatrix}
\quad\text{where}\quad \tanh 2\theta = -\gamma
[/tex]

Don't take this on trust, you should check that the transformation preserves bosonic commutation relations, and that it all checks out.

After checking that I haven't made any errors and am not lying to you, simply massage your hamiltonian into the form on the left and read off the answer on the right!
genneth
#12
Apr12-09, 07:23 AM
P: 981
A Bogoliubov transform is just a way to take some quadratic Hamiltonian and solve it in terms of non-interacting particles. Those particles are then mixtures of the original ones (including mixing of particles and holes). That's it; nothing more. It's quite helpful to remember that there aren't any difference between "real" or "fundamental" particles and "quasi" particles. It's not even helpful, and possibly wrong.

A Bogoliubov approximation is when you replace an operator with a c-number because the expectation value of that operator is much greater than one, so it almost commutes. Minich: this is *not* what the OP asked about.
Rilwen
#13
Jun4-09, 07:46 PM
P: 1
Hmm, the physical meaning of Bogoliubov transform... The initial Hamiltonian describes some unstable interacting excitations of the system, which will shortly decay. They are often impossible to investigate experimentall. You do the Bogoliubov transformation to find the stable independent eigenmodes. Not sure how it works for fermions, though.
(Correct me if I'm wrong :)
Petar Mali
#14
Oct10-09, 08:30 AM
P: 290
What is c-number?

uv transformation is also important because
reduces energy of ground state!
Bob_for_short
#15
Oct10-09, 09:28 AM
P: 1,160
Quote Quote by Minich View Post
...Physisists prefer name it as "reduction to normal coordinates'. Bogoliubov transformation is one of many such reductions (transformations), mainly used for reduction special hamiltonians in superfluid theory of interacting particles (helium, electron gas) to almost non-interacting quasiparticles (normal coordinates).
Consider two interacting classical particles, for simplicity, bound with an elastic force:

m1a1 = F(r1-r2)

m2a2 = -F(r1-r2)

These two equations are essentially coupled and are hard to solve. The dynamics of particles is rather complicated.

Then make the following variable changes: introduce R and r = r1 - r2. You obtain:

MtotA = 0,

a = F(r)

The equations for R and r are decoupled and simple, both describe quasi-particles with different masses. When you measure the system inertia and proper frequency, you obtain the quasi-particle characteristics.

Such a variable change is a transition from very "fluctuating" variables to less "fluctuating" ones, with simple dynamics. Both descriptions are equivalent but the physical sense is easily visible and explainable in the quasi-particle formulation.
Minich
#16
Oct11-09, 06:57 AM
P: 91
To Bob_for_short
Why didn't You write down definition for R?

It is a very interesting question:
R=(r1+r2)/2 OR
R=r1*m1/(m1+m2)+r2*m2/(m1+m2) OR
..........
?????
Why You should select your selection?????
In two body problem we have INFINITE number of cases to have r1-r2 for one DF and second DF for R.
Bob_for_short
#17
Oct11-09, 07:32 AM
P: 1,160
Quote Quote by Minich View Post
To Bob_for_short
Why didn't You write down definition for R?

It is a very interesting question:
R=(r1+r2)/2 OR
R=r1*m1/(m1+m2)+r2*m2/(m1+m2) OR
..........
?????
Why You should select your selection?????
In two body problem we have INFINITE number of cases to have r1-r2 for one DF and second DF for R.
Sorry, I thought the definition of the center of inertia coordinate was well known. It is the second expression: R = (r1m1 + r2m2)/Mtot. It follows from the original Newton equations for r1 and r2: you add them and the forces cancel.
Minich
#18
Oct19-09, 05:56 AM
P: 91
Sorry, I thought the definition of the center of inertia coordinate was well known.
Yes, a center of mass definition is well known.

But why do we must use center of mass? May be another choice is more simple and we would have more simple quasiparticles? And may be new quasiparticles are more appropriate to generalize interaction between quasiparticles?

We know that there are INFINITE number of cases to decouple the equtions in two body problem.

Do You know, that if you choose center of mass representation, then you DON'T find ALL the solutions of the two body problem in quantum mechanics?


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