Bogoliubov transformation / Interpretation of diagonalized Hamiltonian

In summary, the conversation is about a diagonalized Hamiltonian involving fermionic creation and annihilation operators. The solution for E_k is known to be \sqrt{\Delta^2 +\epsilon_{k}^2}, but the speaker is unsure of how to arrive at this result. They mention two equations involving the unknowns u and v, and suggest solving for them and plugging them into the equation for E_k.
  • #1
Abigale
56
0
Hey,

I consider a diagonalized Hamiltonian:

[itex]H=\sum\limits_{k} \underbrace{ (\epsilon_{k} u_{k}^2 -\epsilon_{k} v_{k}^2 -2\Delta u_{k} v_{k} )}_{E_{k}}(d_{k \uparrow}^{\dagger}d_{k \uparrow} + d_{k \downarrow}^{\dagger}d_{k \downarrow}) +const [/itex]
with fermionic creation and annihilation operators.

From solution I know that: [itex]E_{k} =\sqrt{\Delta^2 +\epsilon_{k}^2}[/itex] but how can I get this result?





Things I even know is that: [itex]u_k^2 + v_k^2 =1 [/itex] and:
[itex]\sum\limits_k

\underbrace{(
-2\epsilon_k u_k v_k +\Delta v_k^2 -\Delta u_k ^2
)}_{\stackrel{!}{=}0}

(d_{k \uparrow}^{\dagger}d_{k \downarrow}^{\dagger} + d_{k \downarrow}d_{k \uparrow})[/itex].

Thank you guys!
 
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  • #2
Abigale said:
Things I even know is that: [itex]u_k^2 + v_k^2 =1 [/itex] and:
[itex](
-2\epsilon_k u_k v_k +\Delta v_k^2 -\Delta u_k ^2
)=0
[/itex].

This are two equations for the two unknowns u and v. Solve for them and put into the defining equation for E_k!
 

1. What is a Bogoliubov transformation?

A Bogoliubov transformation is a mathematical tool used in quantum mechanics to transform the creation and annihilation operators of a system into a new set of operators that describe a different basis. This transformation is often used to diagonalize the Hamiltonian of a system, making it easier to solve and interpret.

2. How does a Bogoliubov transformation relate to the diagonalization of the Hamiltonian?

A Bogoliubov transformation is used to transform the creation and annihilation operators of a system into a new basis that diagonalizes the Hamiltonian. This means that the new operators are independent and decoupled from each other, making the Hamiltonian easier to solve and interpret.

3. What is the importance of diagonalizing the Hamiltonian?

Diagonalizing the Hamiltonian allows us to find the energy states of a system and the corresponding eigenvalues. This is important because the eigenvalues represent the possible energy states that a system can have, and the eigenvectors represent the corresponding wavefunctions. This information is crucial in understanding and predicting the behavior of a quantum system.

4. Can a Bogoliubov transformation be used for any quantum system?

Yes, a Bogoliubov transformation can be applied to any quantum system as long as it satisfies certain conditions. These conditions include the system being in a stable state and having a Hamiltonian that is quadratic in the creation and annihilation operators.

5. What is the physical interpretation of a diagonalized Hamiltonian?

The diagonalized Hamiltonian represents the energy levels of a quantum system in a new basis. The eigenvalues of the Hamiltonian correspond to the energies of the system, while the eigenvectors represent the possible wavefunctions that describe the system. This physical interpretation allows us to understand the behavior of a quantum system and make predictions about its properties.

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