This answer has been edited to fix the grammar and punctuation errors.

In summary, The approach taken to solve the problem is correct, but it is important to use the correct notation and keep terms up to first order only. The Hermitian conjugate of ##\hat{U}## can be easily obtained since ##\hat{H}'(t)## is a Hermitian operator. The notation ##\langle \rangle_0## denotes an equilibrium average with respect to the Hamiltonian ##H_0##, and is time independent.
  • #1
aaaa202
1,169
2

Homework Statement


I am supposed to show equation 6.6 on page 97 of http://www.phys.lsu.edu/~jarrell/CO...hysics Henrik Bruus and Karsten Flensberg.pdf
I have tried to plug A and the expression for ln(t)> into 6.3 a) on page 96 but end up with a double integral after doing the linear approximation of the time evolution operator. Is my method correct and if so, how do I conjugate transpose the expression for U when given by the first order integral? Second of all I am quite confused by the notation <A>_0. Does that simply mean <A(t0)> or am I missing something?


Homework Equations





The Attempt at a Solution


I have attached my attempt.
 

Attachments

  • attempt.png
    attempt.png
    15.3 KB · Views: 989
Physics news on Phys.org
  • #2
aaaa202 said:

I have tried to plug A and the expression for ln(t)> into 6.3 a) on page 96 but end up with a double integral after doing the linear approximation of the time evolution operator. Is my method correct


Yes, you have the correct approach. But I think you should be careful with the notation and use a carat ^ to denote any operator or state in the interaction picture (as introduced in section 5.3 of the notes.) Also, there is no operator ##\hat{H}''## with a double prime.

You only want to keep terms up to first order in ##\hat{H}'##. So, you should not get a double integral. You will have two single integrals that you can combine into one single integral.

how do I conjugate transpose the expression for U when given by the first order integral?

It should be easy to take the Hermitian conjugate of ##\hat{U}## since ##\hat{H}'(t)## is a Hermitian operator.

Second of all I am quite confused by the notation <A>_0. Does that simply mean <A(t0)> or am I missing something?

See just below equation (6.6). The notation ##\langle \rangle_0## means an equilibrium average with respect to the Hamiltonian ##H_0##. Thus ##\langle A \rangle_0## means exactly the expression defined by equation (6.1a). Note that ##\langle A \rangle_0## is time independent, so you don't need to worry about the time for this quantity.
 

FAQ: This answer has been edited to fix the grammar and punctuation errors.

What is the Kubo formula?

The Kubo formula is a mathematical formula used in statistical mechanics to calculate the linear response of a system to an external perturbation. It relates the response of a system, such as the electrical conductivity, to the system's equilibrium fluctuations.

How is the Kubo formula derived?

The Kubo formula is derived from the linear response theory, which states that the response of a system to an external perturbation is directly proportional to the perturbation itself. This theory is then applied to the quantum mechanical description of a system, resulting in the Kubo formula.

What are the assumptions made in the derivation of the Kubo formula?

The derivation of the Kubo formula assumes that the system is in thermal equilibrium and that the perturbation is small enough to be considered a linear response. It also assumes that the system is described by quantum mechanics and that the perturbation is time-dependent.

What are the applications of the Kubo formula?

The Kubo formula has various applications in condensed matter physics, including the calculation of transport coefficients such as electrical and thermal conductivity, and the characterization of phase transitions. It is also used in the study of quantum many-body systems and in the development of new materials with specific properties.

Are there any limitations to the Kubo formula?

Yes, the Kubo formula has some limitations. It is only applicable to systems in thermal equilibrium, and it assumes that the perturbation is small enough to be considered a linear response. Additionally, it is not always easy to apply in practice due to the complex mathematical calculations involved.

Similar threads

Back
Top