# Why is There a Discrepancy in the Mastubara Transform Conventions?

• a2009
In summary, the Mastubara transform convention is a mathematical tool used in quantum mechanics to simplify calculations involving thermodynamic quantities at different temperatures. It involves transforming equations from the time domain to the imaginary frequency domain and using contour integration to express them in terms of Matsubara frequencies. This convention has benefits such as avoiding singularities and allowing for the use of Feynman diagrams, but it may not be as accurate in non-equilibrium systems or at low temperatures. Other techniques, like perturbation theory, may be more suitable in those cases.
a2009
Hello,

When calculating the dynamic structure factor I need to express $$\rho(q)$$ in terms of creation and annihilation operators. I tried a direct calculation of the form

$$\rho_q = \int e^{i q x} \bar{\psi}(x) \psi(x)$$
$$= \int e^{iqx} \sum_{k k'} e^{ikx} \bar{\psi}_k e^{-ik'x} \psi_{k'}$$
$$= \sum_k \bar{\psi}_k \psi_{k+q}$$

however in all of the literature the result is stated

$$\rho_q = \sum_k \bar{\psi}_{k+q} \psi_k$$

I've thought of the following possible causes for this discrepancy:
1. I'm using the wrong sign convention for the transform: WRONG because even if I switch all of the signs of the exponents we get back the same result.
2. ONLY the sign of rho's exponent is wrong: possible, because it seems that it is arbitrary to decide that rho transforms like psi and not like psi bar.

This is where I'm at. I would really like to understand the source of the mistake.

Thanks for any help!

Thank you for your post and for sharing your thoughts on the discrepancy in the calculation of the dynamic structure factor. I understand your confusion and I would be happy to provide some insight into the matter.

Firstly, it is important to note that the dynamic structure factor, \rho(q), is defined as the Fourier transform of the density-density correlation function, \chi(q,\omega). This means that in order to calculate \rho(q), we need to express the density operator, \hat{\rho}(q), in terms of creation and annihilation operators. This is where the discrepancy in your calculation lies.

In your calculation, you have correctly expressed the density operator as \hat{\rho}(q) = \int e^{iqx} \bar{\psi}(x) \psi(x), but the next step is where the mistake occurs. The correct expression for the density operator is \hat{\rho}(q) = \sum_k e^{iqx_k} \hat{n}_k, where \hat{n}_k = \bar{\psi}_k \psi_k is the number operator.

Now, using the commutation relations between creation and annihilation operators, we can rewrite \hat{\rho}(q) as \hat{\rho}(q) = \sum_k \bar{\psi}_{k+q} \psi_k.

Substituting this into the definition of \rho(q), we get \rho_q = \int e^{i q x} \langle \hat{\rho}(q) \rangle = \sum_k \langle \bar{\psi}_{k+q} \psi_k \rangle = \sum_k \bar{\psi}_{k+q} \psi_k, which is the correct result.

In summary, the discrepancy in your calculation was due to not properly expressing the density operator in terms of creation and annihilation operators. I hope this clarifies things for you and helps you understand the source of the mistake. Keep up the good work in your research!

## 1. What is the Mastubara transform convention?

The Mastubara transform convention is a mathematical tool used in quantum mechanics to simplify the calculations of the properties of particles at different temperatures. It involves transforming the equations from the time domain to the imaginary frequency domain, making it easier to solve for thermodynamic quantities.

## 2. Why is the Mastubara transform convention used?

The Mastubara transform convention is used because it allows for the simplification of calculations involving thermodynamic quantities at different temperatures. It also helps to avoid singularities in the equations and makes it easier to apply certain mathematical techniques, such as contour integration.

## 3. How is the Mastubara transform convention performed?

To perform the Mastubara transform convention, the equations are first transformed from the time domain to the imaginary frequency domain by replacing the time variable with an imaginary frequency variable. The equations are then integrated over a contour in the complex plane, after which the limits of the integration are taken to infinity. This results in the equations being expressed in terms of the Matsubara frequencies.

## 4. What are the benefits of using the Mastubara transform convention?

One of the main benefits of using the Mastubara transform convention is that it simplifies calculations involving thermodynamic quantities at different temperatures. It also helps to avoid singularities in the equations and makes it easier to apply certain mathematical techniques, such as contour integration. Additionally, the Mastubara transform convention allows for the use of Feynman diagrams, which are useful in visualizing and understanding particle interactions.

## 5. Are there any limitations to the Mastubara transform convention?

While the Mastubara transform convention is a useful tool, it does have some limitations. It is mainly applicable to systems in thermal equilibrium and does not take into account non-equilibrium effects. Additionally, it may not be as accurate for systems with strong interactions or at low temperatures. In these cases, other techniques such as perturbation theory may be more suitable.

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