Why is There a Discrepancy in the Mastubara Transform Conventions?

[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!

 

[AndyW]

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!