0
$\begingroup$

I am following Richard Martin on interacting electrons. For independent electrons at zero temperature he finds that the time-ordered Green's function is given by $$ G(x_1,x_2;\omega) = \sum_{l} \frac{\psi_l(x_1) \psi^*(x_2)}{\omega - \varepsilon_l + i \eta \, \mathrm{sgn}(\varepsilon_l - \mu) }.$$ You could in principle use any other function of $k \in \mathbb{N}$ variables. He says that in the single particle basis $\{\psi_l\}_l$ the Green’s function is diagonal: $$ G_{ll}(\omega) = \frac{1}{\omega - \varepsilon_l + i \eta \, \mathrm{sgn}(\varepsilon_l - \mu) }. $$ I don't understand this. I know what it means to write an operator of $k$ variables in a basis where the elements themselves have $k$ arguments. Suppose for example that $\Phi_j : \mathbb{R}^3 \times \mathbb{R}^3 \to \mathbb{C}$ are linearly independent for every $j$, then $$ G_{ij}(\omega) = \langle \Phi_i , G \Phi_j\rangle = \int_{\mathbb{R}^3 \times \mathbb{R}^3} \Phi_i^*(x_1,x_2)G(x_1,x_2,\omega) \Phi_j(x_1,x_2) \, dx_1 dx_2. $$ The problem here is that the basis $\psi_l$ has exactly 1 argument, so I don't know how to do it. Could you show me please?

$\endgroup$

1 Answer 1

1
$\begingroup$

A minor note: I think there may be a few notational flubs in the original question, and so I have filled in what I think is the intended notation. Please let me know if I misunderstood something.

We start with

\begin{equation} G(x_1,x_2;\omega) \equiv \langle x_1 |G(\omega)|x_2 \rangle = \sum_\ell \frac{\psi_\ell (x_1) \psi_\ell^\star(x_2)}{D_\ell(\omega)} \end{equation} where, just to save writing, I defined the denominator $D_\ell(\omega) = \omega - \epsilon_\ell + i \eta {\rm sgn}(\epsilon_\ell-\mu)$.

We assume the states are normalized \begin{equation} \int dx \psi_a^\star(x) \psi_b(x) = \delta_{ab} \end{equation}

Now we take the inner product of this function with $\psi_a$ and $\psi_b$, and make use of the orthogonality condition to obtain:

\begin{eqnarray} G_{ab}(\omega) \equiv \langle a |G(\omega)|b \rangle & = & \int d x_1 d x_2 \langle a | x_1 \rangle \langle x_1 | G(\omega) | x_2 \rangle \langle x_2 | b \rangle \\ &=& \sum_\ell \frac{1}{D_\ell(\omega)} \left[\int dx_1 \psi_a^\star(x_1) \psi_\ell(x_1)\right] \left[\int dx_2 \psi_\ell^\star(x_2) \psi_b(x_2)\right] \\ &=& \sum_\ell \frac{\delta_{a \ell} \delta_{\ell b}}{D_\ell(\omega)} \\ &=& \frac{\delta_{ab}}{D_a(\omega)} \end{eqnarray}

This result matches your intuition that the Green's function should be an operator in any basis.

Now, your text essentially says to look only at the diagonal elements, with $a=b$, in which case we use $\delta_{aa}=1$ and obtain

\begin{equation} G_{aa}(\omega) = \frac{1}{D_a(\omega)} \end{equation}

This is the result you wanted to confirm.

$\endgroup$
4
  • $\begingroup$ So if I want it in a different basis, say an atomic orbital basis $\theta_j$, I would write: $G_{ij} = \langle i|G|j\rangle = \int \langle i|x\rangle\langle x|G|y \rangle\langle y|j\rangle dx dy = \sum_{l} D_l(\omega)^{-1} \int \theta_i^*(x) \psi_l (x) dx \int \theta_j(y) \psi_l^* (y) dy = \sum_{l} D_l(\omega)^{-1} \langle i | l \rangle \langle l | j \rangle $. $\endgroup$
    – Mikkel Rev
    Nov 5, 2020 at 18:45
  • $\begingroup$ @MikkelRev Exactly! That is the general rule for changing the basis of an operator. $\endgroup$
    – Andrew
    Nov 5, 2020 at 21:01
  • $\begingroup$ Thanks. Could you provide a reference on this general rule; I havn't been able to find a reference for it. Thanks $\endgroup$
    – Mikkel Rev
    Nov 5, 2020 at 21:33
  • $\begingroup$ The fundamental trick that's being used is to insert $1=\int dx | x \rangle \langle x |$ after a bra or before a ket, if you look carefully you can see that this is being used two different times in relating $\langle i | G | j \rangle$ with $\langle x | G | y \rangle$ (ie, the trick is used twice in relating the same operator $G$ in two different bases, $x,y$ vs $i,j$). This identity is sometimes called "resolution of the identity." Every quantum mechanics book will introduce and use this trick. $\endgroup$
    – Andrew
    Nov 5, 2020 at 23:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.