# What does it mean for a function in one Hilbert space to be to be diagonal in a basis for a Hilbert space that is not a subspace?

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?

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.

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

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

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

$$$$G_{aa}(\omega) = \frac{1}{D_a(\omega)}$$$$

This is the result you wanted to confirm.

• So if I want it in a different basis, say an atomic orbital basis $\theta_j$, I would write: $G_{ij} = \langle i|G|j\rangle = \int \langle i|x\rangle\langle x|G|y \rangle\langle y|j\rangle dx dy = \sum_{l} D_l(\omega)^{-1} \int \theta_i^*(x) \psi_l (x) dx \int \theta_j(y) \psi_l^* (y) dy = \sum_{l} D_l(\omega)^{-1} \langle i | l \rangle \langle l | j \rangle$. Nov 5, 2020 at 18:45
• @MikkelRev Exactly! That is the general rule for changing the basis of an operator. Nov 5, 2020 at 21:01
• Thanks. Could you provide a reference on this general rule; I havn't been able to find a reference for it. Thanks Nov 5, 2020 at 21:33
• 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. Nov 5, 2020 at 23:22