Some basic problems with Green's function So, I am trying to calculate and understand the whole process for obtaining Green's function for free particle. For this, I am using phys.libtext.org.
So, the propagator is a sollution of PDE:
$$ \langle\textbf{r}|(E-\hat{H}_0)|\hat{G}_0|\textbf{r}'\rangle=\delta(\textbf{r}-\textbf{r}').$$
To solve for $\langle\textbf{r}|\hat{G}_0|\textbf{r}'\rangle$ they're using momentum eigenstates described by wave vector as follows
$$
\langle\textbf{r}|\hat{G}_0|\textbf{r}'\rangle=\langle\textbf{r}|\hat{G}_0\int d\textbf{k}'|\textbf{k}'\rangle\langle\textbf{k}'|\textbf{r}'\rangle.
$$
Here is my first question. On the left side we have Green's function in position representation. As we used above $\textbf{k}$ vectors, does it mean that we switched into momentum representation or not? I think, I am not completely understand representation here.
Next, it is used formula for Green's function: $\hat{G}_0=(E-\hat{H}_0)^{-1}$, which for free particle can be expressed by $\hat{G}_0=(E-\frac{\hat{p}^2}{2m})^{-1}$. So, we can substitude this and get
$$
\langle\textbf{r}|\hat{G}_0|\textbf{r}'\rangle = \int d \textbf{k}' \langle\textbf{r}|\textbf{k}'\rangle \frac{1}{E-\frac{p^2}{2m}} \langle\textbf{k}'|\textbf{r}'\rangle =\int d \textbf{k}' \langle\textbf{r}|\textbf{k}'\rangle \frac{1}{E-\frac{\hbar^2|\textbf{k}'|^2}{2m}} \langle\textbf{k}'|\textbf{r}'\rangle.
$$
Here is my second question. Can we just express here $\hat{p}=\hbar k$? And are we still in position representation? And then, they are using $\textbf{k}$ eigenvectors and obtain
$$
\langle\textbf{r}|\hat{G}_0|\textbf{r}'\rangle = \frac{2m}{\hbar^2}\frac{1}{(2\pi)^3}\int d\textbf{k} \frac{\exp[i\textbf{k}' \cdot(\textbf{r}-\textbf{r}')]}{k^2-|\textbf{k}'|^2}
.$$
And here is my third question. As I understand, here we used $E=\frac{\hbar^2 k^2}{2m}$. So, how should I interprete $k$ and $k'$? And then, if I calculate this integral, I got expression for free particle's Green's function in position representation? So, if I use Fourier transform on my result, I can switch into momentum representation?
 A: I will try to answer by points:

*

*When we talk about "position representation" or "momentum representation" of some operator $\hat{\mathcal{O}}$ we are implicitly considering matrix elements of this operator, therefore $ \langle \mathbf{k} | \hat{\mathcal{O}} | \mathbf{k}^\prime \rangle$ or $\langle \mathbf{r} | \hat{\mathcal{O}} | \mathbf{r}^\prime \rangle$, what is being meant is that you choose a complete basis for the calculation of the matrix element. You can also have mixed representation $\langle \mathbf{k} | \hat{O} | \mathbf{r} \rangle$, as you probably know whenever you have a vector you can change basis with a unitary transformation, to change from momentum to position this transformation is the well known
$$
| \mathbf{k} \rangle \rightarrow | \mathbf{r} \rangle = \frac{1}{\sqrt{N}} \sum_{\mathbf{k}} e^{i \mathbf{k} \mathbf{r}} | \mathbf{k} \rangle = \hat{U} | \mathbf{k} \rangle
$$
Now, the operator per se does not have "position" or "momentum" represetation but depending on the states you do the braketing you can get one or the other, for example if you want to compute  the momentum representation but you know the position one the logic is the following:
$$
\langle \mathbf{k} | \hat{\mathcal{O}} | \mathbf{k}^\prime \rangle = O_{k k^\prime} \ \rightarrow \ O^\prime_{r r^\prime} =\langle \mathbf{r} | \hat{\mathcal{O}} | \mathbf{r}^\prime \rangle = \sum_{\mathbf{k} \mathbf{k}^\prime} \langle \mathbf{r} | \mathbf{k} \rangle \langle \mathbf{k} | \hat{\mathcal{O}} | \mathbf{k}^\prime \rangle \langle \mathbf{k}^\prime  |\mathbf{r}^\prime \rangle = \sum_{k k^\prime} U^\dagger_{r k} O_{k k^\prime}  U_{r^\prime k^\prime}
$$
Here $O$ and $O^\prime$ are different representations,it's not a matter of the operator but of matrix element computed on a specific basis.


*As I told you it's not a matter of operator but its basis representation. In this sense you should now understand that $(E-H_0)^{-1}$ is diagonal in momentum representation since $E$ is proportional to indentity and $H_0$ is diagonal, it is better to express it in that basis:
$$
\langle \mathbf{k} | \hat{G_0} | \mathbf{k}^\prime \rangle = \frac{1}{E-\frac{\hbar^2 k^2}{2m}} \delta_{k k^\prime}
$$
and you get want you have in your second question where the elemtn you are computing $\langle \mathbf{r} | \hat{G}_0 | \mathbf{r}^\prime \rangle $ is still in position representation but expressed in terms of his momentum representation $ \frac{1}{E-\frac{\hbar^2 k^2}{2m}} \delta_{k k^\prime}$. Indeed all the momentum variables are integrated while the position ones are not.


*Try not to mess up with the variables' names. You are integration over $\mathbf{k}^\prime$ and the meaning of this variable is just that you are considering all the possible contributions of all momenta, while $k$ is the momentum, or the better the modulus of the momentum, for which you are computing the green function. When you are inverting $(E-\hat{H}_0)$ you have $E$ fixed and is the energy for which you are computing the GF, when $k^\prime = k$ the GF has a pole and that means it is an eigenvalue of the hamiltonian $H_0$ and all the analitical properties of green functions you can find in many books. And yes by Fourier transforming you get the momenum representation.
I hope this can clarify your doubts!
