In the famous double-slit experiment, if I inserted two identical sources (i.e electron guns) at the locations of the slits, would I still see an interference pattern on the screen?
I would allow only one particle to be fired out of the two electron guns in turns. In short, I replace the double slits with electron guns.
The situation is entirely different from the double slit experiment!
In the double slit experiment, one electron propagates through the slits, its parts interfere, thus we have a density matrix like (this prepares a pure state $\lvert\psi\rangle = \frac{1}{\sqrt{2}} (\lvert 1 \rangle + \lvert 2 \rangle)$):
$$ \rho = \frac 1 2 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} $$
This has coherent terms, therefore it will be possible to observe interference effects.
The case with two electron guns. Here you have one electron from one source, this can be achieved best by a low emission rate (and then neglecting the few events with multiple electrons). This amounts to a density matrix of:
$$ \rho = \frac 1 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $$
This cannot show interference as the different electrons are incoherent. This is the picture for electron guns (which are macroscopic, therefore distinguishable, and so on).
Now if you build clever "electron guns" such that they will emit one electron, but it is not sure where (not classically, but in a quantum sense), then sure, you will have interference. (So the answer will get arbitrarily complicated, if you specify the situation in more detail).
Here I assume very clever electron guns each emitting one electron.
The single-particle density (which is measured on the screen) is given by (with the shorthand for arguments: $1 := \vec r_1, \sigma_1$, $\int d1 := \sum_{\sigma_1} \int d^3r_1$):
$$n(\vec r_1) = \sum_{\sigma_2} \int d1\, \sum_\sigma \lvert\psi(1, 2)\rvert^2$$
I neglect electron-electron interaction, as they will be far from each other.
If the emitted electrons have the same spin, the real-space part of the wave function must be anti-symmetric due to the Pauli exclusion principle. The spatial part of the wave function is then given by:
$$\phi(\vec r_1, \vec r_2) = \phi_1(\vec r_1)\phi_2(\vec r_2) - \phi_1(\vec r_2)\phi_2(\vec r_1)$$
This leads to:
\begin{align*} n(\vec r) &= \int d^3r'\, \left| \phi_1(\vec r)\phi_2(\vec r') - \phi_1(\vec r')\phi_2(\vec r) \right|^2\\ &= \int d^3r'\, \Big( \big\lvert \phi_1(\vec r)\phi_2(\vec r') \big\rvert^2 + \big\lvert \phi_1(\vec r')\phi_2(\vec r) \big\rvert^2 - 2 \Re \phi_1(\vec r)\phi_2(\vec r')\phi_1^*(\vec r')\phi_2^*(\vec r) \Big) \\ &= \big\lvert \phi_1(\vec r) \big\rvert^2 + \big\lvert \phi_2(\vec r) \big\rvert^2 - 2 \Re \phi_1(\vec r) \phi_2^*(\vec r) \underbrace{\int d^3r' \phi_1^*(\vec r')\phi_2(\vec r')}_{=\,0}. \end{align*}
The integral is zero, because the states $\phi_1, \phi_2$ must be orthogonal (because the initial states are orthogonal).
That means, we do not observe an interference pattern.
We will not observe interference either (as the seperate spin channels do not interfere).
The general (multi-electron, density matrix preparation) case is left as an exercise for the reader ;).
