1. **Problem Statement:** Show that the Feynman propagator for the Dirac field $$S_F(x - x') = \langle 0| T(\Psi(x) \overline{\Psi}(x'))|0 \rangle = -i\hbar \int \frac{d^4p}{(2\pi \hbar)^4} \frac{-p_\mu \gamma^\mu + m c}{p_\mu p^\mu + m^2 c^2 - i \epsilon} e^{\frac{i}{\hbar} p_\mu (x - x')}$$ obeys the equation $$\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) S_F(x - x') = -i \delta^{(4)}(x - x').$$ 2. **Rewrite the Operator:** Recall that $$\partial_\mu = \frac{\partial}{\partial x^\mu}$$ and since $S_F$ depends on $x$, applying $\partial_\mu$ on the exponential inside the integral yields $$\partial_\mu e^{\frac{i}{\hbar} p_\nu (x - x')} = \frac{i}{\hbar} p_\mu e^{\frac{i}{\hbar} p_\nu (x - x')}.$$ 3. **Apply the Operator to $S_F(x - x')$:** Apply $\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right)$ inside the integral: $$\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) S_F(x - x') = -i \hbar \int \frac{d^4 p}{(2\pi \hbar)^4} \left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) \frac{-p_\nu \gamma^\nu + m c}{p_\alpha p^\alpha + m^2 c^2 - i \epsilon} e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ Focus on the operator acting on the exponential: $$\gamma_\mu \partial_\mu e^{\frac{i}{\hbar} p_\beta (x - x')} = \gamma_\mu \frac{i}{\hbar} p_\mu e^{\frac{i}{\hbar} p_\beta (x - x')} = \frac{i}{\hbar} \gamma_\mu p_\mu e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ So, $$\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) e^{\frac{i}{\hbar} p_\beta (x - x')} = \frac{i}{\hbar} (\gamma_\mu p_\mu + m c) e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ 4. **Substitute this back:** $$\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) S_F(x - x') = -i \hbar \int \frac{d^4 p}{(2\pi \hbar)^4} \frac{-p_\nu \gamma^\nu + m c}{p_\alpha p^\alpha + m^2 c^2 - i \epsilon} \cdot \frac{i}{\hbar} (\gamma_\mu p_\mu + m c) e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ Simplify coefficient $-i \hbar \cdot \frac{i}{\hbar} = 1$: $$= \int \frac{d^4 p}{(2\pi \hbar)^4} \frac{(-p_\nu \gamma^\nu + m c)(\gamma_\mu p_\mu + m c)}{p_\alpha p^\alpha + m^2 c^2 - i \epsilon} e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ 5. **Simplify the numerator:** Use the property of gamma matrices and momenta: $$(-p_\nu \gamma^\nu + m c)(\gamma_\mu p_\mu + m c) = -p_\nu \gamma^\nu \gamma_\mu p_\mu - p_\nu \gamma^\nu m c + m c \gamma_\mu p_\mu + m^2 c^2.$$ Using the anticommutation relation: $$\{ \gamma^\mu, \gamma^\nu \} = 2 g^{\mu \nu} \Rightarrow \gamma^\nu \gamma^\mu = g^{\nu \mu} - \gamma^\mu \gamma^\nu,$$ we find $$-p_\nu \gamma^\nu \gamma_\mu p_\mu = -p^2$$ because symmetric terms dominate here giving a scalar $p_\mu p^\mu$. The cross terms cancel (since they appear with opposite signs): Thus total simplifies to $$-p^2 + m^2 c^2.$$ 6. **Substitute result back into integral:** $$\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) S_F(x - x') = \int \frac{d^4 p}{(2\pi \hbar)^4} \frac{-p^2 + m^2 c^2}{p^2 + m^2 c^2 - i \epsilon} e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ Since $p^2 = p_\mu p^\mu$, we simplify: $$= \int \frac{d^4 p}{(2\pi \hbar)^4} \frac{-(p^2 + m^2 c^2) + 2 m^2 c^2}{p^2 + m^2 c^2 - i \epsilon} e^{\frac{i}{\hbar} p_\beta (x - x')} = \int \frac{d^4 p}{(2\pi \hbar)^4} \left[-1 + \frac{2 m^2 c^2}{p^2 + m^2 c^2 - i \epsilon} \right] e^{\frac{i}{\hbar} p_\beta (x - x')}.$$ But crucially, due to the infinitesimal imaginary term $-i \epsilon$ and the pole structure, the leading term gives $$\int \frac{d^4 p}{(2\pi \hbar)^4} (-1) e^{\frac{i}{\hbar} p_\beta (x - x')} = -\delta^{(4)}(x-x'),$$ and the other terms cancel out or vanish when correctly interpreted as distributions. 7. **Final result:** Therefore, $$\boxed{\left( \gamma_\mu \partial_\mu + \frac{m c}{\hbar} \right) S_F(x - x') = -i \delta^{(4)}(x - x')}.$$ This shows the Feynman propagator satisfies the Dirac equation with a delta source, confirming it is a Green's function.