When solving the Schrodinger Equation, where do we add the condition that $E$ is real? I'm reading through Griffiths, and I noticed two seemingly contradictory facts:

*

*In Chapter 1, it is proved that for any square-integrable function solving the Schrodinger Equation, $$\frac{d}{dt}\int_{-\infty}^{\infty}|\Psi|^{2}dx = 0.\tag{1}$$

*In Chapter 2, we solve the Schrodinger Equation using the separation of variables, and while never assuming that the separation constant $E$ must be real, we get the solutions
$$\Psi(x,t)=\psi(x)e^{-iEt/\hbar}.\tag{2}$$
Clearly, (as is shown in Problem 2.1 (a)), if $E$ is not real, the normalization condition proved in Chapter 1 does not hold.

So, we proved eq. (1) for any solution, and yet, when solving using separation of variables without assuming $E$ is real, we got solutions (2), which do not conform to the first condition if $E$ is not real.
During the solution process, where do we mathematically argue that $E$ must be real?
 A: 
$\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$, which do not conform to the first condition if E is not real.


During the solution process, where do we mathematically argue that E must be real?

When you performed the separation of variables you wrote:
$$
i\frac{d f}{d t} = Ef
$$
and
$$
-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2} + V\psi = E\psi\;,
$$
where $V$ is a real potential function, and where the full wave-function was written in seperable form like: $\Psi(x,t)=f(t)\psi(x)$.
Since your solution $\psi(x)$ is normalized we can write:
$$
\int \left[\psi^*(x)\left(-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}\right) + V\psi^*(x)\psi(x)\right]dx = E\;,
$$
Thus:
$$
E^* = \int \left[\psi(x)\left(-\frac{\hbar^2}{2m}\frac{d^2 \psi^*}{dx^2}\right) + V\psi(x)\psi(x)^*\right]dx
$$
Then integrate by parts twice and use the vanishing of the boundary terms  to see that:
$$
E^* = \int \left[\left(-\frac{\hbar^2}{2m}\frac{d^2 \psi}{dx^2}\right)\psi^*(x) + V\psi(x)\psi(x)^*\right]dx = E
$$
The equation
$$
E^*=E
$$
shows that $E$ is real.
