# When solving the Schrodinger Equation, where do we add the condition that $E$ is real?

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

• Wikipedia: “The spectrum of a self-adjoint operator is real.” Dec 21, 2022 at 0:21
• I'm sure that at some point, somewhere, Griffiths must have written that the Hamiltonian is an "observable" that satisfies: $\hat H = \hat H^\dagger$... All observables have real eigenvalues... otherwise how would we observe them?
– hft
Dec 21, 2022 at 1:44
• The formalism of linear operators is introduced in chapter 3. If you are still in chapter 2, you can instead show the realness of E in other ways, for example, using the explicit form of the Hamiltonian in positions space and integrating by parts twice.
– hft
Dec 21, 2022 at 1:58
• There is a practice problem in chapter 2 that might have prompted your concern. The practice problem asks you to show that if the wave-function is to remain normalized for all time then E must be real.
– hft
Dec 21, 2022 at 1:59

$$\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$$, 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.

• I see, so the fact that we are restricting ourselves to square-integrable solutions (meaning boundary terms are zero) implies that for those solutions, E is real? Dec 21, 2022 at 8:23
• This would make sense because the proof that the time derivative of the normalization constant is zero in Chapter 1 uses the fact that the wave function is square-integrable. But based on your answer, being square-integrable implies E is real, so the proof only holds for real values of E. Okay, this feels much better! Thank you for your explanation, it's really helped me square this logic in my brain! Dec 21, 2022 at 8:26
• @MatthewSmith Yes, the fact that the solutions are square-integrable is important. You will sometime see people use the notation $L^2$ or $L^2(X)$ to describe the set of square-integrable functions on the space $X$. Square integrability is important because otherwise we couldn't interpret $|\psi(x)|^2$ as a probability density (we couldn't normalize it). One caveat: you will also find that sometime it is convenient to make use of basis functions that are not square integrable (like plane waves), but the physical wave function will be square integrable.
– hft
Dec 21, 2022 at 17:28