# Is the conjugate of a solution to the time-independent Schroedinger equation also a solution?

Was reading the second chapter of Griffith's Introduction to Quantum Mechanics and have failed to understand why the conjugate of a solution to the Time-Indepedent Schrodinger Equation (hence TISE) is a solution itself.

It's at the end of Section 2.1 for reference (contained in Problem 2.1). I have the second edition.

Consider the time-independent Schrodinger Equation (TISE) in position space:

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V(x)\psi = E\psi.$$

Suppose we take the complex conjugate of both sides. This then gives us

$$-\frac{\hbar^2}{2m}\frac{d^2\psi^*}{dx^2} + V(x)\psi^* = E\psi^*,$$

since the potential and energy are clearly real, and the derivative is unaffected by conjugation. Therefore it is plain to see that if $\psi$ is a solution to the TISE, then so will its complex conjugate $\psi^*$.

• Thanks heaps @Jared Dziurgot ! That is so simple! I think I was reading into it too much. – Alister Chew Dec 19 '17 at 23:29
• Actually the potential could be imaginary if absorbtion or decay of particles were possible. – ZeroTheHero Dec 20 '17 at 0:48