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.


1 Answer 1


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 $\psi^*$.

  • $\begingroup$ Thanks heaps @Jared Dziurgot ! That is so simple! I think I was reading into it too much. $\endgroup$ Dec 19, 2017 at 23:29
  • $\begingroup$ Actually the potential could be imaginary if absorbtion or decay of particles were possible. $\endgroup$ Dec 20, 2017 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.