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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.

Qmechanic
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Alister Chew
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  • More on Problem 2.1: https://physics.stackexchange.com/q/53374/2451 , https://physics.stackexchange.com/q/44003/2451 and links therein. – Qmechanic Dec 19 '17 at 23:19
  • @Alister you can find the answer implicitly here (https://math.stackexchange.com/questions/878258/showing-that-derivative-of-conjugate-is-conjugate-of-derivative-using-chain-rul) – maha Jul 27 '20 at 02:57
  • Then you can read this ( https://math.stackexchange.com/questions/3770417/is-the-complex-conjugate-of-a-solution-of-a-differential-equation-is-a-solution/3770455?noredirect=1#comment7759584_3770455) – maha Jul 27 '20 at 03:03

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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 $\psi^*$.

Jared Dziurgot
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