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I have done the Problem 2.1 in Griffiths' quantum mechanics, and it seems not making sense to me.

What if the wave function isn't symmetric at all? Then obviously the proof doesn't work. The solution confuses me.

If $V(x)=V(-x)$ then the "position" wave function can always be taken to be either even or odd.

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    $\begingroup$ It might be helpful to give a short outline of the proof here, as not all of us have Griffith’s QM at hand. $\endgroup$ – Claudius Nov 15 '12 at 20:01
  • $\begingroup$ You can prove this exercise 2.1c in exactly the same way as 2.1b. In 2.1b, under the conditions assumed, whenever $\psi$ is a solution, so are the real and imaginary parts, $\mathrm{Re}\, \psi$ and $\mathrm{Im}\,\psi$, of the same energy (as Griffiths completely gives away in the hint). Similarly, in 2.1c, the even and odd parts are solutions of the same energy, and can be constructed in a directly analogous way. Assuming the usual Hamiltonian with real, even $V(x)$. $\endgroup$ – Stan Liou Nov 16 '12 at 20:02
  • $\begingroup$ Related: physics.stackexchange.com/q/13980/2451 $\endgroup$ – Qmechanic Dec 11 '13 at 13:49
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Griffiths, Introduction to QM, uses $\Psi$ and $\psi$ to denote a solution to the time-dependent and the time-independent Schrödinger eq., respectively.

For a fixed energy $E$, rather than considering a general solution $\psi$ to the TISE, the book is at this point only interested in finding a generating set $\psi_n$ of solutions, so that a general solution is a linear combination $\psi=\sum_nc_n\psi_n$, cf. the superposition principle.

The purpose of Exercise 2.1.b is to show that it is no loss of generality to assume that the generating element $\psi_n\in\mathbb{R}$ is a real function.

The purpose of Exercise 2.1.c is to show in the case of an even potential $V$ that it is no loss of generality to assume that the generating element $\psi_n$ is an even or an odd function.

The book is not claiming that the general solution $\psi$ has to respect such symmetry.

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  • $\begingroup$ As Qmechanic said, Griffiths isn't claiming that every $\psi$ exhibits even or odd symmetry, but you can show that every even potential will produce wavefunctions that necessarily do exhibit either even or odd symmetry. $\endgroup$ – Fire Feb 18 '16 at 22:07

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