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.

  • 1
    $\begingroup$ 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
    Commented Nov 15, 2012 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
    Commented Nov 16, 2012 at 20:02
  • $\begingroup$ Related: physics.stackexchange.com/q/13980/2451 $\endgroup$
    – Qmechanic
    Commented Dec 11, 2013 at 13:49

1 Answer 1


Griffiths, Introduction to QM, uses $\Psi$ and $\psi$ to denote a solution$^1$ 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.


$^1$ Griffiths is implicitly only talking about normalizable solutions in 1D. For unnormalizable solutions the given boundary conditions at $x=\pm\infty$ might violate reality/parity.

  • $\begingroup$ 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
    Commented Feb 18, 2016 at 22:07

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