# Schrödinger Equation

I am reading up on the SchrÃ¶dinger equation and I quote

Because the potential is symmetric under $x\to-x$, we expect that there will be solutions of definite parity.

Could someone kindly explain why this is true? And perhaps also what it means physically?

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Good question! First you need to know that parity refers to the behavior of a physical system, or one of the mathematical functions that describe such a system, under reflection. There are two "kinds" of parity:

• If $f(x) = f(-x)$, we say the function $f$ has even parity
• If $f(x) = -f(-x)$, we say the function $f$ has odd parity

Of course, for most functions, neither of those conditions are true, and in that case we would say the function $f$ has indefinite parity.

Now, have a look at the time-independent SchrÃ¶dinger equation in 1D:

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

and notice what happens when you reflect $x\to -x$:

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

If you have a symmetric (even) potential, $V(x) = V(-x)$, this is exactly the same as the original equation except that we've transformed $\psi(x) \to \psi(-x)$. Since the two functions $\psi(x)$ and $\psi(-x)$ satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words,

$$\psi(x) = a\psi(-x)$$

Normalizing $\psi$ requires that $|a| = 1$, which leaves two possibilities: $a = +1$ (even parity) and $a = -1$ (odd parity).

As for what this means physically, it tells you that whenever you have a symmetric potential, you should be able to find a basis of eigenstates which have definite even or odd parity (though I haven't proved that here,* only made it seem reasonable). In practice, you get linear combinations of eigenstates with different parities, so the actual state may not actually be symmetric (or antisymmetric) around the origin, but it does at least tell you that if your potential is symmetric, you could construct a symmetric (or antisymmetric) state. That's not guaranteed otherwise. You'd probably have to get input from someone else as to what exactly definite-parity states are used for, though, since that's out of my area of expertise (unless you care about parity of elementary particles, which is rather weirder).

*There is a parity operator $P$ that reverses the orientation of the space: $Pf(x) = f(-x)$. Functions of definite parity are eigenfunctions of this operator. I believe you can demonstrate the existence of a definite-parity eigenbasis by showing that $[H,P] = 0$.

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David, thank you so much! Your explanation is superb! :) – bra-ket Aug 26 '11 at 7:16
States with a definite parity are eigenstates possessing certain energy $E_n$. A perturbed system becomes a superposition with no certain energy and with no certain parity $\Psi(t)=\sum_n C_n \psi_n e^\frac{-iE_n t}{\hbar}$, i.e., it is time-dependent. A tme-independent superposition (without those oscillating exponentials) arise only if you want to represent some time-independent function $\Phi$ as a spectral sum $\sum_n A_n \psi_n$. – Vladimir Kalitvianski Aug 26 '11 at 7:49
Ooh, btw why can't $a\in\mathbb{C}$ and $|a|=1$? – bra-ket Aug 26 '11 at 7:52
Doing reflection twice, you get $a^2 =1$ so it is a real number. – Vladimir Kalitvianski Aug 26 '11 at 7:54
Aha! Thanks! :) – bra-ket Aug 26 '11 at 8:05