# Parity of a bound state determined by potential

Some time ago in my QM class, we were working with an infinite well potential, and my professor told us we could know beforehand the bound states we were going to obtain for said potential would have a well-defined parity (that is to say, would be either even or odd in all of the domain). Nonetheless, he never told us how we could know this.

My best guess is, when writing down the Schrödinger's time-independent wave equation for an arbitrary potential with some parity:

$$H\Psi(x)=E\Psi(x)$$

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

we can do $$x\to-x$$ in the equation, and so the left hand side remains unchanged (if the potential is an even function of $$x$$. Therefore, the right hand side should also remain unchanged, from where we can conclude that $$\Psi(x)$$ must also be an even function of $$x$$. Nonetheless, this does not account for the fact that an infinite potential well also has odd functions of $$x$$, so my argument, even if it is valid for the even case, still doesn't explain the odd one.+

• Compute $[H,P]$, where $P$ is the parity operator, and draw the appropriate conclusion(s). Commented Mar 17 at 16:46
• $[\frac{P^2}{2m}+V(x),P]=[V(x),P]=V(x)(-i\hbar \frac{d}{dx})-(-i\hbar \frac{d}{dx})V$, I don't quite see how this will help me determine the parity of my solutions Commented Mar 17 at 16:54
• $P$ is meant to be the parity operator. I should've written $\Pi$... Commented Mar 17 at 16:57
• Careful for the conflicting notation. Tobias Fünke’s parity operator is not your momentum operator. In the Schrödinger representation, it acts on wavefunctions as $(x\to\psi(x))\to(x\to\psi(-x))$ or abstractly, you can define it as acting on position $x$ and momentum $p$ as $$PxP^{-1}=-x \\ PpP^{-1}=-p$$
– LPZ
Commented Mar 17 at 17:03
• Oh, terribly sorry! I didn't know this operator even existed, I'll do some research on the topic and I'll come back to his answer. Thanks for the heads up! Commented Mar 17 at 17:04

The Hamilton operator with a symmetric potential $$V(-x)=V(x)$$,$$H =-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} +V(x),$$ defined on a suitable dense domain $$D(H) \subset L^2(\mathbb{R})$$ commutes with the parity operator $$\Pi$$ defined by $$(\Pi \psi)(x)=\psi(-x), \qquad \psi \in L^2(\mathbb{R}),$$ i.e. $$[\Pi, H]=0$$ on $$D(H)$$. As a consequence, the two operators can be diagonalized simultaneously.
The spectrum of the parity operator $$\Pi$$ consists of the two eigenvalues $$\pm 1$$, as can be seen from $$\Pi^2=\mathbb{I}$$, the corresponding eigenfunctions being even, $$\phi(-x)=\phi(x)$$, (parity $$+1$$) or odd, $$\phi(-x) =-\phi(x)$$, (parity $$-1$$).
The above arguments can easily be adapted to the infinite well problem with $$L^2(\mathbb{R}) \to L^2 ([-L,L])$$, the Hamilton operator $$H=-\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$ defined on a suitable domain of differentiable functions with boundary conditions $$\psi(-L)=\psi(L)=0$$ and $$(\Pi\psi)(x)=\psi(-x)$$ for $$\psi \in L^2([-L,L])$$. Further details are left as an exercise.
Write the Schrodinger equation for $$\Psi$$ in the well and for $$\Psi$$ outside the well. For the case inside the well, $$V=0$$, so no problems there. For the case outside the well, the Schrodinger equation is simply $$0=0$$, so no problems there either. Even though the equation for $$\Psi$$ in the well looks like the free particle equation, it is not, because the energy eigenvalues are restricted to discrete values by way of the boundary conditions. For example, for the case inside the well: $$\Pi{\hat P^2\over 2m}\Psi(x)=\Pi E\Psi(x)=E\Pi\Psi(x)=E\Psi(-x),$$ and since the infinite square well Hamiltonian commutes with $$\Pi$$, we have: $$\Pi{\hat P^2\over 2m}\Psi(x)={\hat P^2\over 2m}\Pi\Psi(x)={\hat P^2\over 2m}\Psi(-x).$$ So if $$\Psi(x)$$ is even: $${\hat P^2\over 2m}\Psi(-x)=E\Psi(-x)\implies {\hat P^2\over 2m}\Psi(x)=E\Psi(x),$$ and for odd $$\Psi(x)$$, \begin{align}{\hat P^2\over 2m}\Psi(-x)=E\Psi(-x)&\implies -{\hat P^2\over 2m}\Psi(x)=-E\Psi(x)\\ &\implies {\hat P^2\over 2m}\Psi(x)=E\Psi(x). \end{align} Thus, we are able to see that parity is a symmetry of the infinite square well problem, i.e. the Schrodinger equation is invariant under the operation of parity.