When does an even potential give both even and odd solutions to Schrodinger's Eqn?

[jstrunk]

TL;DR Summary

When does an even potential energy result in an even and an odd solution to Time Independent Schrodinger Equation

I don't understand this statement about potential energy V(x) from Griffiths Intro to Quantum Mechanics, 3rd Ed.
Problem 2.1c: If V(x) is an even function (that is V(-x)=V(x)) then psi(x) can always be taken to be either even or odd.
psi(x) refers to a solution of the Time Independent Schrodinger Equation.

It is not clear what "psi(x) can always be taken to be either even or odd" means.

From working the examples and problems in the text, it seems what it means in practice is
a) For bound states (energy < 0), there is an even solution and an odd solution.
b) For scattering states (energy > 0), there is just one solution.
Of course, that 'one solution' may be a superposition of wave functions, I just mean you don't have to solve for
an even case and an odd case.

Is this correct?

 

[vanhees71]

It's about "parity". If $V(-x)=V(x)$ the Hamiltonian is invariant under the space-reflection symmetry $x \rightarrow -x$, $p \rightarrow -p$. The corresponding unitary operator $\hat{P}$ thus commutes with the Hamiltonian and can be diagonalized simultaneously with the Hamiltonian. Thus you can always find a complete set of orthonormal simultaneous eigenvectors of $\hat{H}$ and $\hat{P}$. Now $\hat{P}^2=\hat{1}$ and thus the space-reflection operator has the two eigenvalues $\pm 1$. Which means the corresponding eigenvectors $U_{EP}(x)$ fulfill
$$\hat{H}U_{EP}(x) = E U_{EP}(x), \quad \hat{P} U_{EP}(x)=U_{EP}(-x)=P U_{EP}(x), \quad P \in \{1,-1 \}.$$
Thus in short: You can always choose the complete set of energy eigenfunctions as simultaneous eigenfunctions of $\hat{P}$, i.e., as even or odd functions.

 

[jstrunk]

I'm sorry but I don't understand any of that.
I am doing self study, but the level I am at is like that of someone in the first few weeks of their first undergraduate QM course. At this point, the most important thing I need to understand is this:
When solving Time Independent Schrodinger Equations with even potential energy functions, in all the examples I have seen, Griffiths solves for even and for odd psi for bound states, but doesn't for scattering states. He never states that is the correct procedure in general, he just does it as if it is self explanatory.

On the one hand, he seems to say that there are always even and odd solutions when the potential energy is an even function, and on the other had he only looks for even and odd solutions for bound states.

I am just trying to figure out how to apply the statement "psi(x) can always be taken to be either even or odd"
in solving problems.

 

[vanhees71]

Well, after a view weeks of introductory QM you know that you can find a complete orthonormal set (CONS) of simultaneous eigenvectors of a set of self-adjoint and/or unitary operators if all of them mutually commute. Griffiths is not a very well written book. I don't understand, why it is so widely used.

Again: If the potential in the Hamiltonian
$$\hat{H}=\frac{1}{2m} \hat{\vec{p}}^2 + V(\hat{x})$$
is even, i.e., if $V(\hat{x})=V(-\hat{x})$ the Hamiltonian commutes with the space-reflection (or "parity") operator $\hat{P}$ and thus there is a CONS of simultaneous eigenvectors of $\hat{H}$ and $\hat{P}$. So you can always find simultaneous eigenvectors of $\hat{H}$ and $\hat{P}$. That's what I tried to summarize in #2.

On the other hand it also depends on what you want to describe physically, which set of eigenvectors you want to choose to work with. For a scattering problem you have a potential $V(x)$ which goes to $0$ for $x \rightarrow \pm \infty$, and you want to describe the situation, where a particle comes from, say, $x \rightarrow -\infty$, where they are free particles, and is moving to the right, and you want to know how many particles after a long time are found at $x \rightarrow \infty$ and how many are reflected back to $x \rightarrow -\infty$. So here you look for "asymptotic free states", which are only moving to the right for $x \rightarrow +\infty$. So there you choose energy eigenstates which go to right-moving plane waves $u_E \propto \exp(+\mathrm{i} p x/\hbar)$ with $p>0$ for $x \rightarrow \infty$. Then to towards $-\infty$ you have a wave function, which is a superposition of left- and right-moving plane waves, $u_E \propto T \exp(+\mathrm{i} p x /\hbar) + R \exp(-\mathrm{i} p x/\hbar)$, and these states are for sure not even or odd, i.e., no parity eigenstates.

Nevertheless you can always use a CONS of energy-parity eigenstates to build any energy eigenstate you like (that's why it's a CONS). The corresponding solutions go asymptotically like $u_{E+} \propto \cos(p x/\hbar)$ and $u_{E-} \propto \sin(p x/\hbar)$ (the cos is even and sin is odd under space reflections and thus eigenstates of the parity operator with eigenvalues +1 or -1, respectively).

It's a good exercise to think about the free-particle states first and the two-fold degeneracy of the energy eigenvalues (for the here considered 1D case) and that you can choose a unique CONS of energy eigenvectors by making them simultaneous eigenstates of the parity operator.