## Why is the solution of the schrödinger equation always symmetric or antisymmetric?

Hi!

I read that in an 1D potential, the solution for the Schrödinger equation is always either symmetric or antisymmetric if the potential is a symmetric function: V(x) = V(-x).

How can I proof this?

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 The proof for this is typically done using parity operators. If your hamiltonian is given by $$\mathcal{H} = \hat{p}^2/2m + V(\hat{x})$$ Write down the time-independent Schrodinger equation, then flip the signs on all the x-coordinates and see what this imposes on the wave function. If you want a better discussion of this, check out Chapter 4 of Sakurai.
 I'm not familiar with the word "degenerated". If you mean "having degenerate eigenstates", degeneracies usually arise when you have another observable $$\mathcal{O}$$ such that $$\left [ \mathcal{H}, \mathcal{O} \right ] - 0$$. This implies that an eigenstate of the hamiltonian is also an eigenstate of your new observable (I leave it to you to figure out why). What frequently happens in this case is that there are multiple values of $$\mathcal{O}$$ for a given energy eigenvalue, and so you end up with degeneracies in the energy spectrum. Of course, sometimes things end up being more degenerate than they should be. For example, in hydrogen, the energy levels don't depend on the $$\ell$$ quantum number, although in general a spherically symmetric potential leads to an $$\ell$$ dependent energy spectrum. This is called an "accidental degeneracy". In the case of the hydrogen atom, the degeneracy arises because the angular momentum operators aren't the only ones that commute with the hamiltonian, and the underlying group symmetry of the hydrogen hamiltonian is SO(4).