1
$\begingroup$

Why shouldn't the groundstate wavefunction for symmetric potentials vanish at the origin?

$\endgroup$
2
$\begingroup$

For a symmetric potential (in 1D at least), even if you have an even function as the ground state solution, the fact that it is even says nothing about the value it should take at the origin. An even function is defined as one that satisfies $$f(-x)=f(x)$$ so for $x=0$ this says nothing non-trivial. If the wave-function is non-zero elsewhere than the origin, then if it is zero at the origin it must have turning points elsewhere, and will be more "wiggly" as a function. Roughly speaking, the more wiggly a wave-function the more kinetic energy that system will have, so the ground state is unlikely to have many wiggles!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.