Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We should all fondly remember this basic undergraduate problem: A quantum particle is incident (from the left) upon a potential barrier of height V and width L. Compute the transmission and reflection coefficients.

However, this isn't the full story. To completely solve the potential barrier, we need a complete set of orthogonal and normalized energy eigenstates. The wavefunctions of the leftward traveling particles are just some of the energy eigenstates. The wavefunctions of the rightward traveling particles are some more energy eigenstates. Are there more possible states? What is a complete energy eigenbasis for the Hilbert Space of a single particle in a step potential?

(I'm thinking about a positive potential barrier +V, but one could ask this question for a finite potential well too.)

share|improve this question

1 Answer 1

For a given allowed energy $E$ (greater than both values of $V$ in the step-like potential), the subspace of the Hilbert space associated with this eigenvalue is two-dimensional. It's easily seen because Schrödinger's equation is a differential equation with up to second derivatives with respect to $x$. So any basis that will contain two basis vectors for such an $E$ will be OK. For example, one may pick a particle traveling from the left to the right, being partially reflected to the left, together with a particle traveling from the right to the left, being partially reflected to the right, is an OK basis.

For energies $E$ between $V_1$ and $V_2$, the two values of the potential, there are still eigenstate but the particle can't be transmitted to the region with the higher potential which is classically impenentrable. There will still be eigenstates for these energies $E$ but only one for each value of $E$ in this interval. A legitimate energy-based basis will contain real solutions that look like a (shifted) sine on the "allowed side" and exponentially decrease on the other side.

There are no eigenstates with $E$ smaller than both values of the potential $V_1,V_2$.

If one is really picky, one should discuss the special cases $E=V_1$ and $E=V_2$ separately. The spaces will be one- or two-dimensional here but it isn't really important for physics because these special eigenvalues are of measure zero in the continuum, the corresponding eigenstates aren't normalizable, and their contribution to the normalizable states may be neglected.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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