**Required math: calculus **

**Required physics: **Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.34.

A variant of the finite square well is the finite step, which has the potential

where is a positive constant energy.

There are two distinct cases here:

- Energy below the barrier:
- Energy greater than the barrier:

We’ll consider first the case where .

In this case, the Schrödinger equation for is

where

This has solution

To keep the solution finite as we must have so the solution is an exponentially decaying wave function:

To the left of the barrier, the Schrödinger equation is

Assuming particles coming in from the left, we have

where

Since the potential is finite everywhere, both and are continuous everywhere, which gives us two boundary conditions at .

This has solution

The reflection coefficient is

That is, the probability of an incoming particle being reflected is 1. This is because the wave function for is exponentially decaying, so the probability of a particle reaching infinity is zero, thus no particles can be transmitted.

For the Schrödinger equation for is

where

We now get travelling wave solutions instead of exponentially decaying ones:

Assuming incoming particles arrive only from the left, we can set . Applying the boundary conditions, we get

with solutions

In this case, the reflection coefficient is

Substituting the expressions for and we get

From this we can get the transmission coefficient

Note that this is *not* equal to . Have we done something wrong?

The answer lies in the fact that the wave for is not the same as the wave for , since the net energy on the right is while on the left it is just . One way of looking at it is in terms of the probability current for the free particle. The probability current must be conserved; this is just a way of saying that particles cannot vanish into, nor arise from, thin air. Since the probability current for a free particle with stationary state

is

the conservation law implies, for the case of the step potential

That is, the influx of particles from the left minus the reflected beam must equal the transmitted beam. Dividing through by we get

The first term is the reflection coefficient we calculated in 26. The second term is the transmission coefficient, which works out to

which is what we got earlier.

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ChristianIn solving for Phi double in the first part, for the constant mu wouldn’t it be 2m(E-Vo) under the radical?

growescienceNo, since in the first part, we’re looking at , so for to be real we need .

Laura Schmidshouldn’t

it be kappa/k * (C^2/A^2) in Eq.32?

growescienceQuite right – fixed now. Thanks.

MacThere seems to be a factor -2m/h² missing from (7)

gwrowePost authorFixed now. Thanks.

PHANUEL ALUOMUwhy cant we get the ratio of D/A as 1+exp(alpha*i)

gwrowePost authorWhich equation are you referring to and what is ? (I don’t use anywhere.)

LiseHow to understand the change in velocity of wave in physics？How V0 influence the wave function in physics instead of in math？Thanks！

soniwhy do equation 5 has no “¡” as in line with equation 8,19?

gwrowePost authorBecause it solves equation 3. Plug (5) into (3) to check.

AngusConcerning (b), is this answer correct? Perhaps it is just the form, but it doesn’t match with what seems the intuitive form of [sqrt(E)-sqrt(E-V)]^2 / V^2

gwrowePost authorWhich equation number are you referring to?

AngusEq. 26, apologies

gwrowePost authorPretty sure it’s right. You can simplify it a bit: