# Why does only e^ikx+Re^-ikx solve the TISE for a potential step?

• timn
In summary, the conversation discusses solving the time-independent Schrödinger equation for a potential step in one dimension. The solution involves using a second derivative of the wave function and considering the direction of the waves. The solution for x<0 involves a sum of two terms representing the incident and reflected waves, while for x>0, only the transmitted wave is considered. The momentum operator is also applied to the wave function to determine the momentum of the particle.
timn

## Homework Statement

Solve the time-independent Schrödinger equation in one dimension for a potential step, i.e. V(x) = 0 for x<0 and V(x) = V_0 for x>0.

## Homework Equations

$$- \frac{\hbar^2}{2m} \frac{d^2u(x)}{dx^2} + V(x)u(x) = Eu(x)$$

## The Attempt at a Solution

Rewrite as (4-3)

$$\frac{d^2u(x)}{dx^2} + \frac{2m}{\hbar^2}( E - V(x) )u(x) = \frac{d^2u(x)}{dx^2} + k^2( 1 - V(x)/E )u(x) = 0$$

Looking at x<0, i.e. V(x)=0:

$$\frac{d^2u(x)}{dx^2} + k^2u(x) = 0 \Leftrightarrow u(x) = Ae^{ikx} + Be^{ikx}$$

The derivation in my textbook claims, at this step:

Is it implicit that all multiples of the suggested u(x) are solutions (for x<0)? Furthermore, from the paragraph on the probability flux, it looks like |R(x)| < 1. Why?

It seems like I'm missing some assumption. Why would the x<0 case be different from the case of a free particle?

Edit: I'm having a similar problem with x>0.

Gasiorowicz talks about the waves having a direction. What is meant by this? Why is the term e^-iqx dismissed?

Last edited:
im assuming here that you have done a previous course on wave physics?
ok in very simple terms, the solution describes an unbound particle moving in some direction with wavenumber k, where k is some vector in n-dimensions, describing the direction of the motion of the particle. if you look closely at your original equation:
−ℏ22md2u(x)dx2+V(x)u(x)=Eu(x)
u have a second derivative of a function on the left and on the right the same function with coefficient E-V. you know that V is positive or zero, so you will find that for an incoming particle upon the barrier (x=0) should E>V, then the solution will be an imaginary complex exponential. in this case your wavefunction u is not normalisable (find out why!) and hence you must describe the particle by a dirac-delta function.
so basically the only way the wavefunction will make physical sense is to say it is like a bullet moving on some well defined trajectory. hence you can assume the particles are like waves incident on some boundary, with some probability of reflection (R) and some probability of transmission (T) which should complement one another to give 1 (100%).
hope this will help

Last edited:
Thank you ardie, that makes it clearer.

If you apply the momentum operator
$$\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$$
to the state ψ(x)=eikx, you get
$$\hat{p}\psi(x) = \frac{\hbar}{i}\frac{\partial}{\partial x}e^{ikx} = (\hbar k)\psi(x)$$
This tells you ψ(x) is a momentum eigenstate, where the momentum of the particle is equal to $\hbar k$.

In both regions, the general solution is of the form Aeikx+Be-ikx, which you can, in light of the above, interpret as particles moving in the positive direction and negative direction. Now it's essentially a matter of applying boundary conditions. In this particular problem, you're interested in what happens if a particle is incident on the potential step when coming from the left. You would expect a reflected wave, in the x<0 region, and a transmitted wave, in the x>0 region. Consequently, you keep both terms for x<0, one representing the incident wave and one representing the reflected wave. For x>0, you only have the transmitted wave, so you drop the e-ikx term.

Much clearer! Looking at the momentum makes it more concrete. Thank you very much.

## 1. Why is the solution to the Time Independent Schrödinger Equation (TISE) for a potential step expressed as eikx + Re-ikx?

The solution to the TISE for a potential step is expressed as eikx + Re-ikx because it satisfies the boundary conditions of the problem. In quantum mechanics, the wave function must be continuous and differentiable at the potential step. This solution, known as the plane wave solution, satisfies these conditions and allows for a smooth transition between the two regions of the potential step.

## 2. What does eikx and Re-ikx represent in the solution to the TISE for a potential step?

In the solution eikx represents the incident wave, which is the wave approaching the potential step from the left side. Re-ikx represents the reflected wave, which is the wave that is reflected off the potential step and travels back to the left side. Together, these waves form a standing wave pattern that describes the behavior of the wave at the potential step.

## 3. How does the wave function change at the potential step in the TISE solution?

At the potential step, the wave function undergoes a sudden change in amplitude and phase. This is due to the sudden change in potential energy at the step. The incident wave is partially transmitted through the step and partially reflected, resulting in the standing wave pattern described by the solution eikx + Re-ikx.

## 4. Can the solution to the TISE for a potential step be applied to other potential barriers?

Yes, the solution eikx + Re-ikx can be applied to other potential barriers as long as the potential is constant in space and time. This solution is also known as the scattering solution and can be used to describe the behavior of a wave encountering any type of potential barrier.

## 5. How does the solution to the TISE for a potential step relate to the probability of finding a particle in a certain region?

The square of the absolute value of the wave function, |eikx + Re-ikx|2, represents the probability density of finding a particle at a certain position. This means that the solution to the TISE for a potential step not only describes the behavior of the wave, but also the probability of finding a particle at different positions in the system.

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