Potential barrier. Schroedinger equation.

[LagrangeEuler]

Homework Statement

Schroedinger equation for potential barrier.
What if $V_0=E$
First region. Particles are free.
$\psi_1(x)=Ae^{ikx}+Be^{-ikx}$
In third region
$\psi_3(x)=Ce^{ikx}$

Homework Equations

$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}(V_0-E)\psi=0$
where $V_0$ is height of barrier.
For region II

The Attempt at a Solution

In second region
$\frac{d^2 \psi}{dx^2}=0$
from that
$\frac{d\psi}{dx}=C_1$
$\psi(x)=C_1x+C_2$
Boundary condition
$A+B=C_2$
$C_1a+C_2=Ce^{ika}$
$ikA-ikB=C_1$
$C_1=ikCe^{ika}$
System 4x4
Is this correct?
Could you tell me in this case do I have bond state?

 

[dauto]

Yes, it seems correct. Bond states have negative energy.

 

[LagrangeEuler]

Bound states have negative energy? Can you explain me this. In case of this problem.

 

[dauto]

The oscillatory solutions of regions 1 and 3 happen because energy is positive and the wave is free to propagate to infinity. (unbounded particle). If the energy is negative you get a exponentially decaying wave function and the wave does not propagate to infinity (bounded particle).

 

[paranormal]

I would like to clarify a few things about the content provided and offer a response to the question posed.

Firstly, the content provided does not specify the potential barrier, so it is difficult to accurately discuss the solution to the Schroedinger equation. However, I will assume that the potential barrier is a step function with height V_0 and extends from x=0 to x=a.

In this case, the solution to the Schroedinger equation in the first region (x<0) is given by $\psi_1(x)=Ae^{ikx}+Be^{-ikx}$, as provided in the content. This solution represents a plane wave traveling in the positive and negative x-directions. However, the content does not specify the energy of the particles in this region, so I cannot comment on whether they are truly free or not.

In the third region (x>a), the solution provided is $\psi_3(x)=Ce^{ikx}$. This represents a plane wave traveling in the positive x-direction.

Now, the content asks what happens if V_0=E, which implies that the energy of the particles is equal to the height of the potential barrier. In this case, the second region (0<x<a) becomes a potential well, and the solution to the Schroedinger equation in this region is given by $\psi_2(x)=C_1\cos(kx)+C_2\sin(kx)$. This solution represents a standing wave with nodes at x=0 and x=a.

The boundary conditions provided in the content are not sufficient to determine the coefficients A, B, C, C_1, and C_2. However, assuming that we have a particle with energy E>V_0, we can use the continuity of the wavefunction at x=0 and x=a to determine the coefficients. This results in four equations and four unknowns, which can be solved to find the values of A, B, C, and C_2.

In this case, we do not have a bound state, as the particle can tunnel through the potential barrier if it has enough energy. The solution in the second region is not a standing wave, but rather a combination of incoming and outgoing waves.

In conclusion, the provided solution is not entirely correct and more information is needed to accurately solve the Schroedinger equation for a potential barrier.