Unbound particle through potential

In summary, the problem involves a particle with mass m moving in the positive x-direction and encountering a potential step of height V0 at x=0. The potential experienced by the particle is V(x) = 0 for x < 0 and V(x) = V0 for x ≥ 0. The task is to determine the time-independent wave function for the particle when its energy E is greater than V0. The solution involves using complex notation and defining the amplitudes of the incident, reflected, and transmitted waves as CI, CR, and CT respectively. The wavenumbers k and k' are also defined in their respective regions. The final solution involves finding the values of the constants A, B, C, and
  • #1
omiros
30
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1. Homework Statement
A particle with mass m moving in the positive x -direction (i.e. from left to right) is incident on a potential step of height V0 at x = 0 so that the potential experienced by the particle is;
V(x) = 0 for x < 0 and V(x) = V0 for x ≥ 0
2. Homework Equations

Determine the time-independent wave function for the particle in the case where the particle energy, E, is greater than V0. This case corresponds to the solution for an ‘unbound’ particle (E > V0). Write your wave functions using complex notation; let the amplitudes of the incident, reflected and transmitted waves be CI, CR and
CT respectively. Define the wavenumber, k, in the region x < 0 and the wavenumber
k' in the region x ≥ 0 .

3. The Attempt at a Solution
ψ(x) = CI*eikx + CR*e-ikx for x < 0 (is probably the first part of the equation).

My main problem is what to do with the second one, as the particle is constantly 'under the influence' of the potential V0 and at the same time I have to find CT when the wave has not been exactly transmitted so the equation can't just be CT*eikx(in my point of view)
 
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  • #2
The solution on the left side of the step is ##\psi(x)=Aexp\left(\frac{i\sqrt{2mE}x}{\hbar}\right)+ Bexp\left(-\frac{i\sqrt{2mE}x}{\hbar}\right)## and the solution on the right side of the step is ##\psi(x)=Cexp\left(\frac{i\sqrt{2m(E-V_{0})}x}{\hbar}\right)+ Dexp\left(-\frac{i\sqrt{2m(E-V_{0})}x}{\hbar}\right)##. Start by determining the conditions that the constants ##A,B,C,D## must satisfy so that ##\psi## and its first derivative are continuous at ##x=0##.
 

What is an unbound particle?

An unbound particle is a particle that does not have a fixed position or bound to a specific region. It can move freely through space without being confined by any potential barriers.

What is a potential in physics?

In physics, potential refers to the energy of a system or particle in relation to its position or configuration. It is a measure of the work that can be done by a force on an object.

How does an unbound particle behave in a potential?

An unbound particle in a potential will experience a change in its energy as it moves through the potential. The particle will accelerate towards regions of lower potential and slow down in regions of higher potential.

What are some examples of potentials?

Some examples of potentials include gravitational potential, electric potential, and magnetic potential. These potentials govern the behavior of particles in their respective fields.

How does the energy of an unbound particle change through a potential?

The energy of an unbound particle will change as it moves through the potential. It will have a higher energy in regions of lower potential and a lower energy in regions of higher potential. This change in energy is what causes the particle to accelerate and decelerate as it moves through the potential.

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