[Basic QM] Potential step & particle flow, my answer is inconsistent ?

Thank you so much!In summary, the conversation is about a homework problem involving a flow of particles colliding with a potential step. The potential is zero for x < 0 and a certain value V_0 for x > 0. The question is to find the general form of the wavefunction in the area x > 0 and an expression for the wavenumber k. The attempt at a solution involves writing the Schrodinger equation in a more general form and finding that there is an error with a minus sign, leading to incorrect solutions. After correcting this error and substituting back into the differential equation, the correct solution is found.
  • #1
Nick89
555
0

Homework Statement


A flow of particles collides with a potential step. The potential is zero for x < 0 and a certain value [tex]V_0[/tex] for x > 0.
The flow of particles comes in from the left (x<0 area).

One of the questions is to give the general form of the wavefunction in the area x > 0 and to find an expression for the wavenumber k.

Homework Equations


[tex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + V_0 \psi = E \psi[/tex]
(For x > 0)

The Attempt at a Solution



I first write the schrodinger eq. into a more general form:
[tex]-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} + (V_0 - E) \psi = 0[/tex]
[tex]- \frac{d^2 \psi}{dx^2} + \frac{2m (V_0-E)}{\hbar^2} \psi = 0[/tex]
[tex]\frac{d^2 \psi}{dx^2} - \frac{2m (V_0-E)}{\hbar^2} \psi = 0[/tex]
[tex]\frac{d^2 \psi}{dx^2} - k^2 \psi = 0[/tex]

So
[tex]k^2 = \frac{2m (V_0-E)}{\hbar^2}[/tex]

And the solution are complex exponents due to the minus-sign (they would be real exponents if it was a plus sign):
[tex]\psi (x) = A e^{ikx} + B e^{-ikx}[/tex]Now, when I look at the answer, I noticed that they also use complex exponents, but their k was different:
[tex]k^2 = \frac{2m (E-V_0)}{\hbar^2}[/tex]
(E-V instead of V-E)

This is a pretty important error because:

In my case: If E > V, the particle is 'above the potential step' and thus it should be a complex exp. (sin/cos). But, if E > V then k becomes complex, and the exponents become real!
If E < V, the particle is 'inside the potential step' and thus it should be a real exponent (as usual with penetration in classically forbidden area). But, if E < V then k is real and the exponents stay complex!...

Obviously I made an error with a minus sign somewhere, but I honestly can't find it..!?I also tried switching the minus sign for a plus sign (and swapping the E and V again obviously) to get:
[tex]\frac{d^2 \psi}{dx^2} + \frac{2m (E-V_0)}{\hbar^2} \psi = 0[/tex]
So now, my k is in accordance with the answer (E-V), but because of the + I get REAL exponents, and if E > V, k is real and the exponents stay real (wrong), and if E < V, k is complex and the exponents become complex (wrong)...?

I can't figure it out...
 
Physics news on Phys.org
  • #2
Nick89 said:
...
[tex]\frac{d^2 \psi}{dx^2} - k^2 \psi = 0[/tex]

So
[tex]k^2 = \frac{2m (V_0-E)}{\hbar^2}[/tex]

And the solution are complex exponents due to the minus-sign (they would be real exponents if it was a plus sign):
[tex]\psi (x) = A e^{ikx} + B e^{-ikx}[/tex]


This is where things went backwards. The solution to
[tex]\frac{d^2 \psi}{dx^2} = k^2 \psi[/tex]
is
[tex]\psi (x) = A e^{kx} + B e^{-kx}[/tex]
with no [tex]i[/tex] in the exponents. In order to get the other form we should define
[tex]k^2 = \frac{2m (E-V_0)}{\hbar^2}[/tex]
so that
[tex]\frac{d^2 \psi}{dx^2} = -k^2 \psi.[/tex]

This is the answer "they" get (and has the right behavior for [tex]E > V_0[/tex], [tex]E < V_0[/tex]), but you can easily check the signs just by substituting [tex]\psi[/tex] back into the differential equation.
 
  • #3
Omg... You're completely right of course. I checked and re-checked that, and I was actually under the impression that the solution to
[tex]y'' - k^2y = 0[/tex]
had complex exponents and the solution to
[tex]y'' + k^2y = 0[/tex]
had real exponents... It's the other way around of course. Can't believe I mixed them up, must have been the holidays hehe...
 

1. What is a potential step in quantum mechanics?

A potential step in quantum mechanics refers to a sudden change in the potential energy of a particle. This can occur when a particle encounters a barrier or potential well, causing its energy level to change abruptly.

2. How does a potential step affect particle flow in quantum mechanics?

A potential step can either increase or decrease the flow of particles in quantum mechanics, depending on the direction of the step. For instance, if the potential step is towards higher potential energy, it will decrease particle flow, while a step towards lower potential energy will increase flow.

3. What is the Schrödinger equation and how does it relate to potential steps in quantum mechanics?

The Schrödinger equation is a fundamental equation in quantum mechanics that describes the evolution of a particle's wave function over time. It is used to determine how a particle will behave in the presence of potential steps and other barriers or wells.

4. Can a particle's behavior at a potential step be predicted accurately in quantum mechanics?

Yes, the Schrödinger equation allows us to predict the behavior of particles at potential steps with a high degree of accuracy. However, there are certain scenarios where the behavior may be unpredictable, such as when dealing with very small particles or at extremely high energies.

5. How does the uncertainty principle relate to potential steps in quantum mechanics?

The uncertainty principle in quantum mechanics states that it is impossible to know both the position and momentum of a particle with complete precision. This means that the behavior of particles at potential steps cannot be predicted with absolute certainty, but rather with a certain degree of probability.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
1K
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
458
  • Advanced Physics Homework Help
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
30
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
873
  • Advanced Physics Homework Help
Replies
2
Views
766
Replies
16
Views
394
  • Advanced Physics Homework Help
Replies
1
Views
2K
Back
Top