Solving Schrod. Eqn. for Potential Barrier with E > V

In summary: Yes, and that's an oscillating solution. Do you know the equation: exp(ikx) = cos(kx) + i sin(kx)?Yes, I do. It's just that in the case of the potential barrier, I was expecting k2 to be negative, which would give an oscillating solution.
  • #1
mindarson
64
0
I have a question that's driving me insane and I'm sure there's a simple answer that I'm missing for some reason, but I'm not getting my a-ha moment.

Consider 2 cases from intro QM:

Infinite square well

Potential barrier with E > V0

For the infinite square well, the Schrodinger eqn gives

d2ψ/dx2 + k2ψ = 0

Since k2 > 0, this gives oscillating solutions (some combination of sines and cosines). Correct?

For the potential barrier with E > V0, the Schrodinger eqn gives (with the potential jumping from 0 to V0 at x = 0)

d2ψ/dx2 + k2IE = 0 (where kI = √2mE/(hbar2)

prior to the potential step (i.e. for x < 0) and

d2ψ/dx2 + k2IIE = 0 (where kII = √2m(E-V0)/(hbar2)

after the potential step (for x > 0)

My problem is this: As far as I can tell, both kI and kII are positive numbers (since E > V0 > 0). Why, then, do they give non-oscillating solutions? Whereas, for the infinite square well, a positive k gave oscillating solutions?

Am I making a sign error here or just forgetting something from differential equations? What is different about these equations that one of them gives oscillating, and the other non-oscillating, solutions?

Thanks for any help!
 
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  • #2
mindarson said:
I have a question that's driving me insane and I'm sure there's a simple answer that I'm missing for some reason, but I'm not getting my a-ha moment.

Consider 2 cases from intro QM:

Infinite square well

Potential barrier with E > V0

For the infinite square well, the Schrodinger eqn gives

d2ψ/dx2 + k2ψ = 0

Since k2 > 0, this gives oscillating solutions (some combination of sines and cosines). Correct?

For the potential barrier with E > V0, the Schrodinger eqn gives (with the potential jumping from 0 to V0 at x = 0)

d2ψ/dx2 + k2IE = 0 (where kI = √2mE/(hbar2)

prior to the potential step (i.e. for x < 0) and

d2ψ/dx2 + k2IIE = 0 (where kII = √2m(E-V0)/(hbar2)

after the potential step (for x > 0)

My problem is this: As far as I can tell, both kI and kII are positive numbers (since E > V0 > 0). Why, then, do they give non-oscillating solutions? Whereas, for the infinite square well, a positive k gave oscillating solutions?

Am I making a sign error here or just forgetting something from differential equations? What is different about these equations that one of them gives oscillating, and the other non-oscillating, solutions?

Thanks for any help!

Why do you think the solution is non-oscillating? What do you think the solution is?
 
  • #3
Well, I think it all hinges on the sign of k2.

If I have a differential equation

d2f(x)/dx2 = αf(x),

then if α < 0, the equation is satisfied by f(x) = sin(x√α) (or f(x) = cos(x√α) or combination of the two) (oscillating). Whereas if α > 0, the equation is satisfied by f(x) = exp(x√α) (non-oscillating). Which I verify by actually differentiating the functions.
 
  • #4
stevendaryl said:
Why do you think the solution is non-oscillating? What do you think the solution is?

As far as what is leading me to go against my own reasoning, every text I've consulted, both physical and online, tells me that in the region x > 0 where V0 > 0 (but E > V0), the solutions to the appropriate Schr. Eqn are of the form

Cexp(ikx) + Dexp(-ikx)

where k is the wave number appropriate to the region (kII in the original post).
 
  • #5
Ok, I seriously feel like a heel. I see my mistake. The complex exponentials DO describe an oscillating function, according to Euler's equation.

Ugh. Sorry about that :(
 
  • #6
mindarson said:
As far as what is leading me to go against my own reasoning, every text I've consulted, both physical and online, tells me that in the region x > 0 where V0 > 0 (but E > V0), the solutions to the appropriate Schr. Eqn are of the form

Cexp(ikx) + Dexp(-ikx)

where k is the wave number appropriate to the region (kII in the original post).

Yes, and that's an oscillating solution. Do you know the equation: exp(ikx) = cos(kx) + i sin(kx)?
 

Related to Solving Schrod. Eqn. for Potential Barrier with E > V

1. What is the Schrodinger Equation?

The Schrodinger Equation is a mathematical formula used in quantum mechanics to describe the behavior of a quantum particle over time. It takes into account the particle's energy, mass, and potential energy.

2. What is a potential barrier?

A potential barrier is a region of space where a particle's potential energy is higher than its kinetic energy. This creates a barrier that the particle must overcome in order to move through the region.

3. How does the Schrodinger Equation help solve for a potential barrier with energy greater than the potential?

The Schrodinger Equation allows us to calculate the probability of a particle with energy greater than the potential to tunnel through the barrier. By solving the equation, we can determine the wave function of the particle and its probability of tunneling.

4. What factors affect the probability of tunneling through a potential barrier?

The probability of tunneling through a potential barrier is affected by the energy and mass of the particle, the height and width of the barrier, and the shape of the potential function. The larger the energy and lower the mass of the particle, the higher the probability of tunneling.

5. Can the Schrodinger Equation be applied to other potential barriers besides the one with energy greater than the potential?

Yes, the Schrodinger Equation can be applied to a variety of potential barriers, including those with energy less than the potential and those with varying potential functions. It is a versatile equation that is used to study the behavior of quantum particles in various scenarios.

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