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

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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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stevendaryl
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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?
 
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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.
 
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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
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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
stevendaryl
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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)?
 

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