Wave function in Infinite/Finite Potential Wells

[lee_sarah76]

Homework Statement

What is the functional form of the wave function in the ground state in the
five regions x<0, 0<x<a, a<x<b, b<x<L, and x>L?

I've attached the picture of the potential well as well here:

Homework Equations

Schrodinger time independent equation

The Attempt at a Solution

I understand that for x < 0 and x > L, ψ = 0 because there's no possibility of the particle being there.

And then between a and b: ψ = A*sin(kx) + B *cos(kx) where k = √(2mE)/(hbar)²

It's between 0 < x < a and b < x < L that I'm having issues. Can anyone point me in the right direction?

Thanks!

 

[smithhs]

So remember how you get k in the first place in the region where the potential is 0:

\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}=E\psi

The solution comes from 'guessing' ψ = A\sin(kx) + B\cos(kx), and plugging this into the differential equation and algebraically solving for k.

Now consider the regions where V=U_0


\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}+U_0\psi=E\psi

\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}=(E-U_0)\psi

Notice how this is the same equation as above, the only difference being E-U_0 instead of E.

At this point you just assume a similar solution for \psi in the two regions with potential U_0 as you did with in the 0 potential region. Note however that the coefficients infront of the sines and cosines in each of the 3 regions are different in each region; to solve for each one continuity conditions, boundry values and normalizationmust be applied.

 

[TSny]

You should consider if the the ground state wavefunction will be sinusoidal or exponential in regions 0<x<a and b<x<L.

 

[lee_sarah76]

So I get that it will be the same as the function for a < x < b except that the constants in front of the sine/cosine will be different and k = √2m(E - U_o)/ (hbar)²?

Is that right?

 

[vela]

Maybe, maybe not. Did you consider what TSny said? How did you conclude the wave function would sinusoidal and not exponential in that region?

 

[smithhs]

The others have a point; the solutions you 'should' assume had the form

\psi(x)=Ce^{zx}+De^{-zx}


For some z that may end up being real, imaginary or complex; this is how you get the original ('particle in a box') solution in the first place.

 

[lee_sarah76]

Okay I think I see that. But how do we determine if the wavefunction will be sinusoidal or exponential?

 

[TSny]

Okay I think I see that. But how do we determine if the wavefunction will be sinusoidal or exponential?

Recall the finite square well problem. What is the condition to get exponential behavior of the wavefunction outside the well?

 

[lee_sarah76]

Oh I see. So if the particle is bound by the potential, we know V > E, so we "guess" the exponential solution. But if the particle is free and E > V, then we guess the sinusoidal solution? Is that the correct interpretation?

 

[TSny]

Yes, although I wouldn't say it involves "guessing". If you study the form of the Schrodinger equation for the case where V_o > E, then you can see that the solution must be exponential. (But maybe you're using "guessing" more loosely.) Anyway, you're on the right track!

 

[lee_sarah76]

Ah, yes, I'm using "guessing" in the the sense that as I'm learning differential equations, the book tells us to "guess" solutions to the second order differential equations. It's not really guessing, but I've gotten used to the term.

Thanks so much!