Finite Well Solutions: Understanding d and k in Potential Wells

[physicsjock]

Hey,

I've been trying to work out how, for a finite well of high Vo and width L, the interior solution has the form L Sin(kx + d),

I see that if d=0 then the solution resembles an infinite well, so that implies d depends inversely on the wells potential. But I can't work out what d comes from, and why the constant at the front is the length of the well (why is the amplitude of the wave the length of the well)

d is just the phase of the wave, so does it represent the reflection of any particles which moved past the well?

Ive also been trying to work out what the k is, when I go through the process of finding the wave function inside I end up with (s/k)Asinkx + Acoskx after applying the boundary conditions.

where k²=2mE/$\hbar^{2}$

and s²=2m(Vo-E)/$\hbar^{2}$

i've been trying to work this out because the question also says that the solution inside the well should also satisfy

ka = n$\pi - 2Sin^{-1}(\frac{k\hbar}{\sqrt{2mV_{o}}})$

which makes me think the d resembles that arcsin the above equation

Anyone have any ideas?

 

[Redbelly98]

It looks like you're trying to solve the time-independent Schrödinger equation, which should have been listed as a "Relevant Equation" in the homework template. It also puts this in the Advanced Physics category, so I'll move the thread there.

Hey,

I've been trying to work out how, for a finite well of high Vo and width L, the interior solution has the form L Sin(kx + d),

I see that if d=0 then the solution resembles an infinite well, so that implies d depends inversely on the wells potential. But I can't work out what d comes from, and why the constant at the front is the length of the well (why is the amplitude of the wave the length of the well)

I don't think the constant is simply L -- the units are wrong -- but at any rate it would be found by normalizing the wavefunction.

d is just the phase of the wave, so does it represent the reflection of any particles which moved past the well?

It represents the fact that the potential is not zero at the well boundary x=0. This is related to the fact that the potential is nonzero beyond the boundary as well, i.e. the particle's wavefunction penetrates beyond the boundary. I wouldn't really call that reflection though.

Ive also been trying to work out what the k is, when I go through the process of finding the wave function inside I end up with (s/k)Asinkx + Acoskx after applying the boundary conditions.

where k²=2mE/$\hbar^{2}$

and s²=2m(Vo-E)/$\hbar^{2}$

i've been trying to work this out because the question also says that the solution inside the well should also satisfy

ka = n$\pi - 2Sin^{-1}(\frac{k\hbar}{\sqrt{2mV_{o}}})$

where a = ?

which makes me think the d resembles that arcsin the above equation

Anyone have any ideas?

It's hard to tell exactly what you are stuck on. Is it just in trying to relate the two forms of solution you have given,
$$(s/k)A\sin kx + A\cos kx \ \small \text{ and } \normalsize \ B \sin(kx + d) \text{,}$$
to each other?

 

[physicsjock]

It's hard to tell exactly what you are stuck on. Is it just in trying to relate the two forms of solution you have given,
$$(s/k)A\sin kx + A\cos kx \ \small \text{ and } \normalsize \ B \sin(kx + d) \text{,}$$
to each other?

Yea that's what I'm stuck on,

Sorry that ka = ... formula I wrote was from a set of notes where the the width of the well is a,

so a = L,

I've tried substituting kl into the interior solution, ψ(L)
where
ψ(x)=(s/k)Asinkx + Acoskx
so ψ(L)=(s/k)AsinkL + AcoskL
and ended up with
ψ(L)=-A cos(n$\pi$)
assuming that the k in that formula is the same as the k i posted before

 

[Redbelly98]

Yea that's what I'm stuck on,

Try using trig angle-sum identities on sin(kx+d). I'm pretty sure that will work out to a form that you can more easily equate with

(s/k)Asinkx+Acoskx​

Sorry that ka = ... formula I wrote was from a set of notes where the the width of the well is a,

so a = L,

Okay.

I've tried substituting kl into the interior solution, ψ(L)
where
ψ(x)=(s/k)Asinkx + Acoskx
so ψ(L)=(s/k)AsinkL + AcoskL
and ended up with
ψ(L)=-A cos(n$\pi$)
assuming that the k in that formula is the same as the k i posted before

I'm not following you here, mainly because I can't do this in my head and don't have time right now to work it out on paper to verify what you are saying.

But, if you are satisfied with the "(s/k)Asinkx+Acoskx" form of the solution, we should probably just concentrate on seeing how "Sin(kx + d)" is equivalent. Will you also be needing to normalize the wavefunctions, and finding the energy eigenvalues, or are you good with how to do that?

 

[physicsjock]

Hey,

I worked it out, the d in the sign was just to compensate for the lack of cos in the solution,

I found the d was something like arcTan(k/s) and it was consistent with the boundaries and stuff,

Thanks for your help!