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Triangular potential Well

by QuantumJG
Tags: potential, triangular
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QuantumJG
#1
Apr8-10, 07:06 AM
P: 32
1. The problem statement, all variables and given/known data
Show that the TISE expression you found in part (a):

I found it (a) to be:

[tex]\frac{d^{2} \psi}{dx^{2}} - \frac{2m}{\hbar ^{2}}(e \xi x - E) \psi (x) = 0[/tex]

Show it (a) can be simplified to:

[tex]\frac{d^{2} \psi}{dw^{2}} - w \psi = 0[/tex]

2. Relevant equations

w = z - z0

where:

[tex]z = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } x[/tex]

[tex]z_{0} = \frac{2mE}{\hbar ^{2}} \left( \frac{ \hbar ^{2} }{ 2me \xi } \right)^{ \frac{2}{3} } [/tex]

3. The attempt at a solution

So far I have found:

[tex]w = \left( \frac{2m}{ \hbar ^{2} e^{2} \xi ^{2} } \right)^{ \frac{1}{3} } (e \xi x - E) [/tex]

But my problem is:

how do I convert:

[tex] \frac{d^{2} \psi}{dx^{2}} [/tex] to [tex] \frac{d^{2} \psi}{dw^{2}} [/tex], because I'm totally stuck on that.
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nickjer
#2
Apr8-10, 03:08 PM
P: 675
Well, you know w in terms of x. So what is dw in terms of dx? Then solve for dx^2, and plug it into your Schrodinger Eq.
QuantumJG
#3
Apr8-10, 05:20 PM
P: 32
well,

[tex] dw = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } dx[/tex]

[tex] \therefore dw = z dx [/tex]

But I'm confused as to how thats going to help.

vela
#4
Apr8-10, 05:21 PM
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Triangular potential Well

Quote Quote by QuantumJG View Post
1. The problem statement, all variables and given/known data
Show that the TISE expression you found in part (a):

I found it (a) to be:

[tex]\frac{d^{2} \psi}{dx^{2}} - \frac{2m}{\hbar ^{2}}(e \xi x - E) \psi (x) = 0[/tex]

Show it (a) can be simplified to:

[tex]\frac{d^{2} \psi}{dw^{2}} - w \psi = 0[/tex]

2. Relevant equations

w = z - z0

where:

[tex]z = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } [/tex]

[tex]z_{0} = \frac{2mE}{\hbar ^{2}} \left( \frac{ \hbar ^{2} }{ 2me \xi } \right)^{ \frac{2}{3} } [/tex]
Are you sure this is right? Where did x go? And where did the cube roots come from?
3. The attempt at a solution

So far I have found:

[tex]w = \left( \frac{2m}{ \hbar ^{2} e^{2} \xi ^{2} } \right)^{ \frac{1}{3} } (e \xi x - E) [/tex]
This seems unnecessarily complicated. Perhaps I'm missing something obvious, but if you compare the original and simplified equations, can't you see what w is equal to by inspection?
But my problem is:

how do I convert:

[tex] \frac{d^{2} \psi}{dx^{2}} [/tex] to [tex] \frac{d^{2} \psi}{dw^{2}} [/tex], because I'm totally stuck on that.
Use the chain rule:

[tex]\frac{df}{dx} = \frac{dw}{dx}\frac{df}{dw}[/tex]
nickjer
#5
Apr8-10, 05:25 PM
P: 675
Quote Quote by QuantumJG View Post
well,

[tex] dw = \left( \frac{2me \xi }{ \hbar ^{2} } \right)^{ \frac{1}{3} } dx[/tex]

[tex] \therefore dw = z dx [/tex]

But I'm confused as to how thats going to help.
Knowing this and squaring it to get dx^2, you can replace the denominator in

[tex]\frac{d^2 \psi}{dx^2}[/tex]

with that and get the 2nd derivative in terms of w. Then you will want to divide out the factor in front of your new 2nd derivative to get the 2nd equation you listed in your first post.

EDIT: The chain rule will also get you the same thing. But since w is linearly proportional to x, I just take the proportionality constant out and square it.

EDIT2: Also, the cube roots look right. They are needed to cancel out all the coefficients in the final form of the equation. If QuantumJG finishes this last step, then he should get a simple ODE in terms of w only and no other terms.
QuantumJG
#6
Apr8-10, 05:37 PM
P: 32
Quote Quote by vela View Post
Are you sure this is right? Where did x go? And where did the cube roots come from?
I just realised I forgot to put x into the equation for z

Quote Quote by vela View Post
This seems unnecessarily complicated. Perhaps I'm missing something obvious, but if you compare the original and simplified equations, can't you see what w is equal to by inspection?

Use the chain rule:

[tex]\frac{df}{dx} = \frac{dw}{dx}\frac{df}{dw}[/tex]
but then wouldn't:

[tex] \frac{d}{dx} \left( \frac{df}{dx} \right) = \frac{d}{dx} \left( \frac{dw}{dx}\frac{df}{dw} \right) [/tex]

[tex] = \frac{df}{dw} \frac{d}{dx} \left( \frac{dw}{dx} \right) + \frac{dw}{dx} \frac{d}{dx} \left( \frac{df}{dw} \right) [/tex]
nickjer
#7
Apr8-10, 05:39 PM
P: 675
Use the chain rule on the last term:

[tex]\frac{dw}{dx}\frac{dw}{dx}\frac{d}{dw}\left(\frac{df}{dw}\right)[/tex]
QuantumJG
#8
Apr8-10, 05:47 PM
P: 32
Quote Quote by nickjer View Post
Use the chain rule on the last term:

[tex]\frac{dw}{dx}\frac{dw}{dx}\frac{d}{dw}\left(\frac{df}{dw}\right)[/tex]
Wait thats confusing, why did you put in that extra [tex] \frac{dw}{dx} [/tex]?
vela
#9
Apr8-10, 05:51 PM
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Quote Quote by nickjer View Post
EDIT2: Also, the cube roots look right. They are needed to cancel out all the coefficients in the final form of the equation. If QuantumJG finishes this last step, then he should get a simple ODE in terms of w only and no other terms.
Oh, okay, I did miss something obvious.
vela
#10
Apr8-10, 05:56 PM
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Quote Quote by QuantumJG View Post
Wait thats confusing, why did you put in that extra [tex] \frac{dw}{dx} [/tex]?
That's the one from the first differentiation. Because dw/dx is a constant, you get

[tex]\left(\frac{d}{dx}\right)^2 = \frac{dw}{dx} \frac{d}{dw}\left(\frac{dw}{dx} \frac{d}{dw}\right) = \left(\frac{dw}{dx}\right)^2 \frac{d^2}{dw^2}[/tex]


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