
#1
Apr810, 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  z_{0} 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. 



#2
Apr810, 03:08 PM

P: 676

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.




#3
Apr810, 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. 



#4
Apr810, 05:21 PM

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PF Gold
P: 11,524

Triangular potential Well[tex]\frac{df}{dx} = \frac{dw}{dx}\frac{df}{dw}[/tex] 



#5
Apr810, 05:25 PM

P: 676

[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. 



#6
Apr810, 05:37 PM

P: 32

[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] 



#7
Apr810, 05:39 PM

P: 676

Use the chain rule on the last term:
[tex]\frac{dw}{dx}\frac{dw}{dx}\frac{d}{dw}\left(\frac{df}{dw}\right)[/tex] 



#8
Apr810, 05:47 PM

P: 32





#9
Apr810, 05:51 PM

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#10
Apr810, 05:56 PM

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P: 11,524

[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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