(adsbygoogle = window.adsbygoogle || []).push({}); 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.

**Physics Forums - The Fusion of Science and Community**

# Triangular potential Well

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Triangular potential Well

Loading...

**Physics Forums - The Fusion of Science and Community**