Solving the Schrodinger Equation Using Substitution Method: Tips and Tricks

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I tried simply substituting z and epsilon into the original equation. I managed to get the second term of the left hand side correct but not the first term as I don't know how to turn z into d^2/dz^2. Can you please give me suggestions as to how I can approach this question.

Thanks
 
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You have a d/dz that you've ultimately got to express in terms of d/dx, right? Try thinking back on your elementary calculus formulas, and see if you can find one that allows you to convert a derivative with respect to one variable into a derivative with respect to a different one.
 
The only thing that I can think of is implicit differentiation. But trying that doesn't seem to work. Am I on the right track?
 
Having the \frac{d}{dz} out by itself can sometimes make things confusing. Try expanding out the parentheses so the \psi can stick onto the derivative. Also, for a moment, let's ignore the fact that it's a second derivative, and pretend it's only a first derivative. So we now have

\frac{d\psi}{dz}

We're assuming this matches up to the original problem, so this term must turn into some kind of d/dx. That term just has constants on it, so the translation between one and the other must just be a simple coefficient. So if we call that A, then we have

\frac{d\psi}{dz}=A\frac{d\psi}{dx}

You should now be able to find a calculus rule that tells you how to calculate A.
 
I used the chain rule to get dS/dz = dS/dx * (h/mw)^0.5 . Can I simply square this to get the second derivative?

S = wavefunction

Thanks for the help!
 
Yup. Now that you know the trick, you can back up and do it with the operators, which will make it a little clearer what's going on:

\frac{d}{dx} = \frac{d}{dz}\frac{dz}{dx}
\frac{d}{dz} = \frac{1}{\frac{dz}{dx}}\frac{d}{dx}
\frac{d^2}{dz^2} = \frac{d}{dz}\frac{d}{dz}
= \left(\frac{1}{\frac{dz}{dx}}\frac{d}{dx}\right)\left(\frac{1}{\frac{dz}{dx}}\frac{d}{dx}\right)
= \left(\frac{1}{\frac{dz}{dx}}\right)^2\frac{d}{dx}\frac{d}{dx}
= \left(\frac{1}{\frac{dz}{dx}}\right)^2\frac{d^2}{dx^2}
 
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