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I was reading this article and I'm trying to follow this author's proof. The author jumps from

$$\psi_1(x)\frac{\partial^2\psi_2(x)}{\partial x^2}-\psi_2(x)\frac{\partial^2\psi_1(x)}{\partial x^2}=0$$ to

$$\psi_1(x)\frac{\partial\psi_2(x)}{\partial x}-\psi_2(x)\frac{\partial\psi_1(x)}{\partial x}=Constant$$

I've tried integrating by parts but it just seems to get mucky fast. Can anyone help me clarify how this is such a simple procedure?

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You may get further by looking up the term "first integral of motion", which is what the second line is. – BobaFret Oct 21 '12 at 5:24
up vote 2 down vote accepted

For clarity, let $\partial$ stand for $\frac{\partial}{\partial x}$ and $\partial^2$ for the iterated $\partial$. Now by the product rule run slightly in reverse

$$\psi_1 \partial^2 \psi_2 = \partial(\psi_1 \partial\psi_2) - \partial\psi_1\partial\psi_2$$


$$\psi_2 \partial^2 \psi_1 = \partial(\psi_2 \partial\psi_1) - \partial\psi_2\partial\psi_1$$

Finally, we have

$$\psi_1 \partial^2 \psi_2 - \psi_2 \partial^2 \psi_1 = \partial(\psi_1 \partial\psi_2) - \partial\psi_1\partial\psi_2 - (\partial(\psi_2 \partial\psi_1) - \partial\psi_2\partial\psi_1)$$

which equals

$$\partial(\psi_1 \partial\psi_2) - \partial(\psi_2\partial\psi_1) = \partial(\psi_1 \partial\psi_2-\psi_2\partial\psi_1) $$

which means that

$$\psi_1 \partial\psi_2-\psi_2\partial\psi_1 = \text{constant}$$

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Thanks so much Aleks. That's a neat trick. – tquarton Oct 21 '12 at 5:48

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