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The stationary states of a linear potential well are given by the solutions of the following differential equation:

$$\dfrac {\partial^2 f} {\partial x^2} - 2(x - c)f(x) =0$$

On Sakurai's Modern Quantum Mechanics, he claims the substitution $z = 2^{1/3}(x-c)$ transforms this into:

$$\dfrac {\partial^2 f} {\partial z^2} - zf(z) =0$$

Wolfram alpha confirms this solution, but that is not at all evident to me. I would write:

$$\dfrac {\partial^2 f(x)} {\partial z^2} - 2^{1/3}zf(x^2) =0, $$ where $x$ is evaluated at $x = 2^{-1/3}z-c$. I would like to write the equation in the form prescribed by Sakurai, since then the solution is given by Airy's function. What did he do to transform the equation?

RicardoMM
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1 Answers1

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By the chain rule, $$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = 2^{1/3}\frac{\partial f}{\partial z}$$ $$\frac{\partial^{2}f}{\partial x^{2}} = \frac{\partial}{\partial x}\left[\frac{\partial f}{\partial x}\right] = \frac{\partial}{\partial x}\left[2^{1/3}\frac{\partial f}{\partial z}\right] = \frac{\partial z}{\partial x}\frac{\partial}{\partial z}\left[2^{1/3}\frac{\partial f}{\partial z}\right] = 4^{1/3}\frac{\partial^{2} f}{\partial z^{2}}$$ This turns $$\frac{\partial^{2} f}{\partial x^{2}} - 2(x-c)f = 0$$ Into $$4^{1/3}\frac{\partial^{2} f}{\partial z^{2}} - 4^{1/3}zf = 0$$ And dividing by $4^{1/3}$ gives $$\frac{\partial^{2} f}{\partial z^{2}} - zf = 0$$

conan
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  • But aren't the functions valuated at x? Meaning, you proved $\frac{\partial^{2} f(x) }{\partial z^{2}} - zf(x) = 0$ – RicardoMM Apr 30 '23 at 16:56
  • @RicardoMM : Strictly spoken yes. The function $f$ that depends on $x$ is different from the modified function that depends on $z$. And strictly spoken is the denominator in a differential fraction a position indicator, the actual function arguments can be different. But people are lazy, ... – Lutz Lehmann Apr 30 '23 at 18:08
  • @LutzLehmann But what startles me is that the solutions are apparently the same - I wouldn't say $f'(x+1) = x$ has the same solutions as $f'(x) = x$, why is it true here? – RicardoMM Apr 30 '23 at 21:30
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    It is not wrong to abuse notation that way, it can become irritating and the likelihood of errors raises. And the formula in the first comment has to be understood as $\frac{\partial^{2} f(x(z)) }{\partial z^{2}} - zf(x(z)) = 0$, it is the form of the equation after the coordinate change. – Lutz Lehmann May 01 '23 at 04:07
  • @LutzLehmann Thank you, that makes sense. I don't think I've quite understood why it doesn't lead to slightly diferent solutions (I've tried a few simple examples and the differences always get absorved in a constant, so the solutions are the same), perhaps I should revisit my differential equations books. – RicardoMM May 01 '23 at 21:50
  • Discussions of this notation question are for instance in https://math.stackexchange.com/questions/1898306/why-dont-we-use-the-partial-derivative-symbol-for-normal-derivatives, https://math.stackexchange.com/questions/2354032/how-is-it-justified-to-apply-the-chain-rule-to-a-function-when-the-inputs-themse – Lutz Lehmann May 02 '23 at 03:36