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In page number 37 of Weinberg's lectures of quantum mechanics book (2nd edition), After Eq.2.1.17, he states the following:

The Schrödinger equation (2.1.3) then takes the form

$E \psi(x) = -\frac{\hbar^2}{2\mu r^2} \frac{\partial}{\partial r} (r^2 \frac{\partial \psi(x) }{\partial r} ) + \frac{1}{2\mu r^2} L^2\psi(x) + V(x) \psi(x)$. (2.1.17)

Now let us consider the spectrum of the operator $L^2$. As long as $V(r)$ is not extremely singular at $r = 0$, the wave function $\psi$ must be a smooth function of the Cartesian components $x_i$ near $x = 0$, in the sense that it can be expressed as a power series in these components. Suppose that, for some specific wave function, the terms in this power series with the smallest total number of factors of $x_1$, $x_2$, and $x_3$ have $l$ such factors. Here $l$ can be $0$, $1$, $2$, etc. The sum of all these terms forms what is called a homogeneous polynomial of order $l$ in $x$.

What I do not understand is why does he assume that wave functions for a particle in central potential near $r = 0$ must be homogeneous polynomials? Why cannot they be inhomogeneous?

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    $\begingroup$ In the future, please write $x_i$ as $x_i$ rather than x$_i$. $\endgroup$
    – J.G.
    Oct 14, 2021 at 7:02
  • $\begingroup$ @J.G. Thanks for the edit and suggestion :) $\endgroup$ Oct 14, 2021 at 9:36

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He is basically considering a Taylor expansion of $\psi$ around zero and looking at the lowest order, i.e., the smallest $ l = a+b+c$ such that $$\Big(\frac{\partial^{a+b+c}}{\partial x_1^a \partial x_2^b \partial x_3^c} \psi\Big)(\vec 0) \neq 0.$$

For a multi-dimensional polynomial, each order can contain several terms (Wikipedia has the full formula).

Consider, e.g., $\psi(\vec x)$ such that the lowest order is given by $\psi_2(\vec x) = x_1 x_2 + x_1 x_3$. Then, it is homogeneous of order 2, i.e. $\psi_2(\lambda \vec x) = \lambda^2 \psi_2(\vec x)$.

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  • $\begingroup$ Thank you, I did not notice that he was considering only the lowest term in the expansion. Unfortunately I am unable upvote due to low reputation. $\endgroup$ Oct 14, 2021 at 9:23
  • $\begingroup$ @AdityaKurrodu I'm glad I could help! $\endgroup$
    – Cream
    Oct 14, 2021 at 11:45

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