# Question from Weinberg Lectures on quantum mechanics

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?

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

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 $$\Big(\frac{\partial^{a+b+c}}{\partial x_1^a \partial x_2^b \partial x_3^c} \psi\Big)(\vec 0) \neq 0.$$
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. $$\psi_2(\lambda \vec x) = \lambda^2 \psi_2(\vec x)$$.