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For an axisymmetric potential, insert $\Psi (r,\varphi )=R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation and divide by $R\Phi$. The polar Laplacian separates the angular equation ${\Phi }^{\prime \prime }+m^2\Phi =0$ fromSingle-valuedness under $\varphi \mapsto \varphi +2\pi$ gives $m\in \mathbb{Z}$, with an angular eigenbasis $\boxed{\Phi =e^{im\varphi }}$. Linear combinations of the degenerate $m,-m$ modes are also possible.

Inside the circular infinite quantum well, put $\rho =r\sqrt{2\mu E/{\hslash }^2}$. The equation becomesThe physical solution must be regular at the origin, behaving as $r^{|m|}$ for $m\ne 0$, or with finite value and zero radial derivative for $m=0$. At the infinite wall impose $R(r=a)=0$. The printed question's boundary phrase uses capital $R=a$; it means the radial position $r=a$, not a value imposed on the radial wavefunction. The singular Bessel function of the second kind is excluded; mere square-integrability of its logarithmic $m=0$ singularity would not put it in the ordinary finite-energy Hamiltonian domain.

For $m=0$, multiply the equation by ${\rho }^2$ and insert $R={\sum }_{k\ge 0}{A}_k{\rho }^k$. The coefficient of $\rho$ gives $A_1=0$, and subsequent coefficients giveAll odd coefficients vanish, $A_2=-A_0/4$, and $A_4=A_0/64$. The regular solution is a multiple of the Bessel function of the first kind $J_0$. Its requested fourth-order estimate isTherefore the first positive zero of this truncation is $\boxed{\rho \approx \sqrt{8}}$, givingThe ground state is in the $m=0$ sector because nonzero angular modes add nonnegative centrifugal kinetic energy to the radial energy form. This quartic truncation of the Bessel function J0 is only a rough local-series estimate, not an exact eigenvalue or a certified variational bound; its double zero warns that the truncation does not reproduce the simple zero of the full function.