The identity ${\cos }^{2}x=\frac{1}{2}+\frac{1}{4}{e}^{2ix}+\frac{1}{4}{e}^{-2ix}$ gives the momentum eigenstate expansionUsing the orthonormal basis from part (a), the wavefunction normalization isHence $|A|=2/\sqrt{3\pi }$. The wavefunction normalization leaves a global phase free; the positive real choice used in the requested expansion givesThe two nonzero eigenvalues of the momentum operator have the same energy eigenvalue $E_{\star }=2{\hslash }^2/m$. Under unitary time evolution, their coefficients acquire the same dynamical phase. With $\Omega =E_{\star }/\hslash =2\hslash /m$,At $t=0$, $1+\cos (2x)=2{\cos }^{2}x$ recovers the initial wavefunction. Its probability density, useful below, is

The unperturbed $n=2$ level is degenerate. The useful point is that a central perturbation is already diagonal within its usual orbital eigenbasis:Angular orthogonality removes the possible mixing of $2s$ and $2p$, and also the mixing of different magnetic sublevels. Therefore the matrix required by degenerate perturbation theory needs no further diagonalization: its diagonal entries are precisely the expectation values above. It would be incorrect to justify the calculation by claiming that the unperturbed spectrum was nondegenerate.

For $b=0$ the energy needs only $n$. For a nonzero central confinement term the energy generally needs $n,l$, but remains independent of $m$. The magnetic quantum number is still needed to specify an individual orbital eigenstate, rather than just its energy.

The linear confinement correction to Coulomb bound states uses $H^{\prime }=br$, and its first-order energy correction in a normalized Coulomb eigenstate is $b⟨r⟩$. Replace $a_0$ by $a_q$ in the hydrogen wavefunctions. The normalized spherical harmonics remove the angular integral, leavingFor $1s$, the integral of $r^3{e}^{-2r/a_q}$ yieldsFor $2s$, put $u=r/a_q$ and square the full radial polynomial, including its cross term:For $2p$, the same calculation givesConsequently the first two Coulomb levels become, to first order,Thus the second level splits into a singlet orbital level and a triply degenerate orbital level, with $E_{2s}-E_{2p}=b{a}_q+O(b^2)$. For the confining sign $b>0$, $2p$ lies lower than $2s$: its smaller mean radius gives the smaller positive correction. The expansion assumes these corrections and the induced couplings to other principal shells are small compared with the relevant Coulomb energy gaps; $b{a}_q^2/a\ll 1$ is the governing small parameter for fixed low $n$.

Ignoring spin and working in the fixed quarkonium internal sector, the orbital degeneracy is ${\sum }_{l=0}^{n-1}(2l+1)=n^2$. HenceNo extra spin or colour multiplicity is being counted. The degeneracy between different $l$ at fixed $n$ is special to the Coulomb potential.

Separate the center-of-mass motion from the relative motion. Two equal quark masses give reduced mass $\mu =m_q/2$. Replacing the hydrogen Coulomb strength by $a>0$ gives the Coulomb binding energies and length scaleThe factor of two in the reduced mass matters: using $m_q$ directly would double the binding energy. These are relative binding energies measured from the free quark–antiquark threshold; the rest energy $2m_q{c}^2$ is excluded.

Direct integration over solid angle givesThus $|C|=\sqrt{(2l+1)!/(4\pi )}/(2^{l}l!)$. Choosing the usual alternating sign convention,Apply $L_{-}$ successively and divide by its ladder coefficient at each step. Explicitly,The normalization follows because the product of the squared ladder coefficients is ${\hslash }^{2(l-m)}(2l)!(l-m)!/(l+m)!$. This constructs every magnetic sublevel, including negative $m$.