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A hydrogen atom (Coulomb potential) has energy that only depends on $n$ (if we ignore other effects like spin-orbit coupling). In general (not necessarily Coulomb, can be any V), does $E$ depend on both $\ell$ and $n$ (since the radial equation has both $n$ and $\ell$)? What about $m$? Since $m$ doesn't appear in the radial equation, does it mean that energy does not depend on $m$?

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    $\begingroup$ At higher order the energy does depend on m and manifests itself as fine-structure (the spin-orbit coupling term anyway), and at even higher order in the hyperfine structure. $\endgroup$ – J. O'Brien Antognini Jun 1 '16 at 18:11
  • $\begingroup$ Related: physics.stackexchange.com/q/116244/2451 $\endgroup$ – Qmechanic Jun 1 '16 at 18:35
  • $\begingroup$ But there is m because we use perturbation theory to 'add' those higher order effects, right? I wanna ask given a potential $V(r,\phi,\theta)$ in general, what does the energy depend by just solving the Schrodinger Equation (and ignore other higher effects that may appear in reality)? $\endgroup$ – Physicist Jun 1 '16 at 18:53
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The wave function of the hydrogen atom $\psi(r,\theta,\phi)$ is a product of the radial part, the angular part and azimuthal term $$ \psi_{n,\ell,m}(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi). $$ The radial part $R(r)$ obeys Laguerre polynomials or, $$ R_{n\ell}(r) = Ae^{-\rho/2}\rho^\ell L^{2\ell+1}_{n-\ell-1},~ \rho = \frac{2r}{na_0}, $$ and $A$ is a complicated constant, The angular part $\Theta(\theta)$ is given by spherical harmonics $Y^\ell_m(\theta,\phi)$ and the azimuthal is $\Phi(\phi) = e^{im\phi}$, where $m$ is the projection of the angular momentum on the $z$ axis.

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The energy levels of the bound-states of a hydrogen atom only depend on the radial quantum number n .

This is a special property of a (1/r) type of interaction potential .

For a general central potential, V (r ) the quantized energy levels of a bound-state can depend on both n and l values.

The property that the energy levels of a hydrogen atom only depend on n , and not on l and m, points to the feature that the energy spectrum of a hydrogen atom is highly degenerate.

By degeneracy of a state we mean that there are many different states of the system which possess the same energy eigen value.

For a given value of l , there are (2l+1) different allowed values of m (i.e.,-l , -l+1 , ...0 1 ,2, (l-1) ,+l ).

Likewise, for a given value of n, there are n different allowed values of l (i.e. 0 , 1, 2, ,(n-1)).

Now, all states possessing the same value of n have the same energy (i.e., they are degenerate )

The total number of degenerate states for a given value of n comes out to be n^2.

The above degeneracy gets lifted if the interaction potential has such terms which can give rise to splitting of the degenerate states.

A common example is placing the atom in a magnetic field which splits the different m -states. or spin-orbit interaction term in the potential .

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