Physics Friday 60

Let us consider a hydrogen atom: a single electron “orbiting” a single (much heavier) nucleus containing a single proton. The potential energy due to the attractive Coulomb force between these charged particles is , where r is the distance between the particles. The time-independent Schrödinger equation for the electron (ignoring relativistic effects, particle spins, and magnetic moments) is

(Here, the electron mass m should actually be the reduced mass μ of the electron-proton pair; however, the correction involved is small, and even smaller for more massive atomic nuclei). With the way our potential energy is defined (with zero energy at infinite separation), our bound states for the electron (our states of interest for an atom) will have E<0.

Now, we note that the potential is spherically symmetric, and so our Hamiltonian commutes with the angular momentum operators, and we can perform separation of variables in spherical coordinates, with the angular components being the spherical harmonics (see here and here).
As in here, when we perform the spherical coordinate separation , we obtain radial equation

Making the substitution ,

Now, let us define , which has units of length, and dimensionless variable , so that . Then we have

now, let us make the transform (ρ is dimensionless); then we get

We see that as ρ→∞, the equation is approximated by
, which has general solution ; normalizability requires C2=0. Also, examining ρ→0, our equation is approximately , which has general solution ; considering the origin tells us C2=0 again. Combining these, we thus try the substitution , giving radial equation

Now, letting , we get

Which is the associated Laguerre differential equation

with and ; so the solution to our transformed equation is
, where is an associated Laguerre function. Now, the resulting wavefunction can be normalized only if is a polynomial; this is true when is a non-negative integer; thus, we have , where n is a positive integer, and 0≤ln-1. Looking back through our conversions,
we have

where , is the constant needed to normalize the wavefunction; with some work involving the properties of associated Laguerre polynomials (see for example here), we can find the normalized wavefunction to be:

Where n=1,2,3,…, l=0,1,2,…,n-1, and m=-l, –l+1,…,l-1,l (for a given n, we have n2 different angular momentum states).
One should note that is the Bohr radius.

Now, recall that we had ; with the normalization condition , we have
, and the ground state energy of hydrogen is approximately -13.6 eV, and so the ionization energy of hydrogen is approximately 13.6 eV.


Tags: , , , , , , , , ,

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: