The Hydrogen Atom Reading: Shankar Chpt. 13, Griffiths 4.2 The Coulomb potential for a hydrogen atom is Setting, and defining u(r) = r R(r), the radial.

The Hydrogen AtomReading: Shankar Chpt. 13, Griffiths 4.2

The Coulomb potential for a hydrogen atom is

Setting , and defining u(r) = r R(r), the radial

Schroedinger equation is just

(Here, the mass m is not really the electron mass, but rather the reduced mass μ. However, because the proton mass is so large compared to the electron mass, the reduced mass is practically the same as the electron mass: )

We will be interested in the bound states of hydrogen, for which E<0.

We can make the preceding equation look prettier by defining

in which case we have

Let’s look for guidance on how to solve this equation by first examining its asymptotic behavior in the limits and .

These asymptotic results suggest that we try to write the solution as

Substituting yields the following equation for

We try a power series expansion:

Substituting, and matching like powers of yields

This looks problematic! Why? Because for large values of j, this becomes

But observe that this ratio of successive coefficients is essentially the same as that found if you Taylor expand the exponential function

So this suggests that our power series for is roughly going to behave like an exponential function:

This would be bad because

which would mean that

Is there any way around this? Yes! If the power series for terminates, i.e., if there exists a maximum integer jmax such that all coefficients

This will occur if (see recursion relation for the cj’s):

Defining the integer (n is called the principal quantum number), the condition for the series to truncate is

But recall that is related to the energy

so the condition that the power series truncates is really a condition on the allowed values of the energy! -- i.e., only those special values of energy for which will yield a solution that won’t blow up.

The allowed energies are thus given by

These are the energies of the bound states of the hydrogen atom! Not too shabby! Numerically, we find the ground state energy to be E1=-13.6eV.

Moreover, we have also learned a great deal about the radial part of the hydrogen wavefunction:

where is a polynomial of degree In fact, since we have

the recursion relation relating the coefficients cj of the polynomial, we can explicitly determine what these coefficients are and hence construct these polynomials! These polynomials are called the associated Laguerre polynomials, and denoted by

Here are the first few radial wavefunctions for hydrogen:

where for convenience we have defined the “Bohr radius”

(As will be discussed later, the Bohr radius can be regarded as the approximate “radius” of the hydrogen atom in its ground state.)

Visualizing the radial wavefunctions:

The best way is to simply look at the pictures of the Rnl’s in standard quantum textbooks (e.g., Griffiths), or via various applets available on the web. (I’ll show you one of these on the next page.)

Note: in addition to plotting just Rnl, it is also informative to plot the radial probability density, given by r2 (Rnl)2.

So we have thus found the energy eigenfunctions for the hydrogen atom:

http://www.physics.gatech.edu/gcuo/lectures/ModernPhysicsLectures/MP16HydrogenAtom.ppt#308,31,Probability Distribution Functions

Degeneracy

The energy eigenstates are degenerate. Let’s see just how degenerate they are:

Recall that But since , it follows that

So if we label the eigenstates

So for a given energy level (specified by n), there are n-1 possible values of , and for each there are possible values of m. So the total number of states for a given n is just

i.e., there are n2 states with energy En.

The origin of this degeneracy: The fact that the energy of the hydrogen wavefunction |n,l,m> doesn’t depend on the m values has its roots in the rotation symmetry of the problem: The Hamiltonian for the hydrogen atom commutes with (all components of) angular momentum, since the coulomb potential only depends on radial distance. Consequently, the raising and lowering operators L+ and L- must also commute with H. This means we can raise/lower the value of m without changing the energy. In general then, anytime we have rotational symmetry in a physical system, we expect its energy to be independent of m.

But why doesn’t the hydrogen energy depend on the l values? It’s not because of rotation symmetry! Indeed, for most other problems with rotation symmetry, the energy does depend on the l-values. So why not for the case of hydrogen? Because there is an additional symmetry present in the hydrogen

problem associated with the existence of the so-called Runge-Lenz vector. This extra symmetry is responsible for the energy being independent of the l value. (In such cases, we say that there is an “accidental degeneracy” in the problem.) It’s really not an accident though – it’s associated with the presence of an extra symmetry!

Energy levels and spectroscopic notation

In atomic physics it is customary to use the following nomenclature:

So, for example, a state with n=1, =0 is called the 1s state, while the state with n=2, =1 is called the 2p state.

Here’s an energy-level diagram for hydrogen:

continuum

Energy (in eV’s)

-13.6

-3.4

-1.51

0

1s

3s

2p

3p

2s

3d

Note that all p states are three-fold degenerate (since m=-1,0,1), all d states are five-fold degenerate (since m=-2,-1,0,1,2), etc. Note too that, for a given energy level n, the total degeneracy is n2.

Transitions between states

We mention here that when an electron is in a given state, it can drop down to a lower energy state by emitting a photon (whose energy hν equals the energy drop of the electron, En' – En). This is why hydrogen

(and all other atoms for that matter) have their own characteristic emission and absorption spectra.

http://www.physics.gatech.edu/gcuo/lectures/ModernPhysicsLectures/MP16HydrogenAtom.ppt#304,27,7.6: Energy Levels and Electron Probabilities

At first blush, it may seem hard to understand how any such transitions are even possible, since, after all, according to the time-independent Schroedinger equation, if a particle is in some energy eigenstate, it is supposed to remain in that state for all time!

The answer has to do with spontaneous emission and the fact that, even in a vacuum, there is always some electromagnetic energy present (“zero point” radiation) that can induce a transition. You’ll need time-independent perturbation theory to describe this phenomenon. Selection rules tell you which transitions are actually possible.

Selection rules

Δn = anything Δl = -1,+1 Δm = -1,0,+1

The “radius” of a hydrogen atom:How big is a hydrogen atom when it’s in its ground state? We can estimate this by calculating the expectation value of <r>. Here’s how:

So the characteristic radius of a hydrogen atom is roughly 10-10m.