Bound States of the Hydrogen Atom

We have seen that for a free particle, the (linear) momentum p=mv can be used to label the eigenstates since the linear momentum is constant (no forces act on a free particle). As I mentioned in class, whenever there is a symmetry in nature, there is an associated energy degeneracy of some of the eigenstates. Moreover, whenever there is a symmetry in nature, there is usually a dynamical constant of the motion that can be associated with the symmetry. For the free particle, there is translational symmetry in that you can displace the particle in any direction and the potential energy does not change (since there is no potential energy for a free particle). It can be shown that translational symmetry implies that linear momentum is conserved and is also responsible for the infinite degeneracy of the eigenstates of the free particle (recall that for a given energy there are an infinite number of momenta possible, differeing in their direction).

The electron in the hydrogen atom.has a force acting on it owing to the electrostatic interaction with the proton. As a consequence, its velcoity and linear momentum are not constant. We must find constants of the motion to label the eigenstates. One reliable constant of the motion is the energy, but is this sufficient? The potential energy -kee2/r is spherically symmetric since it depends only on the magnitude of r and not its direction. Thus we expect a constant of the motion to be associated with this spherical symmetry and some energy degeneracy to also result. This implies that we will need additional labels other than the energy to completely label the eigenstates of the hydrogen atom and these labels will be provided by the constant of the motion that results from the spherical symmetry.

As is described in the course pack notes on Angular Momentum, Spin, and the Pauli Principle, the constant of the motion associated with the spherical symmetry is angular momentum denoted by the symbol L. Angular momentum is a vector quantity obtained by multiplying two other vector quantities r and p in a special way called a cross or vector product. Formally one writes L=r×p, where the × means that the magnitude of L is equal to rpsin(q) where q is the angle between r and p, and the direction of L is perpendicular to the plane formed by r and p. In quantum mechanics it is not possible to specify all the components of L since this would determine both r and p.which would violate the Uncertainty Principle. It turns out that we can specify the magnitude of the angular momentum and one of the components of angular momentum, commonly taken to be the z component, Lz. Thus each eigenstate of the hydrogen atom can be labeled uniquely in terms of its energy, magnitude of angular momentum, and z-component of angular momentum.

We are concerned here with bound states of the electron only, those for which the electron is bound to the proton (states with total energy less than or equal to zero). Whenever the motion is bounded, as for the particle in the box, only specific waves can be fit into the potential. In other words, the energy levels are always quantized for bound motion. By solving the Schroedinger equation and seeing how to fit in the waves, one finds that the allowed energies are equal to En= - 13.6 eV/n2, with n=1,2,3,…These are thae same energies as in the Bohr theory! The Bohr theory got the energies right.

(It is also possible to use the Uncertainty Principle to estimate the ground state of hydrogen. The energy of the electron is E=mv2/2-kee2/r=p2/2m-kee2/r. Classically the energy approaches minus infinity as the electron gets very close to the proton. Quantum mechanically, however, if we know that the electron is precisely at r=0, then, by the Uncertainty Principle, we know nothing about its momentum. To estimate the ground state energy, let us set r=Ñ /2p in the equation for the energy, giving E= p2/2m-2pkee2/Ñ = p2/2m-2pkeac, where a=kee2/Ñ c. The smallest value the energy can take occurs when p= 2meac, giving a total energy of E= - 2a2mc2 which is 4 times the actual ground state energy, but still gives a reasonable estimate of its order of magnitude.)

Now we must talk about the angular momentum. For each energy, both classically and quantum-mechanically, there is a maximum value for the magnitude of the angular momentum. In quantum mechanics, the magnitude of the angular momentum is quantized. For a given energy En= - 13.6 eV/n2 the magnitude of the angular momentum can be equal to where . Thus, for the ground state (n=1) of hydrogen, one must have { =0, for the first excited state, { can equal 0 or 1, for the n=2 state, { can equal 0, 1or 2, etc. Thus only possible values for the magnitude of the angular momentum are , etc.

As is described in the course pack notes on angular momentum, for a given magnitude of angular momentum L, classically it is possible for the z-component of angular momentu to vary from –L to +L. Quanutum-mechanically, this is not the case. The z-component of angular monetum is quantized. For a given value of { , the z-component of angular momentum can take on the values , where m{ takes on integral values from -{ to { . For example when { =1 (L=), m{ =-1,0,1 (Lz =-Ñ ,0Ñ ,Ñ ); for { =2 (),m{ =-2,-1,,0,1,2 (Lz =-2Ñ ,-Ñ ,0Ñ ,Ñ ,2Ñ ); etc.

The quantities n, { , and m{ are referred to as quantum numbers. Each bound eigenstate of the hydrogen atom is labeled by one set of these numbers, n determines the energy via En= - 13.6 eV/n2, { determines the magnitude of the angular momentum through , and m{ determines the z-component of angular momentum through . For a given n, there are n2 possible states with different and m{ =--{ ,-{ +1,…{ -1,{ . This energy degeneracy, in part (actually the m{ part) is related to the spherical symmetry of the problem.

I will sketch the probability distribution for finding the electron at a given radius for a few of the eigenstaes. Recall that in the Bohr model, the electron would be at the Bohr radius if it were in the n=1 state and four times the Bohr radius if it were in the n=2 state. Of course you know that must be wrong owing to the Uncertainty Principle – we cannot know precisely the radius withou losing all information on the momentum. What we find is that it is most likely to find the electron at the first Bohr radius if it is the n=1 state and it is most likely to find its speed equal to ac. Similarly it is most likely to find the electron at 4 times the Bohr radius if it is in the n=2, { =1 state. If it is in the n=2, { =0 state, there are two maxima in the probability distribution, indicating that the orbit is more elliptical than circular.