D.33 Quantum field derivations

This derivation will find the properties of a system described by a Hamiltonian of the form:

First note that the commutator (2) directly implies the uncertainty relationship, chapter 4.5.2 (4.46):

Also note that the evolution equations for the expectation values of and follow directly from chapter 7.2 (7.4). The commutator appearing in it is readily worked out using the commutator (2) and the rules of chapter 4.5.4. Since energy eigenstates are stationary, according to the evolution equations in such states the expectation values of and will have to be zero.

The equality of the and terms in the Hamiltonian is a simple matter of symmetry. Nothing changes if you swap and , adding a minus sign for one. Then unavoidably the two terms in the Hamiltonian must be equal; it is shown below that the eigenfunctions are unique.

The commutator (2) also implies that , , and all their combinations, do not commute with the Hamiltonian. So they are not conserved quantities of the system. However, there are two combinations,

In other words, and are commutator eigenoperators of the Hamiltonian. The above relations are readily checked using the given commutator (2) and the rules of chapter 4.5.4.

To see why that is important, multiply both sides of the eigenvalue
problems above with a system energy eigenfunction of energy :

After writing out the definitions of the commutators, recognizing as , and rearranging, that gives

These results can be compared to the definition of an energy eigenfunction. Then it is seen that is an energy eigenfunction with one unit less energy than . And is an energy eigenfunction with one unit more energy than . So apparently and act as annihilation and creation operators of quanta of energy . They act as shown to the left in figure A.6.

There are however two important caveats for these statements. If or is zero, it is not an energy eigenfunction. Eigenfunctions must be nonzero. Also, even if the states or are not zero, they will not normally be normalized states.

To get a better understanding of these issues, it is helpful to first find the Hamiltonian in terms of and . There are two equivalent forms,

Now look at the first Hamiltonian first. If would be zero for some state , then that state would have energy . But that is not possible. If you look at the original Hamiltonian (1), the energy must at least be ; square Hermitian operators are nonnegative.

(To be more precise, if you square a Hermitian operator, you square
the eigenvalues, making them nonnegative. It is said that the square
operator is positive definite,

or, if there are zero
eigenvalues, positive semi-definite. And such an operator produces
nonnegative expectation values. And the expectation values of the
operators in the Hamiltonian do add up to the total energy; just take
an inner product of the Hamiltonian eigenvalue problem with the wave
function. See chapter 4.4 for more information on
expectation values.)

It follows that is never zero. This operator can be applied indefinitely to find states of higher and higher energy. So there is no maximum energy.

But there is a possibility that is zero. As the second
form of the Hamiltonian in (5) shows, that requires that the energy
of state equals

Now if you start from any energy state and apply sufficiently many times, you must eventually end up at this energy level. If not, you could go on lowering the energy forever. That would be inconsistent with the fact that the energy cannot be lower than . It follows that the above energy is the lowest energy that a state can have. So it is the ground state energy.

And any other energy must be a whole multiple of higher than the ground state energy. Otherwise you could not end up at the ground state energy by applying . Therefore, the energy eigenstates can be denoted more meaningfully by rather than . Here is the number of quanta that the energy is above the ground state level.

Now assume that the ground state is unique. In that case, there is one unique energy eigenfunction at each energy level. That is a consequence of the fact that if you go down a unit in energy with and then up a unit again with , (or vice versa), you must end up not just at the same energy, but at the same state. Otherwise the state would not be an eigenfunction of the Hamiltonian in one of the forms given in (5). Repeated application shows that if you go down any number of steps, and then up the same number of steps, you end up at the same state. Since every state must end up at the unique ground state, every state must be the result of applying to the ground state sufficiently many times. There is just one such state for each energy level.

If there are two independent ground states, applying on each gives two separate sets of energy eigenstates. And similar if there are still more ground states. Additional symbols will need to be added to the kets to keep the different families apart.

It was already mentioned that the states produced by the operators
and are usually not normalized. For example, the state
will have a square magnitude given by the inner product

Now if you take or to the other side of an inner product, it will change into the other one; the in the definitions (4) will change sign. So the square magnitude of becomes

From the second form of the Hamiltonian in (5), it is seen that gives the number of energy quanta . And since the state is normalized, the square magnitude of is therefore . That means that

where is some number of magnitude 1. Similarly

But note that you can always change the definition of an energy eigenfunction by a constant of magnitude 1. That allows you, while going up from using , to redefine each state so that is 1. And if is always one, then so is . Otherwise would not be .

In the ground state, the expectation values of and are zero, while the expectation values of and are equal to the minimum allowed by the uncertainty relation (3). The derivations of these statements are the same as those for the harmonic oscillator ground state in {D.13}.