Editing Hamiltonian (quantum mechanics) (section)

==Energy eigenket degeneracy, symmetry, and conservation laws==

In many systems, two or more energy eigenstates have the same energy. A simple example of this is a free particle, whose energy eigenstates have wavefunctions that are propagating plane waves. The energy of each of these plane waves is inversely proportional to the square of its [[wavelength]].  A wave propagating in the <math>x</math> direction is a different state from one propagating in the <math>y</math> direction, but if they have the same wavelength, then their energies will be the same. When this happens, the states are said to be ''degenerate''.

It turns out that [[Degenerate energy levels|degeneracy]] occurs whenever a nontrivial [[Unitary matrix|unitary operator]] <math>U</math> [[commutation relation|commutes]] with the Hamiltonian. To see this, suppose that <math>|a\rang</math> is an energy eigenket. Then <math>U|a\rang</math> is an energy eigenket with the same eigenvalue, since

<math display="block">UH |a\rangle = U E_a|a\rangle = E_a (U|a\rangle) = H \; (U|a\rangle). </math>

Since <math>U</math> is nontrivial, at least one pair of <math>|a\rang</math> and <math>U|a\rang</math> must represent distinct states. Therefore, <math>H</math> has at least one pair of degenerate energy eigenkets. In the case of the free particle, the unitary operator which produces the symmetry is the [[Rotation operator (quantum mechanics)|rotation operator]], which rotates the wavefunctions by some angle while otherwise preserving their shape.

The existence of a symmetry operator implies the existence of a [[Conservation law (physics)|conserved]] observable. Let <math>G</math> be the Hermitian generator of <math>U</math>:

<math display="block"> U = I - i \varepsilon G + O(\varepsilon^2) </math>

It is straightforward to show that if <math>U</math> commutes with <math>H</math>, then so does <math>G</math>:

<math display="block"> [H, G] = 0 </math>

Therefore,

<math display="block">
\frac{\partial}{\partial t} \langle\psi(t)|G|\psi(t)\rangle
= \frac{1}{i\hbar} \langle\psi(t)|[G,H]|\psi(t)\rangle
= 0.
</math>

In obtaining this result, we have used the Schrödinger equation, as well as its [[bra–ket notation|dual]],

<math display="block"> \langle\psi (t)|H = - i \hbar {d\over\ d t} \langle\psi(t)|.</math>

Thus, the [[expected value]] of the observable <math>G</math> is conserved for any state of the system. In the case of the free particle, the conserved quantity is the [[angular momentum]].