Quantum Theory - Particle Physics Group

CHAPTER 3. FORMALISM AND INTERPRETATION 68(where we used the identity XY = [X,Y ] + Y X). Thus a † and a are called creation andannihilation operator, respectively. Their collective name is “ladder operators” becausethey bring us up and down the ladder of energy levels. More precisely, since (3.187) impliesthat a|n〉 and a † |n〉 have occupation numbers n±1, these states must be proportional to |n±1〉a † |n〉 = c n+ |n + 1〉, a|n〉 = c n− |n − 1〉. (3.188)Assuming that all states are normalized 〈n|n〉 = 1 we can now compute the normalizationfactors c n± . Since norms are computed by multiplication with the Hermitian conjugate states,a |n〉 = c n− |n − 1〉a † |n〉 = c n+ |n + 1〉conj.−→ 〈n|a † = c ∗ n− 〈n − 1|, (3.189)conj.−→ 〈n|a = c ∗ n+ 〈n + 1|, (3.190)the eigenvalue equation a † a|n〉 = N |n〉 = n|n〉 implies〈n + 1|n + 1〉 = 1|c n+ | 〈n|a 2 a† |n〉 = 1|c n+ | 2( 〈n|a† a|n〉 + 1) = n + 1 = 1,|c n+ |2(3.191)〈n − 1|n − 1〉 = 1|c n− | 2 〈n|a† a|n〉 = n = 1,|c n− |2(3.192)so thatc n+ = √ n + 1, c n− = √ n, (3.193)where the phase ambiguity of the eigenvectors |n〉 has been used to choose c n± positive real.Quantization of occupation number and energy. Now we are ready to determine theeigenvalues n. We assume that at least one eigenstate |n〉 exists for some eigenvalue n ∈ R,which has to be non-negative n ≥ 0 because of the positivity (3.183) of N. Now we act on thisstate k times with the annihilation operator a and obtaina|n〉 = √ n |n − 1〉, (3.194)a 2 |n〉 = √ n(n − 1) |n − 2〉, (3.195)...a k |n〉 = √ n(n − 1)...(n − k + 1) |n − k〉. (3.196)We thus find new energy eigenstates with occupation numbers n − 1, n − 2, ... However, thisprocedure has to terminate because otherwise we would be able to construct energy eigenstatesfor arbitrary n − k, which turns negative for k > n in contradiction to the positivity of theoperator N. Hence there must exist a positive integer K for which a K |n〉 = 0. Choosing Kminimal, so that a K−1 |n〉 ≠ 0, we conclude that a|n −K +1〉 = 0 and hence 〈n −K +1|a † a|n −K +1〉 = n −K +1 = 0. In other words, if a|n ′ 〉 = 0 the normalization factor c n ′ − must vanish,which is the only possibility to avoid the existence of an energy eigenstate with eigenvalue