19.2 PHYSICAL EXAMPLES OF OPERATORS
In a similar way it can be shown thatA†parallels the operatorUof our angular
momentum discussion and creates an additional quantum of energy each time it
is applied:
(A†)m|En〉=dndn+1···dn+m− 1 |En+mω〉. (19.45)
It is therefore known as acreation operator.
As noted earlier, the expectation value of the oscillator’s energy operator
〈ψ|H|ψ〉must be non-negative, and therefore it must have a lowest value. Let
this beE 0 , with corresponding eigenstate| 0 〉. Since the energy-lowering property
ofAapplies to any eigenstate ofH, in order to avoid a contradiction we must
have thatA| 0 〉=|∅〉. It then follows from (19.40) that
H| 0 〉=^12 ω(A†A+AA†)| 0 〉
=^12 ωA†A| 0 〉+^12 ω(A†A+)| 0 〉, using (19.41),
=0+0+^12 ω| 0 〉. (19.46)
This shows that the commutator structure of the operators and the form of the
Hamiltonian imply that the lowest energy (its ground-state energy) is^12 ω;this
is a result that has been derived without explicit reference to the corresponding
wavefunction. This non-zero lowest value for the energy, known as the zero-point
energy of the oscillator, and the discrete values for the allowed energy states
are quantum-mechanical in origin; classically such an oscillator could have any
non-negative energy, including zero.
Working back from this result, we see that the energy levels of the s.h.o. are
1
2 ω,
3
2 ω,
5
2 ω,... ,(m+
1
2 )ω,..., and that the corresponding (unnormalised)
ket vectors can be written as
| 0 〉,A†| 0 〉, (A†)^2 | 0 〉, ... , (A†)m| 0 〉, ....
This notation, and elaborations of it, are often used in the quantum treatment of
classical fields such as the electromagnetic field. Thus, as the reader should verify,
A(A†)^3 A^2 A†A(A†)^4 | 0 〉is a state with energy^92 ω, whilstA(A†)^3 A^5 A†A(A†)^4 | 0 〉is
not a physical state at all.
The normalisation of the eigenstates
In order to make quantitative calculations using the previous results we need to
establish the values of thecnanddnthat appear in equations (19.44) and (19.45).
To do this, we first establish the operator recurrence relation
Am(A†)m=Am−^1 (A†)mA+mAm−^1 (A†)m−^1. (19.47)