Orodruin said:
What is the norm of ##(a^\dagger)^2|0\rangle##?
Ah! I see!
##|(a^\dagger)^2|0\rangle | = \langle 0 | a (a a^\dagger) a^\dagger | 0 \rangle = \langle 0 | a (1 - a^\dagger a) a^\dagger | 0 \rangle = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger (a a^\dagger) | 0 \rangle) = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger (1 - a^\dagger a) | 0 \rangle) = \langle 0 | a a^\dagger | 0 \rangle - \langle 0 | a a^\dagger | 0 \rangle + \langle 0 | a a^\dagger a^\dagger a | 0 \rangle) = 0##.
Hmm ... I get the idea.
A similar calculation for the norm of ##(a^\dagger)^3|0\rangle## shows that it is 0.
Although a formal proof will use proof by induction, my intuition tells me all the higher order norms are zero as well.
Therefore, the state cannot have two or more quanta.
Let me try and write up the solution now.
(a) The generic normalised excited state is ##a_{i}^{\dagger}a_{j}^{\dagger}\dots a_{r}^{\dagger}a_{s}^{\dagger}\lvert 0 \rangle,## where only the species from ##i## to ##s## are in the excited state. This is under the assumption that the ground state ##\lvert 0 \rangle## is already normalised, i.e. ##\langle 0 \lvert 0 \rangle = 1.##
(b) Take the ##i##-th species. The argument can be retraced for all the other species because the creation or annihilation operator for one species commutes with the creation or annihilation operator for another species.
Consider the state with the ##n-##th excitation of the ##i##-th species, and ignore the excitations of all the other species (because of the commutation rule in the last sentence).
Write ##(a_{i}^{\dagger})^{n}(a_{j}^{\dagger})^{m}\dots (a_{r}^{\dagger})^{k}(a_{s}^{\dagger})^{l}\lvert 0 \rangle## as ##(a^{\dagger})^{n}\lvert 0 \rangle## to simplify the notation in the following calculation.
Now, ##|(a^\dagger)^n|0\rangle| = \langle 0 | (a)^{n-1} (a a^\dagger) (a^\dagger)^{n-1} | 0 \rangle = \langle 0 | (a)^{n-1} (1 - a^\dagger a) (a^\dagger)^{n-1} | 0 \rangle = \langle 0 | (a)^{n-1} (a^\dagger)^{n-1} | 0 \rangle - \langle 0 | (a)^{n-1} a^\dagger a (a^\dagger)^{n-1}) | 0 \rangle).##
Therefore, the anticommutation relation allows us to switch the ##a## operator with the ##a^{\dagger}## operator. In this way, the ##a## operator shifts all the way to the right and annihilates the ##\lvert 0 \rangle## state. In the process, we obtain ##n## terms of the form ##\langle 0 | (a)^{n-1} (a^\dagger)^{n-1} | 0 \rangle## in an alternating ##+\ -\ +\ -\ \dots ## series.
Now, when ##n=1##, the norm is ##\langle 0 | 0 \rangle## = 1.
When ##n=2##, the norm is 0 by cancellation of the two identical terms with each other.
When ##n=3##, the norm is 0 because the first two terms are identical and cancel with each other and the third term is zero because the norm for ##n=2## is zero.
By induction, the norm for ##n>1## is zero.
N.B. The inductive step is necessary because the proof for odd values of ##n## greater than 1 (e.g. ##n=3##) requires the use of the norm for the corresponding smaller even value of ##n## (e.g. ##n=2##). For even values of ##n##, cancellation alone shows that the norm if zero, but for odd values of ##n## greater than 1, cancellation of the identical terms leaves remianing one positive term which has to be zero because it is the norm for the smaller even value of ##n##.Are my solutions correct?