| 0 〉F. The Hamiltonian is
H=
1
2
~ω
∑d
j=1
(aF†jaFj−aFjaF†j) =
∑d
j=1
(
NFj−
1
2
)
~ω
whereNFjis the number operator for thej’th degree of freedom, with
eigenvaluesnFj= 0,1.
Putting these two systems together we get a new quantum system with state
space
H=Fd⊗Fd+
and Hamiltonian
H=
∑d
j=1
(NBj+NFj)~ω
Notice that the lowest energy state| 0 〉for the combined system has energy 0,
due to cancellation between the bosonic and fermionic degrees of freedom.
For now, taking for simplicity the cased= 1 of one degree of freedom, the
Hamiltonian is
H= (NB+NF)~ω
with eigenvectors|nB,nF〉satisfying
H|nB,nF〉= (nB+nF)~ω
While there is a unique lowest energy state| 0 , 0 〉of zero energy, all non-zero
energy states come in pairs, with two states
|n, 0 〉 and |n− 1 , 1 〉
both having energyn~ω.
This kind of degeneracy of energy eigenvalues usually indicates the existence
of some new symmetry operators commuting with the Hamiltonian operator.
We are looking for operators that will take|n, 0 〉to|n− 1 , 1 〉and vice-versa,
and the obvious choice is the two operators
Q+=aBa†F, Q−=a†BaF
which are not self adjoint, but are each other’s adjoints ((Q−)†=Q+).
The pattern of energy eigenstates looks like this