Quantum Mechanics for Mathematicians

| 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