Quantum Mechanics for Mathematicians

(lily) #1

solutions to
Q 1 | 0 〉= 0 or Q 2 | 0 〉= 0


The simplification here is much like what happens with the usual bosonic har-
monic oscillator, where the lowest energy state in various representations can
be found by looking for solutions toa| 0 〉= 0.
There is an example of a physical quantum mechanical system that has ex-
actly the behavior of this supersymmetric oscillator. A charged particle confined
to a plane, coupled to a magnetic field perpendicular to the plane, can be de-
scribed by a Hamiltonian that can be put in the bosonic oscillator form (to show
this, we need to know how to couple quantum systems to electromagnetic fields,
which we will come to in chapter 45). The equally spaced energy levels are
known as “Landau levels”. If the particle has spin , there will be an additional
term in the Hamiltonian coupling the spin and the magnetic field, exactly the
one we have seen in our study of the two-state system. This additional term is
precisely the Hamiltonian of a fermionic oscillator. For the case of gyromagnetic
ratiog= 2, the coefficients match up so that we have exactly the supersym-
metric oscillator described above, with exactly the pattern of energy levels seen
there.


33.2 Supersymmetric quantum mechanics with a superpotential


The supersymmetric oscillator system can be generalized to a much wider class
of potentials, while still preserving the supersymmetry of the system. For sim-
plicity, we will here choose constants~ =ω = 1. Recall that our bosonic
annihilation and creation operators were defined by


aB=

1


2

(Q+iP), a†B=

1


2

(Q−iP)

Introducing an arbitrary functionW(q) (called the “superpotential”) with deriva-
tiveW′(q) we can define new annihilation and creation operators:


aB=

1


2

(W′(Q) +iP), a†B=

1


2

(W′(Q)−iP)

HereW′(Q) is the multiplication operatorW′(q) in the Schr ̈odinger position
space representation on functions ofq. The harmonic oscillator is the special
case


W(q) =

q^2
2
We keep our definition of the operators

Q+=aBa†F, Q−=a†BaF

These satisfy
Q^2 +=Q^2 −= 0

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