Quick notes on Supersymmetric Quantum Mechanics I

Bosonic Harmonic Oscillator:

$H = a a^{\dagger} + \frac{1}{2}$

Where  

$a = \frac{1}{\sqrt{2}}(p - i x)$

$a^{\dagger} = \frac{1}{\sqrt{2}}(p - i x)$

with commutation relations :

$[a^{\dagger}, a] =1$

$[a^{\dagger}, a^{\dagger}] =0$

$[a, a] =0$

Fermionic Harmonic Oscillator:

$H = b b^{\dagger} - \frac{1}{2}$

with anticommutation relations

$\{b^{\dagger}, b^{\dagger}\} =0$

$\{b, b\}=0$

Consider:

 $H = aa^{\dagger} + bb^{\dagger}$

with  Fock space:

  $| \#bosons, \#fermions \rangle$

One seeks for example:

$Q|1, 0 \rangle = |1, 1 \rangle$

$Q = a b^{\dagger}$

$Q^{\dagger} = b a^{\dagger}$

$Q$ is called a supercharge.

Bosonic state: Even number of fermions

Fermionic state: Odd number of fermions

Consider $N = 2 $ supersymmetry

$Q_1|boson 1 \rangle = |Fermion 1 \rangle$

$Q_2|boson 1 \rangle = |Fermion 2 \rangle$

$Q_1^{\dagger}|Fermion 1\rangle = |boson 1 \rangle$

$Q_2^{\dagger}|Fermion 1 \rangle = |boson 2\rangle$

Supersymmetry can be broken

     - Spontaneously broken

            - $[H, Q] = 0$   So the Hamiltonian preserves supersymmetry but the ground state does not.

             -$Q| 0 \rangle \neq 0$

             -$Q^\dagger | 0 \rangle  \neq 0$

    - The "Witten Index" tells you if supersymmetry is spontaneously broken

           $W(\beta) =  Tr (-1)^F  e^{-\beta H}$

                          $=\sum n_b (E_i) e^{- \beta E_i} - \sum n_f (E_i) e^{- \beta E_i}$

   

                            

                          Where $n_b$ is the number of bosons;

                                      $n_f$ is the number of fermions.

        

                         If $W(\beta) \neq  0$ supersymmetry is not spontaneously broken

                         If $W(\beta) =  0$ Then supersymmetry is broken

Spontaneously broken supersymmetry  leads to goldstino (fermions)

Spontaneously broken unitary symmetry leads to  goldstone modes (bosons)

If the ground state has zero energy