Quick notes on Supersymmetric Quantum Mechanics II
Getting to sigma models
Consider $\phi(x^i,t)$
Let $\phi(t)$ be $0 + 1D$ QFT
Let $\mathcal{L} = \frac{1}{2} \frac{d \phi(t)}{dt} \frac{d \phi(t)}{dt}$
$\phi: \mathcal{R} \rightarrow \mathcal{R}$
$\phi$: base space $\rightarrow $ target space
Consider now $\mathcal{L} = \frac{1}{2} g_{ij} \frac{d \phi^i}{dt} \frac{d \phi^j}{dt}$
$\phi: \mathcal{R} \rightarrow \mathcal{M}$
$\mathcal{L} = \frac{1}{2} g_{ij} \frac{d \phi^i}{dt} \frac{d \phi^j}{dt}$
Setting $g_{ij}$ to $\delta_{ij}$
$\mathcal{L} = \frac{1}{2} \dot{\phi_1}^2 + . . . . . \frac{1}{2} \dot{\phi_n}^2$
$\phi$ is a scalar
Let Dirac spinors be $\psi^i$
$\mathcal{L_f} = g_{ij}\frac{1}{2} {\psi^i}_\alpha (\gamma^{\mu} \partial_\mu)_{\alpha \beta} {\psi^j} _{\beta}$
where $\alpha$ is the spinor index
$\mathcal{L_f} = \frac{i}{2} \delta_{ij} \psi^i_\alpha \partial_t \psi^j_\alpha$
$\mathcal{L_f} = \frac{i}{2} \psi^i_\alpha \dot{\psi^j_\alpha}$
$\mathcal{L_{susy}} = \mathcal{L_{b}}+ \mathcal{L_{f}}$
$\mathcal{L_{susy}} = \frac{1}{2} \dot{\phi_1}^2 + . . . . . \frac{1}{2} \dot{\phi_n}^2+ \frac{i}{2} \psi^i_\alpha \dot{\psi^j_\alpha}$
$\delta \phi^i = i \epsilon \psi$ and $\delta \psi^i = - \epsilon \dot{\phi}^i$
1 epsilon so 1 supercharge. Finding Noether charge.
$\epsilon Q = \epsilon \sqrt{2} \psi^i \dot{\phi}_i$
Comments
Post a Comment