-2
$\begingroup$

Starting with the general notion of supersymmetry:

$$Q| boson \rangle = | fermion \rangle \\ Q| fermion \rangle = | boson \rangle$$

I want to construct the superalgebra relations. After applying $Q$ twice we get:

$$ Q^2| boson \rangle = | boson \rangle; \\ Q^2| fermion \rangle = | fermion \rangle $$

Now, we can use the identical particle commutation and anti-commutation relations. And since $Q$ is both bosonic and fermionic this must hold:

$$\{Q_\alpha,Q_\beta\} = \left[Q_\alpha,Q_\beta\right] = 0 \\ \{Q_\alpha,Q_\beta ^\dagger \} = \left[Q_\alpha,Q_\beta ^\dagger \right] = \delta_{\alpha \beta}$$

Where we assume $Q$ are elements of Poincaré algebra namely:

$$ [Q_\mu, Q_\nu] = 0 \\ -i ~[M_{\mu\nu}, Q_\rho] = \eta_{\mu\rho} Q_\nu - \eta_{\nu\rho} Q_\mu \\ -i ~[M_{\mu\nu}, M_{\rho\sigma}] = \eta_{\mu\rho} M_{\nu\sigma} - \eta_{\mu\sigma} M_{\nu\rho} - \eta_{\nu\rho} M_{\mu\sigma} + \eta_{\nu\sigma} M_{\mu\rho} $$

The combination of being a Poincaré algebra and obeying the fermions and bosons commutations relations should make it a Super-Poincaré algebra, but I am not sure how to show it.

$\endgroup$
  • 4
    $\begingroup$ 1. Please stop making trivial edits to this question, such as adding a single comma (as in the latest revision). Ponder the changes you want to make for a while and make them substantial. 2. It is not clear what you're asking here: a) The notion of "supersymmetry" is more than "mapping bosons to fermions". Your starting point is wholly insufficient. b) What does the 2 in the subscript denote after you apply $Q$ twice? c) $Q$ is not an element of the Poincaré algebra, but of the super-Poincaré algebra. d) Due to what "spin algebra" and how do your relations after that follow? $\endgroup$ – ACuriousMind Nov 16 '17 at 2:42
  • 1
    $\begingroup$ As pointed out by the above comment, it is incorrect to assume that $Q$ belongs to the Poincare algebra-$Q$ does not have a Lorentz index. However, you can derive the commutator relations in the superalgebra by making the most general Ansatz for various commutators and using the graded Jacobi identity to determine the coefficients. $\endgroup$ – user110373 Nov 22 '17 at 21:03
  • $\begingroup$ @user110373 if you find the time to compile it into an answer that would be nice of you. $\endgroup$ – 0x90 Nov 22 '17 at 22:47
3
$\begingroup$

For me the most satisfying exposition is either Quantum Fields and Strings: A Course for Mathematicians or Wess and Bagger's Supersymmetry and Supergravity. The latter starts off with providing the motivation from the Coleman-Mandula theorem, leading to graded Lie algebras.

In particular, we start off with the general ansatz $\{Q,Q'\} = X, [X,X']= X''$ and $[Q,X] = Q''$. We have that $X$ can only be elements of the Poincaré algebra or elements of a Lorentz invariant compact Lie algebra $\mathscr A = \mathscr A_1 \oplus \mathscr A_2$ where $\mathscr A_1$ is semisimple and $\mathscr A_2$ abelian.

On the other hand, the former book is a masterful approach, starting with a great degree of generality and rigour, starting off with the properties of $\mathbb Z_2$ graded objects, such as vector spaces and manifolds.

$\endgroup$
  • $\begingroup$ Is my approach wrong? $\endgroup$ – 0x90 Aug 25 '18 at 22:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.