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  Degrees of freedom in m(atrix) theory

+ 4 like - 0 dislike

The Hamiltonian for m(atrix) theory is given by $$H=\frac{1}{2\lambda}\text{Tr}\left(P^{a}P_{a}+\frac{1}{2}\left[X^{a},X^{b}\right]^{2}+\theta^{T}\gamma_{a}\left[X^{a},\theta\right]\right).$$ Where $X^a$ are the $9$ bosonic $N\times N$ matrices and $\theta$ are a $16$-component matrix-valued spinor of $SO(9)$. (Given by Taylor on page 8 in "M(atrix) Theory: Matrix Quantum Mechanics as a Fundamental Theory" at http://arxiv.org/abs/hep-th/0101126). This can be derived from an action looking like $$S=\frac{1}{2\lambda}\int\text{d}t\,\text{Tr}\left(\partial_{t}X^{a}\partial_{t}X_{a}-\frac{1}{2}\left[X^{a},X^{b}\right]^{2}+i\theta^{T}\partial_{t}\theta-\theta^{T}\gamma_{a}\left[X^{a},\theta\right]\right)$$.

Since this is a super-symmetric theory the bosonic and fermionic degrees of freedom must match. The equations of motion for the fermions halves the it's degrees if freedom and we are left with only 8. But what about the bosonic degrees, we should have 8 not 9. There is however also a Gauss-constraint given by $\left\{ \partial_{0}X^{I},X_{I}\right\} +\left\{ \theta^{T},\theta\right\} =0$ (which we must impose if we are to have a gauge invariant theory, here the usual gauge field $A$ has been put to zero $A=0$), but this involves both the fermions and bosons and can not only restrict a degree of freedom for the bosonic matrices?
How do I get the bosonic and fermionic degrees of freedom to match?

This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Natanael
asked Nov 10, 2013 in Theoretical Physics by Natanael (75 points) [ no revision ]
See this blog article. You probably neglect an overall $U(N)$ symmetry

This post imported from StackExchange Physics at 2014-03-07 16:47 (UCT), posted by SE-user Trimok

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