Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose you have a theory of maps $\phi: {\cal T} \to M$ with $M$ some Riemannian manifold, Lagrangian $$L~=~ \frac12 g_{ij}\dot\phi^i\dot\phi^j + \frac{i}{2}g_{ij}(\overline{\psi}^i D_t\psi^j-D_t\overline{\psi}^i\psi^j) -\frac{1}{4}R_{ijkl}\psi^i\psi^j\overline{\psi}^k\overline{\psi}^l,$$

where $g_{ij}=g_{ij}(\phi)$ is the metric, $R$ its riemann tensor, and covariant derivative

$$D_t\psi^i~=~ \partial_t \psi^i +\Gamma^i_{jk}\dot{\phi}^j\psi^k.$$

(Details taken from the book Mirror Symmetry, written by Vafa et al., paragraph 10.4.1.)

Taken for granted that the above Lagrangian is classically supersymmetric, with susy transformations given by

$$ \begin{aligned}\delta\phi^i &= \epsilon \overline\psi^i-\overline\epsilon \psi^i \\ \delta\psi^i &= i\epsilon (\dot\phi^i-\Gamma^i_{jk}\overline\psi^j \psi^k)\\ \delta\overline\psi^i &= -i\epsilon (\dot\phi^i-\Gamma^i_{jk}\overline\psi^j \psi^k). \end{aligned}$$

How can I quantise the classical supercharges

$$Q=i\overline\psi_i\dot\phi^i, \qquad \overline Q=-ig_{ij}\psi^i\dot\phi^j$$

in a way such that

$$ \delta F=[\epsilon Q+\overline\epsilon\overline{Q},F]_{\pm}$$

where $F$ is either a fermionic or bosonic field and $\pm$ is used appropriately?

The natural answer would be something like calculate conjugate momenta

$$ p_i=\frac{\partial L}{\partial\dot\phi^i}, \qquad \pi_{i\psi}=\frac{\partial L}{\partial\dot\psi^i},$$

and impose canonical commutation relations

$$ [\phi^i,p_j]=i\delta^i_j, \qquad \{\psi^i,\pi_{\psi,j}\}=\delta^i_j.$$

Since in doing this I face non-trivial ordering issues, which the book doesn't seem to be worried about, and moreover its quantized version of conjugate momentum to $\phi$ seems wrong to me, as well as its quantized $Q$ doesn't seem to reproduce the correct transformations for the fields, I ask if someone could clarify this.

Moreover, looking in the paper Constraints on supersymmetry breaking by Witten, in the neighbourhood of eqs. (90), (91), he seems to claim that the correct definition of conjugate momentum is derivative with respect to covariant derivative instead of time derivative, and this is another thing which leaves me with some doubts.

share|improve this question
    
Thanks for this nice question, it seems to deserve a research-level tag to me. –  Dilaton Jan 29 '13 at 11:27
1  
I should add that Vafa in the book at page 184 makes some comments about operator ordering, even though in a simpler case. The difference here is that, to my understanding, we have to provide an ordering (also in the susy transformations themselves) between $p_i,$ $\phi^j$ $\psi^k$ and $\overline\psi_m$ such that the obtained $Q$'s generate the transformations. –  jj_p Feb 2 '13 at 23:32
add comment

1 Answer

up vote 3 down vote accepted

Stopping short of actually doing the calculation, checking sign conventions, etc., my impression is as follows:

  1. At the classical level, the definition of Bosonic canonical momentum $p_i:=\frac{\partial L}{\partial\dot{\phi}^i}$ is standard.

  2. The Fermionic canonical momentum is more subtle, because the Lagrangian for the Fermions is already on first order form. A construction is feasible via the standard Dirac-Bergmann procedure leading to second-class constraints. A shortcut is offered by the Faddeev-Jackiw procedure. For a simple illustrative (Bosonic) example of these two methods, see e.g. this Phys.SE answer.

  3. In the Bosonic case, one should distinguish between the mechanical/kinetic momentum (which is roughly the velocity), and the canonical/conjugate momentum. In the Schrödinger/position representation, the former corresponds to covariant derivatives, while the latter corresponds to partial derivatives. (A similar situation takes place for a charged particle in an EM field with the role of Christoffel symbols replaced by the EM gauge potential, cf. this Phys.SE answer.)

  4. Note in particular that the commutator between two kinetic momenta will in general not be zero.

  5. In Ref. 1 eq. (92) Witten correctly places the covariant derivative in the super-charge formula, but then proceeds below eq. (92) to wrongly/confusingly refer to the covariant derivative as the momentum conjugate to $\phi^i$.

  6. Ref. 2 forgets to inform its readers that that it in the middle has changed the meaning of $p_i$ to now denote the covariant derivative, cf. eq. (10.217). Moreover, Ref. 2 wrongly/confusingly refers to the new $p_i$ as the conjugate momentum just above eq. (10.210). These mistakes are undoubtedly spurred by the confusing terminology of Ref. 1.

  7. Finally at the quantum mechanical level, operator ordering prescriptions should be chosen such that Hermiticity and SUSY are preserved.

References:

  1. E. Witten, Constraints on supersymmetry breaking, Nucl. Phys B202 (1982) 253.

  2. K. Hori, S. Katz, A. Klemm, R. Pandharipande, R. Thomas, C. Vafa, R. Vakil, and E. Zaslow, Mirror Symmetry, 2003. The pdf file is available here or here.

share|improve this answer
    
thanks for the answer, point 3 gives some insight. –  jj_p Feb 2 '13 at 23:26
    
moreover, in the book they are at least not precise when saying all the other (anti-)commutators vanish, since the real canonical variable is $\overline\psi_k$ so that $[\overline\psi^i,p_m]$ does not vanish –  jj_p Feb 2 '13 at 23:35
    
...if $p_m$ denotes the kinetic Bosonic momentum. The Poisson bracket between $\overline{\psi}^i$ and the canonical Bosonic momentum does indeed vanish. –  Qmechanic Feb 3 '13 at 19:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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