Suppose you have a theory of maps $\phi: {\cal T} \to M$ with $M$ some Riemannian manifold, Lagrangian $$L~=~ \frac12 g_{ij}\dot\phi^i\dot\phi^j + \frac{i}{2}g_{ij}(\overline{\psi}^i D_t\psi^j-D_t\overline{\psi}^i\psi^j) -\frac{1}{4}R_{ijkl}\psi^i\psi^j\overline{\psi}^k\overline{\psi}^l,\tag{10.198}$$

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

$$D_t\psi^i~=~ \partial_t \psi^i +\Gamma^i_{jk}\dot{\phi}^j\psi^k.\tag{10.199}$$

(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

$$ \begin{eqnarray}\delta\phi^i &=& \epsilon \overline\psi^i-\overline\epsilon \psi^i \tag{10.200}\cr \delta\psi^i &=& i\epsilon (\dot\phi^i-\Gamma^i_{jk}\overline\psi^j \psi^k)\tag{10.201}\cr \delta\overline\psi^i &=& -i\epsilon (\dot\phi^i-\Gamma^i_{jk}\overline\psi^j \psi^k). \tag{10.202}\end{eqnarray}$$

How can I quantise the classical supercharges

$$Q=i\overline\psi_i\dot\phi^i, \qquad \overline Q=-ig_{ij}\psi^i\dot\phi^j \tag{10.210}$$

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

$$ p_i=\frac{\partial L}{\partial\dot\phi^i}, \qquad \pi_{i\psi}=\frac{\partial L}{\partial\dot\psi^i}, \tag{10.207b}$$

and impose canonical commutation relations

$$ \begin{eqnarray} [\phi^i,p_j]&=&i\delta^i_j,\tag{10.208} \cr \{\psi^i,\pi_{\psi,j}\}&=&\delta^i_j.\tag{10.209}\end{eqnarray}$$

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.


Comment added: 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.


1 Answer 1


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.


  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.

  3. E. Witten, Supersymmetry and Morse theory, J. Diff, Geom. 17, (1982) 661; Chapter 2.

  • $\begingroup$ 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 $\endgroup$
    – jj_p
    Commented Feb 2, 2013 at 23:35
  • $\begingroup$ ...if $p_m$ denotes the kinetic Bosonic momentum. The Poisson bracket between $\overline{\psi}^i$ and the canonical Bosonic momentum does indeed vanish. $\endgroup$
    – Qmechanic
    Commented Feb 3, 2013 at 19:02
  • $\begingroup$ In the third comment above, don't you mean commutator instead of Poisson bracket? $\endgroup$
    – Mtheorist
    Commented Jan 25, 2017 at 13:25
  • 1
    $\begingroup$ @Mtheorist: I wrote Poisson brackets to stress that the construction already makes sense at the classical level. Of course it also makes sense at the quantum operator level, so you are free to replace Poisson brackets by commutators (divided by $i\hbar$). $\endgroup$
    – Qmechanic
    Commented Jan 25, 2017 at 13:44
  • $\begingroup$ Notes for later: Ref. 1 eq. (90) has 1/8 normalization in front of the curvature term while Ref. 2 eq. (10.198) has 1/2. OP (v5) has a 1/4. $\endgroup$
    – Qmechanic
    Commented Mar 30, 2018 at 14:49

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