pardon if this turns out to be badly phrased or a not well thought out question.

  1. Following the discussion on: Canonical quantization in supersymmetric quantum mechanics and Conjugate Momenta in a Supersymmetric Sigma Model

    Given the follwing supersymmetric Lagrangian for a non-linear sigma model: $$L = \frac12 g_{ij}\dot{\phi}^{i}\dot{\phi}^{j} + \dfrac{i}{2}g_{ij}\left(\overline{\psi}^{i}\nabla_{t}\psi^{j}-\nabla_{t}\overline{\psi}^{i}\psi^{j}\right)-\frac12 R_{ijkl}\psi^{i}\overline{\psi}^{j}\psi^{k}\overline{\psi}^{l}, \tag{1}$$

    where the covariant derivative is of the form $$ \nabla_{t}\psi^{i} = \partial_{t}\psi^{i} + \Gamma^{i}_{jk}\partial_{t}\phi^{j}\psi^{k}, \tag{2}$$

    We are told in Mirror Symmetry by Hori, K. et. al. that the conjugate momenta of the fields are given by

$$\begin{align} &\frac{\partial L}{\partial\dot{\phi}^{m}} = p_{m} = g_{mj}\dot{\phi}^{j},\tag{3}\\ &\frac{\partial L}{\partial\dot{\psi}^{m}} = \Pi_{\psi m} = i g_{mj}\overline{\psi}^{j}\tag{4}. \end{align}$$

  • Similar definitions have also been found in E. Witten, Constraints on supersymmetry breaking. I'm still confused and uncertain about the answers provided by @Qmechanic and hence not certain if the defintion for the conjugate momenta given in Mirror Symmetry by Hori, K. et. al. is incorrect (as commented by @Qmechanic in this SE answer.

  • This is because I'm inclined to believe that the Hamiltonian of the system should be $$H = \frac12 g^{ij}p_{i}p_{j} + \frac12 R_{ijkl}\psi^{i}\overline{\psi}^{j}\psi^{k}\overline{\psi}^{l},\tag{6}$$ where $p_{i}$ is as defiend in $(3)$, following the Faddeev-Jackiw method, which I infer to be the recipe used for the Single Variable Potential Theory case in section 10.2 of Mirror Symmetry.

Is this interpretation correct? And besides taking the operation of $\frac{\partial L}{\partial\dot{\phi}^{m}}$ as a strict definition of the conjugate momenta of the Hamiltonian, are they any reasons one might have to suggest an alternative form for the conjugate momenta?

  1. Also, if we were to see interpret the Lagrangian of the supersymmetric particle as a free massless boson moving on the Riemannian manifold, the second and third term of the Lagrangian should then be interpreted as a source term. Calculations of the corresponding path integral would then be done by taking the perturbative expansion in the action. I believe this means bringing the corresponding terms in the exponent down from the exponential in $$\int \mathcal{D}\phi \mathcal{D}\psi \mathcal{D}\overline{\psi} \left(\exp{i\int dt L}\right).\tag{7}$$

    • Hence to get a consistent theory, we may consider the conjugate momenta to simply just be $g_{ij}\dot{\phi}^{j} = p_{i}$

This seems to support the claim that the definitions in the papers are correct?

  • $\begingroup$ @Qmechanic hi, I've 'heard' much about you. Would you be able to enlighten me on the problems I have? And thanks for highlighting the incomplete sentence. $\endgroup$ – Marvin Mar 30 '18 at 13:46
  • $\begingroup$ Comment to the post (v4): For starters, eq. (3) seems inconsistent with eqs. (1) & (2)? $\endgroup$ – Qmechanic Mar 30 '18 at 16:31
  • $\begingroup$ I agree with you wholeheartedly. It seems that I confused myself. Also, (6) would hold, but more exactly in the form of $$H = \frac12 g_{ij}\phi^{i}\phi^{j} + \frac12 R_{ijkl}\psi^{i}\overline{\psi}^{j}\psi^{k}\overline{\psi}^{l}$$ with the corrected conjugate momenta (with terms carrying the Christoffel symbol). But I suppose (10.204) and (10.205) in Mirror Symmetry would be wrong too and instead be rewritten carrying additional terms with the Christoffel symbol. And that point 2. is redundant. $\endgroup$ – Marvin Mar 30 '18 at 18:08

Your Answer

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

Browse other questions tagged or ask your own question.