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
    Commented Mar 30, 2018 at 13:46
  • $\begingroup$ Comment to the post (v4): For starters, eq. (3) seems inconsistent with eqs. (1) & (2)? $\endgroup$
    – Qmechanic
    Commented Mar 30, 2018 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
    Commented Mar 30, 2018 at 18:08

1 Answer 1


It is surprising how difficult it is to build a coherent story about this because it seems that all of the references have a couple of typos. Rather than an answer, this post has a ton of questions that I came up with trying to build a coherent story. Hopefully it will be a starting point to building one, although I am sure it will also have its own typos.

Let us start with an action whose time derivatives are $$S=\int\text{d}t\left(\frac{1}{2}g_{ab}\dot{x}^a\dot{x}^b+ig_{ab}\bar{\psi}^a\frac{D}{Dt}\psi^b+\cdots\right).$$ In this form, the conjugate momenum to $x^a$ is $$p_a=g_{ab}\dot{x}^b+i\bar{\psi}^b\Gamma_{bac}\psi^c,$$ while $\psi^a$ and $\bar{\psi}_a$ form a fermionic conjugate pair. Here we already have some subtleties worth mentioning.

  1. As mentioned in a comment by @Qmechanic in Canonical quantization in supersymmetric quantum mechanics, we could've started with a different action by integration by parts $$S=\int\text{d}t\left(\frac{1}{2}g_{ab}\dot{x}^a\dot{x}^b+ig_{ab}\left(\lambda\bar{\psi}^a\frac{D}{Dt}\psi^b+(1-\lambda)\psi^a\frac{D}{Dt}\bar{\psi}^b\right)+\cdots\right),$$ with $\lambda\in[0,1]$. All of these actions would've yielded a different canonical conjugate momentum, at least naively.
  2. All of these action differ by boundary terms. However, these boundary terms are important! On the one hand, these boundary terms are precisely what fixes the symplectic structure, as shown in the covariant phase space formalism. On the other, different boundary terms will allow for different initial-final value problems. This is discussed in detail in e.g. Henneaux, Marc, and Claudio Teitelboim. Quantization of Gauge Systems. Princeton University Press, 1992. My guess is that all versions of the action are actually equivalent once these boundary terms are taken care of precisely. But I don't have the energy at the moment to carry out this analysis.
  3. Actually, not all of these actions are equivalent. This is because we needed to integrate by parts a covariant derivative. Accordingly, for they to be equivalent, we need to replace the integral measure $\text{d}t$ by the covariant one $\text{d}t\,\sqrt{g_{ab}\dot{x}^a\dot{x}^b}$. This has new time derivative terms that would modify everything. How does the $p_a$ change? Is the action still supersymmetric if we add that term? Is there a superspace action that yields directly the correct boundary terms and the correct integration measure?
  4. I believe there may be a typo in Tong's lectures regarding the contraction of the Christoffel symbols and the fermions in the formula for $p_a$. But maybe I am overlooking something.

Leaving these questions aside, canonical quantization demands we find a Hilbert space in which even operators $X^a$ and $P_a$, and odd operators $\Psi^a$ and $\bar{\Psi}_a$ act with the only non-trivial supercommutators being $$[X^a,P_a]=i\delta^a_b,\,\{\Psi^a,\bar{\Psi}_b\}=\delta^a_b.$$ The solution proposed in the literature, which is well-inspired by the usual methods in physics, is to take as a Hilbert space the (square integrable) differential forms $\Omega(M,\mathbb{C})$ with its standard inner product, and the operators $$X^a=x^a,\,P_a=-i\partial_a,\,\Psi^a=\text{d}x^a,\,\bar{\Psi}_a=\frac{\partial}{\partial\text{d}x^a}=\iota_\frac{\partial}{\partial x^a}.$$ Again there are some comments to be done at this stage:

  1. In general the coordinates $x^a$ are not globally defined so this does not provide well-defined operators. This is a known problem from, say, the quantization of a particle in the circle, where the angle $\theta$ is not a well-defined operator in the quantum theory, but $\sin(\theta)$ and $\cos(\theta)$ are. It is the latter which should be used for quantization.
  2. Why not use $P_a=-i\nabla_a$? Well, it wouldn't commute with the fermionic operators.
  3. Actually, the momentum $P_a$ is indeed a covariant derivative, but over the sections of a complex line bundle over the target space, as described in Schuller's lectures. Otherwise it wouldn't be self-adjoint. So, we should set $P_a=D_a$, where, for example, when acting on functions $f\in\Omega^0(M)=C^\infty(M)$ we have $$D_af=\partial_a f+\frac{1}{4}f\partial_a\ln\det(g)$$
  4. I have decided to make the unorthodox choice of $\Psi^a$ being creation operators instead of annihilation operators. It just makes more sense to me. This makes $\text{d}x^a$ have the same degree as $\psi^a$, which in my convention is positive, and will eventually also set the de Rham differential in this degree. Moreover, I was only able to build a coherent story in this picture. Other pictures are equivalent of course but the discussion is less streamlined.

Finally, the SUSY QM model is completely fixed once we quantize the generator of supersymmetry transformations $Q$. Classically, Noether's procedure yields $$Q=g_{ab}\dot{x}^a\psi^b-i\psi^a\partial_ah.$$ To quantize this, we could go two routes. First, we could note that classically we can equivalently write $$Q=p_a\psi^a-i\psi^a\partial_a h,$$ due to the lack of torsion $$\Gamma_{bac}\psi^c\psi^a=0.$$ We can then quantize this expression, which has no ordering amibiguities, leaving us with $$Q=P_a\Psi^a-i\Psi^a\partial_a h=-i(\partial_a\text{d}x^a+\text{d}x^a\partial_ah)=-i(\text{d}+\text{d}h)=-ie^{-h}\text{d}e^h.$$ The other method we could use is considering the kinetic momentum $$\pi_a=g_{ab}\dot{x}^b=p_a-i\bar{\psi}^b\Gamma_{bac}\psi^c.$$ Quantizing this has operator ambiguities. However, the following ordering yields a particularly nice result $$\Pi_a=P_a+i\Psi^c\Gamma_{bac}\bar{\Psi}^b=-i\left(\partial_a-{\Gamma^b}_{ac}\text{d}x^c\frac{\partial}{\partial\text{d}x^b}\right).$$ For example, by acting on a 2-form $\omega=\frac{1}{2}\omega_{ab}\text{d}x^a\text{d}x^b$, it is easy to convince one-self that this is in fact the covariant derivative on differential forms $$\Pi_a=-i\nabla_a.$$ We can then use this to quantize $Q$. This again has operator ambiguities. However, if we choose $$Q=\Psi^a\Pi_a-i\Psi^a\partial_a h,$$ the same cancellation of the Christoffel symbols will happen and we recover $$Q=-ie^{-h}\text{d}e^{h}.$$ Note that this agrees with Tong's result but my definition for $Q$ is different! What I call $Q$ would be what Tong calls $Q^\dagger$ (my convention agrees with the Gaiotto, Witten, Moore paper). This has to do with my choice of making the $\Psi^a$ the creation operators. One should compare these choices keeping in mind the comments right after the equations (3.8) in Tong's lectures.

Hopefully this can help set the ground for more discussion! I apologize if there are any typos!

  • 1
    $\begingroup$ I was going through schullers lecture and see you in comment of almost every video. I haven't reached the lecture about these topics but I am sure it'll be helpful then. Thanks. $\endgroup$
    – Babu
    Commented Mar 29, 2022 at 17:50
  • 1
    $\begingroup$ I am a big fan of those videos. I keep revisiting them every once in a while and I always learn something new :). Hope my comments are helpful! $\endgroup$ Commented Mar 29, 2022 at 18:03
  • $\begingroup$ Computing $\bar{Q}$ and the Hamiltonian $H=\frac{1}{2}[Q,\bar{Q}]$ can get pretty confusing. In my answer math.stackexchange.com/questions/2828243/… I clarify how to compute $\bar{Q}$. In the notation used here, the result is $\bar{Q}=\bar{\Psi}^a\Pi_a+i\bar{\Psi}^a\partial_a h$. One should note that $\Pi_a$ is not selfadjoint, which explains why the order is not $\Pi_a\bar{\Psi}^a$. $\endgroup$ Commented Apr 1, 2022 at 23:51
  • $\begingroup$ In my answer I raised the problem of integrating by parts the covariant derivative. This is actually legal $$\int\text{d}t\,g_{ab}\bar{\psi}^a\frac{D\psi^b}{Dt}=-\int\text{d}t\,g_{ab}\frac{D\bar{\psi}^a}{Dt}\psi^b+\left[g_{ab}\bar{\psi}^a\psi^b\right]_{t_i}^{t_f}.$$ This is a classic computation involving the time derivative of the metric tensor obtained when integrating by parts the normal time derivative. $\endgroup$ Commented Jun 6, 2022 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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