First, I must say that I'm not very familiar with the path integral formalism, so maybe I'm missing something very basic.

In Section III.A. of this paper, Toms considers a particle in $D$-dimensional Euclidean space, whose position is specified by $q^i$, constrained to an arbitrary surface given by $f(q)=0$. Its Lagrangian is

$$L=\frac{1}{2}\dot{q}_i\dot{q}^i-V(q)+\sigma f(q),\tag{3.2}$$

where $\sigma$ is a Lagrange multiplier. In order to apply the Faddeev-Jackiw formalism, he writes it in first-order form, defining the canonical momentum in the usual way, so

$$L=p_i\dot{q}^i-\frac{1}{2}p_ip^i+\sigma f(q)-V(q)\tag{3.4}.$$

Now, he treats $\sigma f(q)$ as part of the symplectic part of the Lagrangian. As he explains generally on page 5 and particularly on page 9, this is better motivated if we rename the Lagrange multiplier $\sigma\to\dot \sigma$ and treat $\sigma$ as a new variable. He does not, but let me do it for clarity.

After that, he applies the procedure and finds another constraint $\partial_i f~ p^i=0$, which is included using another Lagrange multiplier $\dot{\lambda}$, so the final Lagrangian is

$$L_f=p_i\dot{q}^i+\dot{\sigma} f(q)+\dot{\lambda}~\partial_if~p^i-\frac{1}{2}p_ip^i-V(q)\tag{3.12'}.$$

Finally, after the whole analysis, he gets the path integral measure

$$d\mu=\left(\prod_i[dq^i]\right)\left(\prod_i[dp_i]\right)[d\sigma][d\lambda]|\nabla f|^2.\tag{3.23}$$

So far, so good. What happens now is that I'm not sure how he gets the partition function

$$Z=\int\left(\prod_i[dq^i]\right)\left(\prod_i[dp_i]\right)|\nabla f|^2~\delta\left(f(q)\right)~\delta(\mathbf p\cdot\nabla f)\\\times\exp\left\{i\int dt\left(p_i\dot{q}^i-\frac{1}{2}p_ip^i-V(q)\right)\right\},\tag{3.24}$$

especially the delta functions of the constraints. I see that it makes sense, but I would like to obtain it mathematically.

I would propose the following

$$Z=\int\left(\prod_i[dq^i]\right)\left(\prod_i[dp_i]\right)[d\sigma][d\lambda]|\nabla f|^2~\exp\left\{i\int dt~L_f\right\}\tag{1},$$


$$\int dt~L_f=\int dt\left(p_i\dot{q}^i-\frac{1}{2}p_ip^i-V(q)\right)+\int dt~\dot{\sigma}f(q)+\int dt~\dot{\lambda}~(\mathbf p\cdot\nabla f).\tag{2}$$

The second integral is

$$\int dt~\dot{\sigma}f(q)=\sigma f(q)-\int dt~\sigma \frac{df(q)}{dt}=\sigma f(q),\tag{3}$$

where I took $df(q)/dt=0$ since the constraint must be preserved. I am not sure at all that this is correct, since the second constraint is precisely the time derivative of the first one and I'm not taking it to be zero in the third integral of $(2)$. However, doing the same for the third integral of $(2)$, I get

$$Z=\int\left(\prod_i[dq^i]\right)\left(\prod_i[dp_i]\right)[d\sigma][d\lambda]|\nabla f|^2~e^{i\sigma f(q)}e^{i\lambda~(\mathbf p\cdot\nabla f) }\\\times\exp\left\{i\int dt\left(p_i\dot{q}^i-\frac{1}{2}p_ip^i-V(q)\right)\right\}\tag{4},$$

which apart from $2\pi$-factors from the Dirac deltas, gives $(3.24)$.

I'm almost sure that this is not correct, but I would like to know exactly why and how to get $(3.24)$.


2 Answers 2


I) More generally Ref. 1 observes that if we start with a Lagrangian $$L_0~=~\sum_{I=1}^{2n}\vartheta_I\dot{z}^I-H\tag{A}$$ on the Faddeev-Jackiw form with a non-degenerate symplectic 1-form potential $$\begin{align}\vartheta~=~&\sum_{I=1}^{2n}\vartheta_I\mathrm{d}z^I,\cr \mathrm{d}\vartheta~=~&\omega~=~\frac{1}{2} \sum_{I,J=1}^{2n}\omega_{IJ}\mathrm{d}z^I\wedge \mathrm{d}z^J,\end{align}\tag{B}$$ and we add $2m$ second-class constraints $f_a$ with Lagrange multipliers $\dot{\lambda}^a$, $$L~=~L_0 +\sum_{a=1}^{2m}f_a\dot{\lambda}^a, \tag{C}$$ we get a non-degenerate symplectic 1-form potential $$\begin{align}A~=~&\sum_{\alpha=1}^{2(n+m)}A_{\alpha}\mathrm{d}\xi^{\alpha}~=~\sum_{I=1}^{2n}\vartheta_I\mathrm{d}z^I +\sum_{a=1}^{2m}f_a \mathrm{d}\lambda^a, \cr \mathrm{d} A~=~&F~=~\frac{1}{2} \sum_{\alpha,\beta=1}^{2(n+m)}F_{\alpha\beta}\mathrm{d}\xi^{\alpha}\wedge \mathrm{d}\xi^{\beta},\cr F_{\alpha\beta} ~=~&\begin{pmatrix}\omega_{IJ} &\frac{\partial f_b}{\partial z^I} \cr -\frac{\partial f_a}{\partial z^J}& 0 \end{pmatrix},\end{align} \tag{D}$$ with variables $\xi^{\alpha}=(z^I,\lambda^a)$. After a matrix trick similar to eq. (3.17) in Ref. 1 the path integral $$\begin{align} Z ~=~&\int\! {\cal D}\frac{\xi}{\sqrt{\hbar}}~{\rm Pf}(F_{\cdot\cdot}) \exp\left(\frac{i}{\hbar}\int\!dt ~L \right) \cr ~=~&\int\! {\cal D}\frac{z}{\sqrt{\hbar}}~{\rm Pf}(\omega_{\cdot\cdot})~{\rm Pf}(\{f_{\cdot},f_{\cdot}\}) \delta[f_{\cdot}] \exp\left(\frac{i}{\hbar}\int\!dt ~L_0 \right) \tag{E} \end{align}$$ becomes on the Senjanovic form. See also e.g. this related Phys.SE post.

II) The Jacobian for the change of integration variable $\lambda^a\to\dot{\lambda}^a$ is an irrelevant normalization constant after taking zero-modes and boundary conditions into account.

It is implicitly understood that the Pfaffian is a functional Pfaffian. Similarly, the delta functional is formally $$ \delta[f_{\cdot}]~=~\prod_t \delta(f(t)), \tag{F}$$ and so forth.

III) In OP's example there are 2 second-class constraints $$f_1~=~f\quad\text{and} \quad f_2~=~{\bf p}\cdot \nabla f,\tag{G}$$ which is also discussed in this Phys.SE post. The Poisson bracket is then $$\{f_1,f_2\}=|\nabla f|^2.\tag{H}$$ Therefore the path integral measure factor is $${\rm Pf}(\{f_{\cdot},f_{\cdot}\})~=~\prod_t |\nabla f(t)|^2 \tag{I}$$ in this example.


  1. D.J. Toms, Faddeev-Jackiw quantization and the path integral, arXiv:1508.07432.
  • $\begingroup$ It is more clear now, thanks. 1. The functional Jacobian would be the same as in this post, right? 2. In that case, is Toms ignoring it? $\endgroup$
    – AFG
    Commented Jun 7, 2021 at 14:15
  • $\begingroup$ 1. Yes. 2. Yes, because it can be ignored. $\endgroup$
    – Qmechanic
    Commented Jun 7, 2021 at 14:22

A path integral version of the delta function identity $$\int\limits_{-\infty}^\infty \frac{\mathrm{d}p}{2\pi}\; \mathrm{e}^{\mathrm{i}\, p\cdot x}=\delta(x)$$ reads $$ \int \mathrm{D}{\phi}\ \mathrm{e}^{\mathrm{i} \left<\phi,\psi\right>}=\delta(\psi),$$ where $\left<\bullet,\tilde{\bullet}\right>$ is an inner product on the space where $\phi$ and $\psi$ live. In your case the inner product is simply $$\left<\bullet,\tilde{\bullet}\right>:= \int \mathrm{d}t \ \bullet\!(t)\ \tilde{\bullet}\!\,(t),$$ so an expression like $$\int \mathrm{D}\sigma\ \exp\!\left(\mathrm{i}\int \mathrm{d}t\ \sigma(t) \ f[q](t) \right)$$ is just $$\int \mathrm{D}\sigma\ \mathrm{e}^{\mathrm{i}\,\left<\sigma,f[q]\right>}=\delta\big(f[q]\big).$$ Similarly $$ \int \mathrm{D}\lambda \exp\left(\mathrm{i}\int \mathrm{d}t\ \lambda(t)\ \big[p\cdot\nabla f\big](t)\right)=\delta\big(p\cdot\nabla f\big).$$ And so you get (3.24).

Note that your conclusion after your equation (4) is not correct, as $\sigma$ and $\lambda$ are functions of $t$ and you are doing a functional integral, whereas what you have there, $\sigma f[q]$, and $\lambda \big[p\cdot\nabla f\big]$, is evaluated at the boundaries of integration (wherever these may be). Hence only the zero mode of that path integral gives you a delta function, but that is a delta function only of the zero modes, whereas in (3.24) and in the way I did it above all the modes contribute. Moreover you should have had a functional Jacobian passing from $\mathrm{D}\dot{\sigma}$ to $\mathrm{D}\sigma$ and similarly for $\lambda$. Finally the rest of the modes find nothing to hit, i.e. the non-zero-mode sectors look (schematically) like $\displaystyle \int \prod\limits_t \mathrm{d}\sigma(t) \ 1$, yielding an infinite result (and similarly for $\lambda$). So there are many subtleties in your computation.

  • $\begingroup$ I appreciate your answer, but there are several things that I still don't understand. First, let's clear up a couple of things. Should I have $D\sigma$ or $D\dot\sigma$ in the measure? Should I have $\dot\sigma$ or $\sigma$ in the integrand of the exponential? The way I wrote it does not match with the delta function identity. $\endgroup$
    – AFG
    Commented Jun 6, 2021 at 16:07
  • $\begingroup$ According to the reference you are reading you should have $\sigma$ and $\mathrm{D}\sigma$ (around equation (2.7) there). If you rename $\sigma$ to $\dot{\sigma}$ as you did in your post (I don't understand why this is more natural, but there must be a reason for it) then you must rename $\mathrm{D}\sigma$ to $\mathrm{D}\dot{\sigma}$ too. If you instead start with $\dot{\sigma}$ from the get-go, you should have $\mathrm{D}\sigma$. $\endgroup$ Commented Jun 6, 2021 at 21:34
  • $\begingroup$ I use $\dot\sigma f$ instead of $\sigma f$ because I need to have that term in the symplectic part of the Lagrangian, that is what Faddeev-Jackiw method is about. Then, I use $D\sigma$ in the measure because $\sigma$ itself is one of the canonical variables, and the general measure is given by $(2.14)$. Apart from that, Toms says "We could choose $\dot{\sigma}$ and $\dot{\lambda}$ in place of $\sigma$ and $\lambda$ here if desired, but as we stated above this is not really necessary" (around eq. $(3.12)$). Well, I think that it $\textit{is}$ necessary, at least a clarification. $\endgroup$
    – AFG
    Commented Jun 7, 2021 at 7:56
  • $\begingroup$ Fair. I don't know much about the Faddeev–Jackiw method, so I can't give clarifications regarding that, but since Toms uses $\sigma$ and $\lambda$, the procedure he uses to get the result is the one I described in my answer. $\endgroup$ Commented Jun 7, 2021 at 8:36

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