Consider the path integral quantisation of a scalar field $\phi$ on flat spacetime. Let the Lagrangian be $\mathcal{L}$. I would like to prove the following equal time commutation relation: \begin{equation} \tag{1} \left[\phi(\mathbf{x},t),\frac{\partial{\mathcal{L}}}{\partial (\partial_0\phi)}(\mathbf{y},t)\right]=i\hbar\delta^{(3)}(\mathbf{x}-\mathbf{y}). \end{equation}

The motivation is hopefully clear: I am aiming to show that the path integral is able to reproduce the CCRs that we impose in canonical quantisation.

So far, I've tried writing commutator matrix elements as follows: \begin{equation} \tag{2} \langle \phi_2|\left[\phi(\mathbf{x},t),\frac{\partial{\mathcal{L}}}{\partial (\partial_0\phi)}(\mathbf{y},t)\right]|\phi_1\rangle =\int D\phi \left(\phi(\mathbf{x},t+\epsilon)\frac{\partial{\mathcal{L}}}{\partial (\partial_0\phi)}(\mathbf{y},t)- \phi(\mathbf{x},t)\frac{\partial{\mathcal{L}}}{\partial (\partial_0\phi)}(\mathbf{y},t+\epsilon)\right)e^{iS[\phi]/\hbar}, \end{equation} where $\epsilon> 0$ is small, and the path integral has some appropriate domain ($\mathbb{R}^3\times[0,\epsilon]$ is my guess) and BCs. Note that the introduction of $\epsilon$ ensures the path integral time-orders the products as desired. However I've been unable to proceed from here.

Am I on the right track, or is there a better way to prove (1)?

[Please note, before you tell me that (1) is simply the CCR, which is an assumption/axiom in QFT: that is indeed true in the canonical framework, but the point of this question is to take the path integral as our starting point. Given the path integral tells us everything about the QFT in principle, it should be able to derive (1).]

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    $\begingroup$ I think you might want to try looking at the Ward identity associated to the transformations $\phi\rightarrow \phi+\delta\phi$. The commutator should arise from the contact terms in the Ward identity. I haven't actually done this calculation so it's just a suggestion. $\endgroup$
    – octonion
    Dec 28, 2021 at 1:20
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    $\begingroup$ Yes the approach using Ward identities (aka the Schwinger-Dyson equation) works $\endgroup$
    – octonion
    Dec 28, 2021 at 2:01
  • $\begingroup$ @octonion Unfortunately even with your hint I'm still stuck. In particular, I'm struggling to use the Ward identities to get anything involving the operator $\frac{\partial \mathcal{L}}{\partial(\partial_0 \phi)}$. Any more detailed hints (or a full solution) would be appreciated. $\endgroup$ Dec 28, 2021 at 17:56

1 Answer 1


Consider the Schwinger-Dyson equation $$\langle\left[\partial_\mu\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi\right)}- \frac{\partial\mathcal{L}}{\partial\phi} \right](x) \,\,\phi(y)\rangle = -i\delta^{(4)}(x-y)$$ This is just the equations of motion inside a correlation function and the presence of the $\phi(y)$ operator leads to the contact term on the right side (I'm setting $\hbar=1$). It can be derived in the path integral approach by considering $\langle\phi(y)\rangle$ and doing a local change of variables $\phi\rightarrow \phi+\delta\phi$, see e.g. Polchinski's String Theory Vol 1 Sec 2.3.

Now integrate this equation over a cylinder where the spatial coordinates $x^i$ are integrated over a region $A$, and the time coordinate $x^0$ is integrated over a tiny interval from $y^0-\epsilon/2$ to $y^0+\epsilon/2$. If we have an ordinary Lagrangian which doesn't involve terms with products of both $\phi$ and $\dot{\phi}$ we would expect the `potential' term involving $\partial \mathcal{L}/\partial \phi$ to involve no extra divergences and after integrating over $x$ this term will vanish in the limit $\epsilon\rightarrow 0$ and we will drop it.

The other `kinetic' term is a total divergence so we may rewrite the integral over $x$ as a flux through the surface of the cylinder. The flux of $\partial \mathcal{L}/\partial \left(\partial_i\phi\right)$ through the sides of the cylinder should likewise vanish in the limit $\epsilon\rightarrow 0$. So we are left with

$$\lim_{\epsilon\rightarrow 0}\int_A d^3 x\left[ \langle\left.\frac{\partial\mathcal{L}}{\partial\left(\partial_0\phi\right)}(x)\,\,\phi(y)\rangle\right|_{x^0=y^0+\epsilon/2} - \langle\phi(y)\left.\frac{\partial\mathcal{L}}{\partial\left(\partial_0\phi\right)}(x) \rangle\right|_{x^0=y^0-\epsilon/2} \right] = -i\,1_A(y^i)$$ where $1_A$ is an indicator function that equals one if $y^i\in A$, and zero otherwise. So the quantity in brackets is clearly $-i\delta^{(3)}(x^i-y^i)$. And also from the relation of the path integral correlation functions to time ordered correlation functions in the operator approach it must be the commutator $$\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_0\phi\right)}(x),\phi(y)\right]=-i\delta^{(3)}(x^i-y^i)$$

  • $\begingroup$ Brilliant, this is exactly what I was looking for. Thank you! $\endgroup$ Dec 29, 2021 at 23:13
  • $\begingroup$ By the way - I take it this only works in Lorentzian signature, due to the use of time ordering in the last line? I suppose in Euclidean signature we'd find that the commutator equals zero, which seems a little odd... $\endgroup$ Dec 29, 2021 at 23:19
  • $\begingroup$ It works in the Euclidean signature too. If you start from a given quantum mechanical system and form the Euclidean path integral you see that correlation functions in the path integral correspond to Euclidean time ordered correlation functions on the operator side. So in the end we get the same canonical commutation relations, just the factor of $i$ comes from different places. $\endgroup$
    – octonion
    Dec 30, 2021 at 3:10
  • $\begingroup$ To be honest I'd always thought that all local operators commute in Euclidean QFT, since they're all spacelike separated. Therefore there's no ordering ambiguity. But perhaps this is only the case for theories which are Euclidean from the get-go, as opposed to the Euclidean section of a Lorentzian QFT. $\endgroup$ Dec 30, 2021 at 10:53
  • $\begingroup$ Btw you answered my original question perfectly, and I have now asked a separate question which focusses more on ordering issues in Euclidean signature. $\endgroup$ Dec 30, 2021 at 10:54

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