As we known, if there is time derivative interaction in $\mathcal L_\mathrm{int}$, then $\mathcal{H}_\mathrm{int}\neq -\mathcal{L}_\mathrm{int}$. For example, Scalar QED, $$ \begin{aligned} \mathcal{L}_\mathrm{int}&= -ie \phi^\dagger(\partial_\mu \phi) A^\mu+ie(\partial_\mu \phi^\dagger) \phi A^\mu +e^2\phi^\dagger \phi A_\mu A^\mu, \\ \mathcal{H}_\mathrm{int}&=-\mathcal{L}_\mathrm{int} -e^2 \phi^\dagger\phi (A^0)^2. \end{aligned} $$ There is the last term breaking Lorentz invariance.

Derivation: \begin{eqnarray} \mathcal{L}&=&(\partial_\mu+i e A_\mu)\phi (\partial^\mu-i e A^\mu)\phi^\dagger-m^2\phi^\dagger \phi\\ &=&\mathcal{L}_0^\mathrm{KG}+\mathcal{L}_\mathrm{int}, \end{eqnarray} where $$\mathcal{L}_0^\mathrm{KG}=\partial_\mu\phi \partial^\mu\phi^\dagger-m^2\phi^\dagger \phi,$$ $$\mathcal{L}_\mathrm{int}= -ie \phi^\dagger(\partial_\mu \phi) A^\mu+ie(\partial_\mu \phi^\dagger) \phi A^\mu +e^2\phi^\dagger \phi A_\mu A^\mu, $$

$$\pi=\frac{\partial \mathcal{ L}}{\partial(\partial_0 \phi)}=\partial^0\phi^\dagger-i e A^0 \phi^\dagger,$$

$$\pi^\dagger=\frac{\partial \mathcal{ L}}{\partial(\partial_0 \phi^\dagger)}=\partial^0\phi+i e A^0 \phi, $$

\begin{eqnarray} \mathcal{H}&=&\pi \dot\phi+\pi^\dagger \dot\phi^\dagger-\mathcal{L} \\ &=&\pi \dot\phi+\pi^\dagger \dot\phi^\dagger-(\dot\phi^\dagger\dot\phi-\nabla\phi^\dagger \cdot\nabla\phi-m^2 \phi^\dagger\phi)-\mathcal{L}_\mathrm{int} \\ &=&\pi(\pi^\dagger-ieA^0\phi)+\pi^\dagger (\pi +ieA^0\phi^\dagger)-((\pi^\dagger-ieA^0\phi)(\pi +ieA^0\phi^\dagger) \\ &&-\nabla\phi^\dagger \cdot\nabla\phi-m^2 \phi^\dagger\phi)-\mathcal{L}_\mathrm{int}\\ &=&(\pi^\dagger \pi + \nabla\phi^\dagger \cdot\nabla\phi+m^2 \phi^\dagger\phi)-\mathcal{L}_\mathrm{int} -e^2 \phi^\dagger\phi (A^0)^2 \\ &=&\mathcal{H}_0^\mathrm{KG}+\mathcal{H}_\mathrm{int}. \end{eqnarray}

My questions:

  1. The Feynman Rules for Scalar QED is here. But we see there is an extra term in interaction Hamitonian $ -e^2 \phi^\dagger\phi (A^0)^2$, according to Wick's theorem, it should have some contribution to Feynman Rule which does not occur in this textbook. I've computed this vertex and I find it's nonzero. Why there is no Feynman rules for such Lorentz breaking term?

  2. As we known, for path integral quantization, the coordinate space path integral: $$Z_1= \int D q\ \exp\left(\int dt\ L(q,\dot q) \right).$$ And phase space path integral: $$Z_2= \int D p\, D q\ \exp\left(\int dt\ p\dot q -H(p,q) \right).$$ Only for this type Lagrangian $L=\dot q^2-V(p)$, then $Z_1=Z_2$. (The Feynman Rules for Scalar QED in textbook is same as what is derived by coordinate space path integral. ) I consider the 2nd method of path integral quantization is always equivalent to canonical quantization. So for Scalar QED, are these two kinds of path integral quantization same? How to prove?

  3. For non-abelian gauge theory, there is derivative interaction even in gauge field itself. It seems that all textbooks use $Z_1$ to get the Feynman rules. Are these two kinds of path integral quantization same in non-abelian gauge field? If not same, why we choose the coordinate space path integral? It's the axiom because it coincides with experiment?

2 Answers 2


Some general remarks:

  1. In the operator formalism, the non-covariant extra term in the Hamiltonian is cancelled by a non-covariant term coming from the naïve time ordering symbol: $$ \mathrm T\sim\mathrm T_\mathrm{cov}-e^2\phi^2A^2_0 $$

    You can find the details in ref.1, section 6-1-4.

    On the other hand, the case of the path-integral formalism is covered by item 2 below.

  2. The formal equivalence $Z_1=Z_2$ can be proven for any Hamiltonian of the form $$ H\sim A^{ij}\pi_i\pi_j+B^i(\phi)\pi_i+C(\phi) $$ of which your Hamiltonian is an example of. For the proof and relevant discussion, see ref.2, Vol.1., section 9.3.

  3. For the discussion of the path-integral quantisation of non-abelian gauge theories, see ref.2, Vol.2, chapters 15.4 -- 15.8. Ref.1, chapter 12-2 is also worth a read. In short, "$Z_1=Z_2$" up to the subtleties introduced by gauge-invariance.


[1] Itzykson & Zuber, Quantum field theory.

[2] Weinberg, Quantum theory of fields.

  • $\begingroup$ Could you elaborate a bit on the first point of yours? What is the reason behind choosing such $\text{T}$? $\endgroup$ May 12, 2017 at 2:51
  • $\begingroup$ Hi @SolenodonParadoxus I think that the first point is very well explained in the reference [1], so it would be useless to summarise it here. The essential point is that $\mathrm T(\partial_\mu\phi\partial_\nu\phi)=\partial_\mu\partial_\nu\mathrm T(\phi\phi)+\delta_\mu^0\delta_\nu^0\delta(t-t')$, which is not covariant. We define $\mathrm T_\mathrm{cov}$ such that $\mathrm T_\mathrm{cov}(\partial_\mu\phi\partial_\nu\phi)=\partial_\mu\partial_\nu\mathrm T_\mathrm{cov}(\phi\phi)$, which is now covariant. The extra term $\delta_\mu^0\delta_\nu^0\delta(t-t')$ cancels the non-covarant term in H_int $\endgroup$ May 13, 2017 at 11:28

AccidentalFourierTransform has already given a good answer. Here we will provide more details & justifications for a class of non-gauge derivative interactions.

  1. We start from a Lagrangian action, $$\begin{align} S[Q]~=~&\int_{t_i}^{t_f}\! dt~ L~=~S_0[Q]+S_{\rm int}[Q], \cr L~=~& L_0(Q,\dot{Q})+L_{\rm int}(Q,\dot{Q}),\cr S_0[Q]~=~&\int_{t_i}^{t_f}\! dt~ L_0(Q,\dot{Q}), \cr L_0(Q,\dot{Q})~=~&\frac{1}{2}\dot{Q}^2~=~\frac{1}{2} \dot{Q}^i G_{ij} \dot{Q}^j,\cr S_{\rm int}[Q]~=~&\int_{t_i}^{t_f}\! dt~ L_{\rm int}(Q,\dot{Q}), \cr L_{\rm int}(Q,\dot{Q})~=~&A_i\dot{Q}^i - V, \cr G_{ij}~=~&G_{ij}(Q), \qquad A_i~=~A_i(Q), \qquad V~=~V(Q), \end{align} \tag{1}$$ which is quadratic in velocities. We shall assume that the Lagrangian action (1) is manifestly Lorentz covariant. [We are using DeWitt condensed notation$^1$ to suppress spatial (but not temporal) dimensions, which may superficially obscure the manifest Lorentz covariance. So e.g. the $\frac{1}{2}\dot{Q}^2$ term in $L_0$ is implicitly accompanied by a $\frac{1}{2}(\nabla Q)^2$ term in $V$, and so forth. Source terms $J_iQ^i$ are also assumed to be inside $V$.]

  2. The canonical momentum read $$ P_i~=~G_{ij}\dot{Q}^j+A_i.\tag{2}$$ We stress that the corresponding Hamiltonian action is also Lorentz covariant, $$\begin{align} S_H[Q,P]~=~&\int_{t_i}^{t_f}\! dt~ L_H~=~S_{H,0}[Q,P]+S_{H,{\rm int}}[Q,P], \cr L_H~=~&P_i \dot{Q}^i-H(Q,P), \cr H(Q,P)~=~&H_0(Q,P)+H_{\rm int}(Q,P), \cr S_{H,0}[Q,P]~=~&\int_{t_i}^{t_f}\! dt~ L_{H,0}, \cr L_{H,0} ~=~&P_i \dot{Q}^i-H_0(Q,P), \cr H_0(Q,P)~=~&\frac{1}{2}P^2~=~\frac{1}{2}P_i G^{ij}P_j,\cr S_{H,{\rm int}}[Q,P] ~=~&-\int_{t_i}^{t_f}\! dt~H_{\rm int}(Q,P), \cr H_{\rm int}(Q,P)~=~& -A^iP_i + \color{red}{\frac{1}{2} A^2}+V,\end{align}\tag{3}$$ despite the non-covariant term $A^2:=A_iG^{ij}A_j$ marked in red in eq. (3). We did not put this in by hand. It actually comes from doing the Legendre transformation properly.

    We mention (for a later instructive comparison with eq. (6) below) that $$ L_{\rm int}(Q,\dot{Q})+H_{\rm int}(Q,P) ~\stackrel{(1)+(2)+(3)}{=}~ - \color{red}{\frac{1}{2} A^2(Q)},\tag{4}$$ although eq. (4) will not be used in what follows. Eq. (4) corresponds to OP's second formula.

  3. So far we have only discussed the classical theory. In the corresponding quantum mechanical operator formulation, the operators $\hat{Q}^i$ and $\hat{P}_j$ are in the Heisenberg picture.

  4. We next consider the interaction picture. Here velocity and momentum are related via $$\dot{q}^i~=~\frac{\partial H_0(q,p)}{\partial p_i}~=~G^{ij}p_j,\tag{5}$$ which should be compared with the corresponding relation (2) in the Heisenberg picture. Eq. (5) has two consequences.

    Firstly, we derive the somewhat surprising relation $$\begin{align} L_{\rm int}(q,\dot{q})+H_{\rm int}(q,p) ~\stackrel{(5)}{=}~&L_{\rm int}(q,\dot{q})+H_{\rm int}(q,G\dot{q})\cr ~\stackrel{(1)+(3)}{=}&+\color{red}{\frac{1}{2} A^2(q)},\end{align}\tag{6}$$ which has the opposite sign of eq. (4)! This sign of eq. (6) will be important in what follows.

    Secondly, eq. (5) implies the equal-time CCR $$\begin{align} [\hat{q}^i(t),\dot{\hat{q}}^j(t)] ~\stackrel{(5)}{=}~&i\hbar~ G^{ij} {\bf 1},\cr [\hat{q}^i(t),\hat{q}^j(t)]~=~&0. \end{align} \tag{7} $$ We derive that the covariant time-ordering is $$\begin{align} T_{\rm cov}\{\hat{q}^i(t_1)\hat{q}^j(t_2)\} ~\equiv~& T\{\hat{q}^i(t_1)\hat{q}^j(t_2)\},\cr T_{\rm cov}\{\hat{q}^i(t_1)\dot{\hat{q}}^j(t_2)\} ~\equiv~&\frac{d}{dt_2} T\{\hat{q}^i(t_1)\hat{q}^j(t_2)\}\cr ~=~&\frac{d}{dt_2}\left\{\theta(t_1\!-\!t_2)\hat{q}^i(t_1)\hat{q}^j(t_2)+ \theta(t_2\!-\!t_1)\hat{q}^j(t_2)\hat{q}^i(t_1)\right\}\cr ~=~&T \{\hat{q}^i(t_1)\dot{\hat{q}}^j(t_2)\} -\delta(t_1\!-\!t_2)[\hat{q}^i(t_1),\hat{q}^j(t_2)]\cr ~\stackrel{(7)}{=}~&T \{\hat{q}^i(t_1)\dot{\hat{q}}^j(t_2)\},\cr T_{\rm cov} \{\dot{\hat{q}}^i(t_1)\dot{\hat{q}}^j(t_2)\} ~\equiv~&\frac{d}{dt_1}\frac{d}{dt_2} T \{\hat{q}^i(t_1)\hat{q}^j(t_2)\}\cr ~=~&\frac{d}{dt_1}T \{\hat{q}^i(t_1)\dot{\hat{q}}^j(t_2)\}\cr ~=~&\frac{d}{dt_1}\left\{\theta(t_1\!-\!t_2)\hat{q}^i(t_1)\dot{\hat{q}}^j(t_2)+ \theta(t_2\!-\!t_1)\dot{\hat{q}}^j(t_2)\hat{q}^i(t_1) \right\} \cr ~\stackrel{(7)}{=}~&T \{\dot{\hat{q}}^i(t_1)\dot{\hat{q}}^j(t_2)\} ~+~\color{red}{i\hbar~ G^{ij} {\bf 1} \delta(t_1\!-\!t_2)}. \end{align}\tag{8}$$ We have marked the non-covariant term in red.

  5. Consider next a Wilson line $$ \exp\left\{ \frac{i}{\hbar}\int\!dt~ A_i(q)\dot{q}^i \right\}. \tag{9}$$ From Wick's theorem, eq. (8) exponentiates to $$\begin{align} T_{\rm cov}\exp & \left\{ \frac{i}{\hbar}\int\!dt~ A_i(\hat{q})\dot{\hat{q}}^i \right\}\cr ~\stackrel{(8)}{=}~&\exp\left\{ \color{red}{ \frac{i\hbar}{2} \iint dt_1 ~dt_2~G^{ij} \frac{\delta}{\delta \dot{\hat{q}}^i(t_1)} \frac{\delta}{\delta \dot{\hat{q}}^j(t_2)} }\right\}\cr &\qquad T\exp\left\{ \frac{i}{\hbar}\int\!dt~ A_i(\hat{q})\dot{\hat{q}}^i \right\}\cr ~=~&T\exp\left\{ \frac{i}{\hbar}\int\!dt\left( A_i(\hat{q})\dot{\hat{q}}^i -\color{red}{\frac{1}{2} A^2(\hat{q})} \right)\right\} .\end{align} \tag{10}$$

  6. The standard derivation$^2$ of the Hamiltonian phase space path integral/partition function from the operator formalism in the Heisenberg picture goes like this $$\begin{align} Z_H~\sim~~~&{}_H\langle Q_f,t_f |Q_i,t_i\rangle_H \cr ~=~~~&{}_H\langle Q_f,0| T_{\rm cov} \exp\left\{ - \frac{i}{\hbar}\int_{t_i}^{t_f}\!dt~ H(\hat{Q},\hat{P})\right\} |Q_i,0\rangle_H \cr ~=~~~&{}_H\langle Q_f,0| T \exp\left\{ - \frac{i}{\hbar}\int_{t_i}^{t_f}\!dt~ H(\hat{Q},\hat{P})\right\} |Q_i,0\rangle_H \cr ~=~~~&\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~{\cal D}P~\exp\left\{ \frac{i}{\hbar}S_H[Q,P]\right\}\cr \stackrel{\text{Gauss. int.}}{\sim}&\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~ {\rm Det}(G_{ij})^{1/2} \exp\left\{ \frac{i}{\hbar} S[Q]\right\} . \end{align} \tag{11}$$

  7. In the interaction picture we have$^3$ $$\begin{align} Z_H~\sim~& {}_H\langle q_f,0| T_{\rm cov} \exp\left\{ - \frac{i}{\hbar}\int\!dt~ H_{\rm int}(\hat{q},\hat{p})\right\}|q_i,0\rangle_H \cr ~=~&{}_H\langle q_f,0| T \exp\left\{ - \frac{i}{\hbar}\int\!dt~ H_{\rm int}(\hat{q},\hat{p})\right\}|q_i,0\rangle_H \cr ~\stackrel{(5)}{=}~&{}_H\langle q_f,0| T \exp\left\{ - \frac{i}{\hbar}\int\!dt~ H_{\rm int}(\hat{q},G\dot{\hat{q}})\right\}|q_i,0\rangle_H \cr ~\stackrel{(6)}{=}~&{}_H\langle q_f,0| T \exp\left\{ \frac{i}{\hbar}\int\!dt\left( L_{\rm int}(\hat{q},\dot{\hat{q}})-\color{red}{\frac{1}{2} A^2(\hat{q})} \right)\right\}|q_i,0\rangle_H \cr ~\stackrel{(10)}{=}~&{}_H\langle q_f,0| T_{\rm cov} \exp\left\{ \frac{i}{\hbar}\int\!dt~ L_{\rm int}(\hat{q},\dot{\hat{q}})\right\}|q_i,0\rangle_H .\end{align}\tag{12}$$ We find that two effects cancel each other, the non-covariant term in the interaction Hamiltonian (6) and the Wick's theorem (10), so that the partition function (12) is Lorentz covariant. This is the main answer to OP's question.

  8. We suggest for completeness an interaction picture phase space path integral $$\begin{align} Z_H~\stackrel{(12)}{\sim}~~~& \int_{q(t_i)=q_i}^{q(t_f)=q_f}\! {\cal D}q~{\cal D}p~\cr &\exp\left\{ \frac{i}{\hbar} S_{H,0}[q,p] +\frac{i}{\hbar}\int_{t_i}^{t_f}\!dt \left(-\frac{1}{2}\dot{q}^2 + L_{\rm int}(\hat{q},\dot{\hat{q}}) \right)\right\}\cr ~\stackrel{\text{Gauss. int.}}{\sim}& \int_{q(t_i)=q_i}^{q(t_f)=q_f}\! {\cal D}q~ {\rm Det}(G_{ij})^{1/2} \exp\left\{ \frac{i}{\hbar} \int_{t_i}^{t_f}\!dt~ L_{\rm int}(\hat{q},\dot{\hat{q}}) \right\}. \end{align}\tag{13}$$

  9. The naive Lagrangian path integral $$Z_L~\sim~\int_{Q(t_i)=Q_i}^{Q(t_f)=Q_f}\! {\cal D}Q~ \exp\left\{ \frac{i}{\hbar} S[Q]\right\} \tag{14}$$ may differ from the Hamiltonian phase space path integral (11) because it lacks the determinant from the Gaussian integration over momenta $P_j$. In practice, it is often implicitly implied that the path integral measure ${\cal D}Q$ in eq. (14) contains this determinant factor by definition. In other words, the definition of $Z_L$ is tweaked to agree with $Z_H$. See also this related Phys.SE post.


  1. M.D. Schwartz, QFT and the Standard Model, 2014; Section 9.2.

  2. C. Itzykson & J.B. Zuber, QFT, 1985; Subsection 6-1-4.

  3. S. Weinberg, Quantum Theory of Fields, Vol. 1, 1995; Sections 7.2, 7.5 & 9.3.

  4. M. Srednicki, QFT, 2007; Chapter 6. A prepublication draft PDF file is available here.


$^1$ Notation: We will suppress spatial (but not temporal) dimensions by using DeWitt condensed notation. Capital letters for fields in the Heisenberg picture and small letters for fields in the interaction picture. The metric $G_{ij}(Q)$ in configuration space should not be confused with the space(time) metric.

$^2$ There is a usual story on how to wash out instantaneous eigenstates and replace them with a vacuum state, which we will not repeat here, see e.g. Ref. 4.

$^3$ It should be stressed that the derivation here is formal & shamelessly focused on the non-covariant term marked in red. We have ignored various higher-order operator-ordering issues, cf. e.g. this & this Phys.SE posts.

  • $\begingroup$ Additional references: 5. L.H. Ryder, QFT, 2nd eds., 1996; Section 5.5. $\endgroup$
    – Qmechanic
    Jan 22, 2022 at 12:40

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