Exact Results in Quantum Field Theory

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 4; 16 worked parts.

Question prompts paraphrased from the official Perimeter PSI Online sheet; the equations and notation follow the source.

1 The Källén--Lehmann Representation

For a free scalar field, recall the Feynman propagator \[D_0^F(x-y) \equiv \langle 0|T[\phi(x)\phi(y)]|0\rangle =-i\int\frac{d^4k}{(2\pi)^4}\, \frac{e^{ik\cdot(x-y)}}{k^2+m_0^2}.\tag{1.1}\] It describes propagation of a particle of mass $m_0$. Its pole at $k^2=-m_0^2$ is understood with the usual $-i\epsilon$ prescription. In an interacting theory, replace the free vacuum by the interacting vacuum and define \[D^F(x-y)=\langle\Omega|T[\phi(x)\phi(y)]|\Omega\rangle.\tag{1.2}\] The lowest pole of this propagator determines the physical mass. To obtain its nonperturbative spectral representation, insert a complete set of interacting states between the two fields.

Assume Poincaré invariance, with four-momentum operator $P^\mu=(H,\mathbf P)$ satisfying \[[P^\mu,P^\nu]=0.\tag{1.3}\] Determine the action of translations on the field and the interacting vacuum: \[e^{-iP\cdot x}\phi(0)e^{iP\cdot x}=? \qquad e^{-iP\cdot x}|\Omega\rangle=?\tag{1.4}\]

The vacuum should be invariant under Poincaré transformation, namely $e^{-iP\cdot x}\ket{\Omega} = \ket{\Omega}$. It can also be shown that \[e^{-iP\cdot x}\phi(0)e^{iP\cdot x}= \phi(x).\tag{1.5}\]

Take the states to be simultaneous energy--momentum eigenstates. Denote a one-particle state of physical mass $m$ by $|\mathbf k\rangle$, with $E_{\mathbf k}=\sqrt{\mathbf k^2+m^2}$, and use $|\mathbf k,n\rangle$ for states containing two or more particles, where $n$ collects the remaining quantum numbers.
Write the spectral decomposition of the identity, including the vacuum, one-particle states, and multiparticle states.

\[1 = \ket{\Omega}\bra{\Omega} + \int\frac{d^3k}{(2\pi)^32E_\mathbf{k}}\ket{\mathbf{k}}\bra{\mathbf{k}} + \sum_n\int\frac{d^3k}{(2\pi)^32E_\mathbf{k}}\ket{\mathbf{k},n}\bra{\mathbf{k},n}.\tag{1.6}\]
For $x^0>y^0$, insert that identity on the right-hand side of the interacting propagator and show that \[\begin{aligned}\langle\Omega|\phi(x)\phi(y)|\Omega\rangle &={} \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2 \notag\\ &\quad+ \sum_n\int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k,n|\phi(0)|\Omega\rangle\right|^2.\end{aligned}\tag{1.7}\] Assume that $\phi(x)|\Omega\rangle$ has a one-particle contribution.

\[\begin{aligned} \langle\Omega|\phi(x)\phi(y)|\Omega\rangle &= \langle\Omega|\phi(0)e^{iP\cdot (x-y)}\phi(0)|\Omega\rangle\\ &=\langle\Omega|\phi(0)e^{iP\cdot (x-y)}\bigg(\ket{\Omega}\bra{\Omega} + \int\frac{d^3k}{(2\pi)^32E_\mathbf{k}}\ket{\mathbf{k}}\bra{\mathbf{k}} \\ &\qquad+ \sum_n\int\frac{d^3k}{(2\pi)^32E_\mathbf{k}}\ket{\mathbf{k},n}\bra{\mathbf{k},n}\bigg)\phi(0)|\Omega\rangle\\ &=\int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2 + \sum_n\int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k,n|\phi(0)|\Omega\rangle\right|^2. \end{aligned}\tag{1.8}\]
Introduce the spectral function \[\rho(s) \equiv \delta(s-m^2) \left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2 +\sum_n \left|\langle\mathbf k,n|\phi(0)|\Omega\rangle\right|^2 \delta(s-M_n^2), \qquad M_n^2=E_{\mathbf k}^2-\mathbf k^2.\tag{1.9}\] Assuming there are no bound states, show that $\rho(s)=0$ for $s<4m^2$ except at $s=m^2$, whereas $\rho(s)$ is nonzero for $s\geq4m^2$. Use this result to establish, for $x^0>y^0$, \[\begin{aligned}\langle\Omega|\phi(x)\phi(y)|\Omega\rangle &={} \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2 \notag\\ &\quad+ \int_{4m^2}^{\infty}ds\,\rho(s) \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}},\end{aligned}\tag{1.10}\] and, for $x^0<y^0$, \[\begin{aligned}\langle\Omega|\phi(y)\phi(x)|\Omega\rangle &={} \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{-ik\cdot(x-y)}}{2E_{\mathbf k}} \left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2 \notag\\ &\quad+ \int_{4m^2}^{\infty}ds\,\rho(s) \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{-ik\cdot(x-y)}}{2E_{\mathbf k}}.\end{aligned}\tag{1.11}\]

If there are no bound states, then it is obvious that there are no additional poles for $s<4m^2$. The result follows trivially.
Use the QFT I identity \[\begin{aligned}\int\frac{d^4k}{(2\pi)^4}\, \frac{e^{ik\cdot(x-y)}}{k^2+m^2-i\epsilon} &={} i\Theta(x^0-y^0) \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{ik\cdot(x-y)}}{2E_{\mathbf k}} \notag\\ &\quad+ i\Theta(y^0-x^0) \int\frac{d^3\mathbf k}{(2\pi)^3}\, \frac{e^{-ik\cdot(x-y)}}{2E_{\mathbf k}} \end{aligned}\tag{1.12}\] to derive the Källén--Lehmann representation \[\widetilde D_F(k^2) =\frac{-i\left|\langle\mathbf k|\phi(0)|\Omega\rangle\right|^2} {k^2+m^2-i\epsilon} +\int_{4m^2}^{\infty}ds\,\rho(s) \frac{-i}{k^2+s-i\epsilon},\tag{1.13}\] where $\widetilde D_F$ is the Fourier transform of $D_F(x)$.

This follows trivially from repeating the same contour integral one used to obtain (1.12).
Regard $\widetilde D_F(k^2)$ as an analytic function of $k^2$. Locate its lowest pole and explain the physical meaning of both the pole and its residue.

The lowest pole is at $k^2 = -m^2$, which is the physical mass of the one-particle state, and the residue is the field renormalisation factor $Z$.
The continuum begins at $s=4m^2$ and produces a branch cut. If the theory also contains bound states, which determine where their contributions would appear in $\rho(s)$.

The bound states would still be single poles,m they would appear as $\delta$-functions in the spectral function.

2 Schwinger--Dyson Equations

The Schwinger--Dyson equations state that correlation functions obey the classical equations of motion up to local contact terms.

In the free Euclidean Klein--Gordon theory, use invariance of the path integral under $\phi(x)\mapsto\phi(x)+\epsilon(x)$. Starting from \[\langle\phi(x)\rangle =\left. \frac{1}{\mathcal Z[0]} \frac{\delta\mathcal Z[J]}{\delta J(x)} \right|_{J=0},\tag{2.1}\] show that \[(-\partial_z^2+m^2) \langle\phi(z)\phi(x)\rangle =\hbar\delta(z-x).\tag{2.2}\]

Consider the integral under an infinitesimal field shifting $\phi \to \phi +\epsilon$ \[\int\mathcal{D}\phi F[\phi]e^{-S[\phi]/\hbar} = \int \mathcal{D}\phi F[\phi+\epsilon]e^{-S[\phi+\epsilon]/\hbar}.\tag{2.3}\] Then, we have \[\begin{aligned} 0 &=\int \mathcal{D}\phi\left[\left(F[\phi] + \frac{\delta F}{\delta\phi}\epsilon\right)\left(1- \frac{1}{\hbar}\frac{\delta S}{\delta\phi}\epsilon\right) - F[\phi]\right]e^{-S[\phi]/\hbar}\\ &=\left\langle\frac{\delta F[\phi]}{\delta\phi }\right\rangle\epsilon -\frac{1}{\hbar} \left\langle F[\phi]\frac{\delta S[\phi]}{\delta \phi}\right\rangle\epsilon \end{aligned}\tag{2.4}\] for all $\epsilon$. Therefore, we find \[\hbar \left\langle\frac{\delta F[\phi]}{\delta\phi }\right\rangle = \left\langle F[\phi]\frac{\delta S[\phi]}{\delta \phi}\right\rangle.\tag{2.5}\] For our case, we take \[S_0[\phi] = \int d^dz \frac{1}{2}\phi(z)(-\partial_z^2 +m^2)\phi(z), \quad F[\phi] = \phi(x),\tag{2.6}\] which gives \[(-\partial_z^2+m^2) \langle\phi(z)\phi(x)\rangle =\hbar\delta(z-x).\tag{2.7}\]
Add the interaction \[S[\phi] =S_0[\phi]+\frac{\lambda}{n!}\int d^4y\,\phi^n(y),\tag{2.8}\] and show that \[(-\partial_z^2+m^2) \langle\phi(z)\phi(x)\rangle =- \left\langle \frac{\lambda}{(n-1)!}\phi^{n-1}(z)\phi(x) \right\rangle +\hbar\delta(z-x).\tag{2.9}\]

Now, we have \[\frac{\delta S[\phi]}{\delta\phi} = (-\partial_z^2+m^2)\phi(z) + \frac{\lambda}{(n-1)!}\phi^{n-1}(z),\tag{2.10}\] and hence, using the master equation, we find \[(-\partial_z^2+m^2) \langle\phi(z)\phi(x)\rangle =- \left\langle \frac{\lambda}{(n-1)!}\phi^{n-1}(z)\phi(x) \right\rangle +\hbar\delta(z-x).\tag{2.11}\]
Use invariance of the generating functional under $\phi\mapsto\phi+\delta\phi$ to prove the general identity \[\left\langle \frac{\delta S}{\delta\phi(x)} \prod_{r=1}^{n}\phi(x_r) \right\rangle =\hbar\sum_{j=1}^{n} \delta^4(x-x_j) \left\langle \prod_{\substack{r=1\\r\neq j}}^{n}\phi(x_r) \right\rangle.\tag{2.12}\] What changes in real time?

This follows directly from substituting \[F[\phi] = \prod_{r=1}^n\phi(x_r).\tag{2.13}\] In real time, the expression becomes \[\left\langle \frac{\delta S}{\delta\phi(x)} \prod_{r=1}^{n}\phi(x_r) \right\rangle =i\hbar\sum_{j=1}^{n} \delta^4(x-x_j) \left\langle \prod_{\substack{r=1\\r\neq j}}^{n}\phi(x_r) \right\rangle.\tag{2.14}\]
Show that \[\langle\phi_1\cdots\phi_n\rangle =\int d^4x\, \Delta_{x1}(-\partial_x^2+m^2) \langle\phi_x\phi_2\cdots\phi_n\rangle,\tag{2.15}\] where $\phi_i=\phi(x_i)$ and $\Delta_{ij}=\Delta_F(x_i-x_j)$.

Using $(-\partial_x^2+m^2)\Delta_{1x} = \delta_{1x}$, we obtain the above equation.
In the free Euclidean Klein--Gordon theory, combine the preceding result with the Schwinger--Dyson equations to obtain \[\langle\phi_1\cdots\phi_4\rangle_0 =\Delta_{12}\Delta_{34} +\Delta_{13}\Delta_{24} +\Delta_{14}\Delta_{23}.\tag{2.16}\]

\[\begin{aligned} \langle\phi_1\phi_2\phi_3\phi_4\rangle_0 &= \int d^dx \Delta_{x1}(-\partial_x^2 +m^2)\langle\phi_x\phi_2\phi_3\phi_4\rangle_0\\ &=\int d^dx \Delta_{x1}(\delta_{x2}\Delta_{34} + \delta_{x3}\Delta_{24} + \delta_{x4}\Delta_{23})\\ &=\Delta_{12}\Delta_{34} +\Delta_{13}\Delta_{24} +\Delta_{14}\Delta_{23}, \end{aligned}\tag{2.17}\] where we will take $\hbar =1$ from now on.
In Euclidean $\phi^3$ theory, apply the same identities recursively and show that \[\begin{aligned}\langle\phi_1\phi_2\rangle &={} \Delta_{12} +(-g)^2\int d^4x\,d^4y\, \left( \frac{1}{2}\Delta_{x1}\Delta_{xy}^2\Delta_{y2} +\frac{1}{4}\Delta_{x1}\Delta_{xx}\Delta_{yy}\Delta_{y2} \right) +O(g^4).\end{aligned}\tag{2.18}\] Does this agree with the result obtained in Tutorial 2?

\[\begin{aligned} \langle\phi_1\phi_2\rangle &= \int d^dx \Delta_{1x}(-\partial_x^2 + m^2)\langle\phi_x\phi_2\rangle\\ &= \int d^dx \Delta_{1x}\left(-\frac{g}{2}\langle\phi_x^2\phi_2\rangle + \delta _{x2}\right)\\ &= \Delta_{12} - \frac{g}{2}\int d^dx d^dy \Delta_{1x}\Delta_{2y}(-\partial_y^2 + m^2)\langle\phi_x^2\phi_y\rangle\\ &= \Delta_{12} - \frac{g}{2}\int d^dx d^dy \Delta_{1x}\Delta_{2y}\left(-\frac{g}{2}\langle\phi^2_y\phi_x^2\rangle + 2\delta_{xy}\langle\phi_x\rangle\right). \end{aligned}\tag{2.19}\] Since we are only interested to order $g^2$, we have \[\langle\phi^2_y\phi_x^2\rangle = \langle\phi^2_y\phi_x^2\rangle_0 + \mathcal{O}(g^2) = 2\Delta_{xy}^2 + \Delta_{xx}\Delta_{yy}+ \mathcal{O}(g^2).\tag{2.20}\] Meanwhile, the tadpole diagram is given by \[\begin{aligned} \langle\phi_x\rangle &= \int d^dz \Delta_{xz}(-\partial_z^2+m^2)\langle\phi_z\rangle\\ &=-\frac{g}{2}\int d^dz \Delta_{xz}\langle\phi_z^2\rangle\\ &=-\frac{g}{2}\int d^dz \Delta_{xz}\Delta_{zz} + \mathcal{O}(g^3). \end{aligned}\tag{2.21}\] Then, we have \[\begin{aligned} \langle\phi_1\phi_2\rangle &= \Delta_{12} +(-g)^2\int d^dx d^dy \Delta_{1x}\Delta_{2y}\left(\frac{1}{2}\Delta_{xy}^2 + \frac14\Delta_{xx}\Delta_{yy}\right)\\ &\qquad+ (-g)^2\int d^dx d^dy\,\frac{1}{2}\Delta_{1x}\Delta_{2x}\Delta_{xy}\Delta_{yy} + \mathcal{O}(g^4). \end{aligned}\tag{2.22}\]
Optional. For a continuous symmetry of the Lagrangian, prove the Ward--Takahashi identity \[\begin{aligned}\partial_\mu \langle j^\mu(x)\phi(x_1)\cdots\phi(x_n)\rangle ={}& \hbar\sum_{j=1}^{n} \bigl\langle \phi(x_1)\cdots\phi(x_{j-1}) \notag\\[-0.4em] &\qquad\times \delta\phi(x)\delta^4(x-x_j) \phi(x_{j+1})\cdots\phi(x_n) \bigr\rangle,\end{aligned}\tag{2.23}\] where, in the notation of the sheet, \[j^\mu(x) =\frac{\delta S}{\delta\phi_{,\mu}(x)}\delta\phi(x) -\delta\mathcal L(x), \qquad \phi_{,\mu}\equiv\partial_\mu\phi.\tag{2.24}\] What changes in real time?

The classical action has some global symmetry, $\phi(x) \to \phi(x) + \epsilon(x) \delta\phi(x)$, which corresponds to a Noether current, $j_\mu(x)$. Now, we promote the symmetry to be a local symmetry $\phi(x) \to \phi(x) + \epsilon(x)\delta\phi(x)$. Again, we consider the path integral \[\int \mathcal{D}\phi F[\phi] e^{-S[\phi]/\hbar} = \int \mathcal{D}\phi (F[\phi]+\delta F[\phi])\left(1 - \frac{1}{\hbar}\delta S[\phi]\right)e^{-S[\phi]/\hbar},\tag{2.25}\] where $F[\phi] \to F[\phi] + \delta F[\phi]$ under the symmetry transformation. Note that $\delta S, \delta F$ are of order $\epsilon$, and for an infinitesimal transformation, we can ignore terms of order $\epsilon^2$. Then, we have \[\hbar \langle\delta F[\phi]\rangle = \langle\delta S[\phi]F[\phi]\rangle.\tag{2.26}\] Recall \[\begin{aligned} \delta S &= \int d^d x\, \left[ \frac{\partial\mathcal L}{\partial\phi(x)} \epsilon(x)\delta\phi(x) + \frac{\partial\mathcal L} {\partial(\partial_\mu\phi(x))} \partial_\mu\!\left( \epsilon(x)\delta\phi(x) \right) \right]\\ &= \int d^d x\,\epsilon(x) \left[ \frac{\partial\mathcal L}{\partial\phi(x)} \delta\phi(x) + \frac{\partial\mathcal L} {\partial(\partial_\mu\phi(x))} \partial_\mu\delta\phi(x) \right] + \int d^d x\, \frac{\partial\mathcal L} {\partial(\partial_\mu\phi(x))} \delta\phi(x)\partial_\mu\epsilon(x)\\ &= \int d^d x\, \left[ \epsilon(x)\partial_\mu \delta\mathcal L^\mu(x) + \frac{\partial\mathcal L} {\partial(\partial_\mu\phi(x))} \delta\phi(x)\partial_\mu\epsilon(x) \right]\\ &= \int d^d x\, \partial_\mu\!\left( \epsilon(x)\delta\mathcal L^\mu(x) \right) + \int d^d x\, \left[ \frac{\partial\mathcal L} {\partial(\partial_\mu\phi(x))} \delta\phi(x) - \delta\mathcal L^\mu(x) \right] \partial_\mu\epsilon(x)\\ &= \int d^d x\, j^\mu(x)\partial_\mu\epsilon(x)\\ &= -\int d^d x\, \epsilon(x)\partial_\mu j^\mu(x). \end{aligned}\tag{2.27}\] and take \[\begin{aligned} F[\phi] = \phi(x_1) \cdots \phi(x_n) \implies \delta F[\phi] &=\sum_{j=1}^{n} \phi(x_1)\cdots\phi(x_{j-1}) \epsilon(x_j)\delta\phi(x_j) \phi(x_{j+1})\cdots\phi(x_n)\\ &=\sum_{j=1}^{n}\int d^dx \delta^d(x-x_j) \phi(x_1)\cdots\phi(x_{j-1}) \epsilon(x)\delta\phi(x) \phi(x_{j+1})\cdots\phi(x_n). \end{aligned}\tag{2.28}\]

Hence, we find the Ward-Takahashi identity \[\begin{aligned}\partial_\mu \langle j^\mu(x)\phi(x_1)\cdots\phi(x_n)\rangle ={}& -\hbar\sum_{j=1}^{n} \bigl\langle \phi(x_1)\cdots\phi(x_{j-1}) \delta\phi(x)\delta^4(x-x_j) \phi(x_{j+1})\cdots\phi(x_n) \bigr\rangle.\end{aligned}\tag{2.29}\] In real time, this becomes \[\begin{aligned}\partial_\mu \langle j^\mu(x)\phi(x_1)\cdots\phi(x_n)\rangle ={}& -i\hbar\sum_{j=1}^{n} \bigl\langle \phi(x_1)\cdots\phi(x_{j-1}) \delta\phi(x)\delta^4(x-x_j) \phi(x_{j+1})\cdots\phi(x_n) \bigr\rangle.\end{aligned}\tag{2.30}\]
More optional. Show that contact terms on the right-hand side of the Schwinger--Dyson and Ward--Takahashi identities do not contribute to $i\mathcal M$. In Euclidean momentum conventions, use \[\langle f|S|i\rangle =\lim_{k_i^2\to-m^2} (k_i^2+m^2)\cdots \langle\Omega|T\widetilde\phi(k_1)\cdots|\Omega\rangle,\tag{2.31}\] with \[\widetilde\phi(k) =i\int d^4x\,e^{-ik\cdot x}\phi(x).\tag{2.32}\] The scattering amplitude and momentum-space correlator contain the same momentum-conserving delta function: \[\langle f|S|i\rangle =(2\pi)^4\delta^4\!\left(\sum k\right)i\mathcal M,\tag{2.33}\] \[\langle\Omega|T\widetilde\phi(k_1)\cdots|\Omega\rangle =(2\pi)^4\delta^4\!\left(\sum k\right)\mathcal F(k_i).\tag{2.34}\] Near the simultaneous poles, \[\mathcal F(k_i) =\frac{i\mathcal M} {(k_1^2+m^2)\cdots(k_n^2+m^2)} +\text{nonsingular terms}.\tag{2.35}\] Argue that the Fourier transform of a contact term lacks the required multivariable pole and hence cannot contribute to $i\mathcal M$.

Consider a contact term \[\delta^d(x_1-x_2)\langle\phi(x_3) \cdots \phi(x_n)\rangle,\tag{2.36}\] which is Fourier transformed to \[(2\pi)^d\delta^d(k_1 +k_2)\langle\tilde \phi(k_3) \cdots \tilde \phi(k_n)\rangle.\tag{2.37}\] Its contribution to the LSZ is \[\lim_{k_i^2\to-m^2}(k_1^2+m^2)\cdots(k_n^2+m^2)(2\pi)^d\delta^d(k_1 +k_2)\langle\tilde \phi(k_3) \cdots \tilde \phi(k_n)\rangle.\tag{2.38}\] Since it contains no poles in $k_1,k_2$, it has no contribution to the LSZ reduction formula.
More optional. In Lorenz gauge, use the photon LSZ formula \[\langle f|S|i\rangle =\epsilon^\mu\int d^4x\,e^{-ik\cdot x} (-\partial^2)\cdots \langle\Omega|T A_\mu(x)\cdots|\Omega\rangle\tag{2.39}\] together with the Ward--Takahashi identity to derive \[k_\mu\mathcal M^\mu=0, \qquad \mathcal M=\epsilon_\mu\mathcal M^\mu.\tag{2.40}\]

In QED, we have the equation of motion for the gauge fields in the Lorenz gauge \[-\partial^2A_\mu = j_\mu = e\bar \psi \gamma_\mu \psi.\tag{2.41}\] The current corresponds to the global $U(1)$ transformation. The LSZ becomes \[\langle f|S|i\rangle=\epsilon^\mu\int d^dx\,e^{-ik\cdot x} \langle\Omega|T j_\mu(x)\cdots|\Omega\rangle = (2\pi)^d\delta^d\left(\sum k\right)\epsilon_\mu \mathcal{M}_\mu.\tag{2.42}\] Now replacing $\epsilon^\mu \to k^\mu$, we have \[\begin{aligned} (2\pi)^d\delta^d\left(\sum k\right)k_\mu \mathcal{M}_\mu &= i\int d^dx (\partial^\mu e^{-ik\cdot x}) \langle\Omega|T j_\mu(x)\cdots|\Omega\rangle\\ &= - i\int d^dx e^{-ik\cdot x}\partial^\mu\langle\Omega|T j_\mu(x)\cdots|\Omega\rangle. \end{aligned}\tag{2.43}\] Now, since we have shown that the contact terms do not contribute to $S$-matrix, and the LHS of the Ward-Takahashi identity is purely contact terms. Therefore, we can conclude that $k_\mu \mathcal{M}^\mu = 0$.