The Klein Gordon Field

2.1 The Necessity of the Field Viewpoint

13P4E5

Consider the amplitude for a free particle to propagate from $\vec{x_{0}}$ to $\vec{x}$: $$ \begin{align*} U(t) = \braket{\vec{x} \vert e^{-i H t} \vert \vec{x}_{0}} \end{align*} $$ In the nonrelativisitic quantum mechanics, we have $E = \frac{\vec{p}^{2}}{2m} \equiv \frac{\vec{p}\cdot \vec{p}}{2m}$ (we will replace Hamiltonian operator by its eigen value), so

$$ \begin{aligned} U(t) &= \braket{\vec{x} \vert e^{-i ( \frac{\vec{p}^{2}}{2m}) t} \vert \vec{x}_{0}} \\ &= \text{Using \href{https://en.wikipedia.org/wiki/Orthonormal_basis}{completeness relation}, } \int \frac{d^{3} \vec{p}}{(2\pi)^{3}} \ket{\vec{p}}\bra{\vec{p}} = \mathbb{\hat{1}} \\ &= \int \frac{d^{3} \vec{p}}{(2\pi)^{3}} \braket{\vec{x} \vert e^{-i ( \frac{\vec{p}^{2}}{2m}) t} \ket{\vec{p}}\bra{\vec{p}}\vert \vec{x}_{0}} \\ &= \frac{1}{(2\pi)^{3}} \int d^{3}\vec{p} ~ e^{-i ( \frac{\vec{p}^{2}}{2m}) t} \braket{\vec{x} \vert \vec{p}} \braket{\vec{x}_{0} \vert \vec{p}}^{\dagger} \\ &\hspace{0.5cm} \text{where, } e^{-i ( \frac{\vec{p}^{2}}{2m}) t} \text{ is just a number.} \\ &= \text{Using } \braket{\vec{x} \vert \vec{p}} = e^{i \vec{p}\cdot \vec{x}}.\\ &= \frac{1}{(2\pi)^{3}} \int d^{3}\vec{p} ~ e^{-i ( \frac{\vec{p}^{2}}{2m}) t} e^{i \vec{p}\cdot (\vec{x} - \vec{x}_{0})}. \\ &= \text{Using Gaussian integral property}, \int_{-\infty}^{\infty} e^{-a x^{2} + bx + c} ~dx = \sqrt{\frac{\pi}{a}} e^{\frac{b^{2}}{4a} + c}. \\ &= \frac{1}{(2\pi)^{3}} \left(\sqrt{\frac{\pi}{\frac{i t}{2m}}}\right)^{3} ~ \exp\left(\frac{i^{2} (\vec{x} - \vec{x}_{0})^{2}}{4 (\frac{it}{2m})}\right) \\ &= (2\pi)^{3/2 - 3} \left(\frac{m}{i t} \right)^{3/2} \exp\left( \frac{i^{\cancel{2}} (\vec{x} - \vec{x}_{0})^{2}~2m}{4 \cancel{i} t}\right) \\ &= \left(\frac{m}{2\pi i t} \right)^{3/2} e^{i m (\vec{x} - \vec{x}_{0})^{2}/2t}. \end{aligned} $$

Some notes on completeness relations from a functional analysis point of view

Completeness relation allows us to define the inner product. Let’s understand it with an example using position space. It can be done in momentum space as well.

$\rightarrow$ For a single free particle moving in $n$-dimensional Euclidean (flat) manifold $\mathbb{R}^{n}$, we can have a complete orthonormal set of position basis vectors $\{\ket{x}\}$. So, we can define

$$ \begin{aligned} \braket{x \vert y} &= \delta(x, y), \\ \int_{\mathbb{R}^{n}} d^{n}x \ket{x}\bra{x} &= \mathbb{\hat{1}}. \end{aligned} $$

We can identify the Hilbert space $\mathcal{H}$ of such a system with the square-integrable complex-valued functions $L^{2}(\mathbb{R}^{n}, \mathbb{C})$ with the inner product

$$ \begin{aligned} \braket{\phi \vert \psi} &= \bra{\phi}\mathbb{\hat{1}}\ket{\psi} \\ &= \bra{\phi} \int_{\mathbb{R}^{n}} d^{n}x \ket{x}\bra{x} \ket{\psi} \\ \therefore \braket{\phi \vert \psi} &= \int_{\mathbb{R}^{n}} d^{n}x ~\phi_x \psi^{*}_{x} \end{aligned} $$

where $\phi_{x} \equiv \phi(x) := \braket{\phi \vert x}$ is the position basis coefficient, and same for $\psi_{x}$. Thus, $\psi^{*}_{x} = (\braket{\psi \vert x})^{*} = \braket{x \vert \psi}$. The unique existence of their spaces is guaranteed by the Riesz representation theorem.

I showed you how the completeness relation is used to define the inner product. It was possible because the position basis localized the state of a particle. But, this naive notion of position basis is not viable for fields (or differential forms) in Riemannian manifold $(\mathcal{M}, g)$ where the metric $g$ is not necessarily flat. It would be out of the scope of this note to further discuss the issue but, I will at least give you short reason.

$\rightarrow$ Suppose our $\phi$ is a $k$-forms in the space of differentials $k$-forms $\Omega^{k}(\mathcal{M})$. Due to the Hodge duality, we can find $(n - k)$-forms $* \phi \in \Omega^{n - k}(\mathcal{M})$ such that $\phi \leftrightarrow *\phi$ where

$$ \begin{aligned} \phi &= \phi_{i_1 \ldots i_k} ~dx^{i_1}\wedge \ldots \wedge dx^{i_k},\\ *\phi &= (*\phi)_{i_{k+1} \ldots i_n} ~ dx^{i_{k+1}}\wedge \ldots \wedge dx^{i_n} \end{aligned} $$

with coefficients

$$ \begin{aligned} \phi_{i_1 \ldots i_p} \leftrightarrow (*\phi)_{i_{k+1} \ldots i_n} = \frac{\sqrt{g}}{(n - k)!} \epsilon_{i_1 \ldots i_n}~ g^{i_1 j _1}\ldots g^{i_k j_k} \phi_{i_1 \ldots i_k}. \end{aligned} $$

The duality is up to a sign. i.e. $\ast\ast\phi = (-1)^{k (n - k)} \phi$. The inner product of $k$-forms is defined as

$$ \begin{aligned} (\phi, \psi) &= \int_{\mathcal{M}} {\color{red}\phi} \wedge * \psi \\ &= \text{Using above definitions, we get}\\ &\quad \int_{\mathcal{M}} {\color{red}\phi_{i_1 \ldots i_k} ~dx^{i_1}\wedge \ldots \wedge dx^{i_k}} \wedge (*\phi)_{i_{k+1} \ldots i_n} ~ dx^{i_{k+1}}\wedge \ldots \wedge dx^{i_n}\\ &= \int_{\mathcal{M}} \phi_{i_1 \ldots i_k} (*\phi)_{i_{k+1} \ldots i_n} ~dx^{i_1}\wedge \ldots \wedge dx^{i_k} \wedge dx^{i_{k+1}}\wedge \ldots \wedge dx^{i_n} \\ &= \int_{\mathcal{M}} \phi_{i_1 \ldots i_k} (*\phi)_{i_{k+1} \ldots i_n} ~dx^{i_1}\wedge \ldots \wedge dx^{i_n} \\ \end{aligned} $$

and has a tensorial structure. It’s not easy to define a localized field. So, it is an open question of how to define it. Fortunately, with the quantum field theory that we learn from this book, our background metric is flat. So, we don’t have to worry about this issue. But, in quantum gravity, it’s an important question!

13P4B2

This expression is nonzero for all $x$ and $t$, indicating that a particle can propagate between any two points in an arbitrarily short time. In a relativistic theory, this conclusion would signal a violation of causality.

In special relativity, any motion of a particle from $\vec{x}_0$ to $\vec{x}$ in a time $t$ faster than the speed of light violates causality (i.e. $t < \frac{||\vec{x} - \vec{x}_0||}{c}$ where $c$ is the speed of light). This is because there is another inertial frame of reference in which the particle arrives at $\vec{x}$ at a time earlier than the time at which it leaves $\vec{x}_0$. Since the propagator $U(t)$ is non-zero for $t < \frac{||\vec{x} - \vec{x}_0||}{c}$, this means that the particle can be in two places at the same time. This is a violation of causality and is not allowed in special relativity theory. We will come back to this point later on pages 28-29 in the book.

13P4E8

… . In analogy with the non-relativistic case, we have

$$ \begin{aligned} U(t) &= \braket{\vec{x} \vert e^{-i t \sqrt{\vec{p}^{2} + m^{2}}} \vert \vec{x}_{0}} \\ &= \text{Using completeness relation, } \int \frac{d^{3} \vec{p}}{(2\pi)^{3}} \ket{\vec{p}}\bra{\vec{p}} = \mathbb{\hat{1}} \\ &= \int \frac{d^{3} \vec{p}}{(2\pi)^{3}} \braket{\vec{x} \vert e^{-i t \sqrt{\vec{p}^{2} + m^{2}}} \ket{\vec{p}}\bra{\vec{p}}\vert \vec{x}_{0}} \\ &= \frac{1}{(2\pi)^{3}} \int d^{3}\vec{p} ~ e^{-i t \sqrt{\vec{p}^{2} + m^{2}}} \braket{\vec{x} \vert \vec{p}} \braket{\vec{x}_{0} \vert \vec{p}}^{\dagger} \\ &= \text{Using } \braket{\vec{x} \vert \vec{p}} = e^{i \vec{p}\cdot \vec{x}}.\\ &= \frac{1}{(2\pi)^{3}} \int d^{3}\vec{p} ~ e^{-i t \sqrt{\vec{p}^{2} + m^{2}}} e^{i \vec{p}\cdot (\vec{x} - \vec{x}_{0})}. \end{aligned} $$

Before we proceed, we do a transformation of momentum space in spherical (also called polar) coordinates $(p, \theta, \phi)$, and the volume element spanning from $\vec{p}$ to $\vec{p} + d\vec{p}$, $\theta$ to $\theta + d\theta$ and $\phi$ to $\phi + d\phi$ is given by the determinant of the Jacobian matrix of partial derivatives. i.e.

$$ \begin{aligned} dV := d^{3} \vec{p} &= \left| \frac{\partial (p_x, p_y, p_z)}{\partial (\vec{p}, \theta, \phi)} \right| d\vec{p} ~d\theta~ d\phi\\ &= p^{2} \sin(\theta) d\vec{p} ~d\theta~ d\phi\\ \therefore d^{3}p &= d\phi \sin(\theta) d\theta ~p^{2} d\vec{p} \end{aligned} $$

Now,

$$ \begin{aligned} U(t) &= \text{Using } \vec{p}.(\vec{x} - \vec{x}_{0}) = p | \vec{x} - \vec{x}_{0}| \cos(\theta), \\ &\hspace{0.5cm} \text{\& spherical coordinate in momentum space}\\ &= \frac{1}{(2\pi)^{3}}\int d\phi \sin(\theta) d\theta ~p^2 d\vec{p} ~ e^{-i t \sqrt{p^2 + m^2}} ~e^{i p | \vec{x} - \vec{x}_{0}| \cos(\theta)} \\ &= \frac{1}{(2\pi)^{3}} \left[ \int_{0}^{2\pi} d\phi \right] \int \sin(\theta) d\theta ~p^2 d\vec{p} ~ e^{-i t \sqrt{p^2 + m^2}} ~e^{i p | \vec{x} - \vec{x}_{0}| \cos(\theta)} \\ &= \frac{\cancel{2\pi}}{(2\pi)^{\cancel{3}}} \int d\vec{p} ~p^2 ~ e^{-i t \sqrt{p^2 + m^2}} \left[ \int_{0}^{\pi}\sin(\theta) d\theta ~e^{i p | \vec{x} - \vec{x}_{0}| \cos(\theta)} \right]\\ &= \text{Do change of variable: } \zeta = \cos(\theta) \implies -d\zeta = \sin(\theta) d\theta\\ &= \frac{1}{(2\pi)^{2}} \int d\vec{p} ~p^2 ~ e^{-i t \sqrt{p^2 + m^2}} \left[ (-1) \int_{1}^{-1} d\zeta~ e^{i p |\vec{x} - \vec{x}_{0}| \zeta} \right]\\ &= \text{Since, } \int dx ~ e^{a x} = \frac{e^{a x}}{a} + \text{constant}.\\ &= \frac{1}{(2\pi)^{2}} \int d\vec{p} ~p^{\cancel{2}} ~ e^{-i t \sqrt{p^2 + m^2}} (-1) \left[ \frac{e^{i p |\vec{x} - \vec{x}_{0}|\zeta}}{i \cancel{p} |\vec{x} - \vec{x}_{0}|} \right]_{-1}^{~~1}\\ &= \frac{1}{(2\pi)^{2} |\vec{x} - \vec{x}_{0}|} \int d\vec{p} ~p ~ e^{-i t \sqrt{p^2 + m^2}} \left( \frac{e^{i p |\vec{x} - \vec{x}_{0}|} - e^{-i p |\vec{x} - \vec{x}_{0}|}}{i}\right)\\ &= \text{Using } \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}.\\ &= \frac{1}{(2\pi)^{2} |\vec{x} - \vec{x}_{0}|} \int d\vec{p} ~p ~ e^{-i t \sqrt{p^2 + m^2}} (2) \sin(p |\vec{x} - \vec{x}_{0}|)\\ &= \frac{1}{2 \pi^{2} |\vec{x} - \vec{x}_{0}|} \int_{0}^{\infty} d\vec{p} ~p \sin(p |\vec{x} - \vec{x}_{0}|) ~ e^{-i t \sqrt{p^2 + m^2}}. \end{aligned} $$

13P4B3S1

This integral can be evaluated explicitly in terms of Bessel functions.

Using the integral formula given in “2007 - Gradshteyn, Rhyzik- Table of Integrals, Series and Products”, page 491, formula number 3.914 (6) to the final expression of 13P4E8: $$ \int_0^\infty x e^{-\beta \sqrt{\gamma^2 + x^2}} \sin(bx) dx = \frac{b \beta \gamma^2}{\beta^2 + b^2} K_{2}( \gamma \sqrt{\beta^2 + b^2}) $$ where $K_{2}$ is the modified Bessel function of the second kind.

Here, $x \implies p, \beta \implies i t, \gamma \implies m, b \implies |\vec{x} - \vec{x}_{0}|$, so we get

$$ \begin{aligned} U(t) &= \frac{1}{2 \pi^{2} \cancel{|\vec{x} - \vec{x}_{0}|}} \frac{\cancel{|\vec{x} - \vec{x}_{0}|} (i t) m^{2}}{(i t)^{2} + |\vec{x} - \vec{x}_{0}|^{2}} K_{2}(m \sqrt{(i t)^{2} + |\vec{x} - \vec{x}_{0}|^{2}}) \\ &= \frac{i t m^{2}}{2 \pi^{2} (|\vec{x} - \vec{x}_{0}|^{2} - t^{2})} K_{2}(m \sqrt{|\vec{x} - \vec{x}_{0}|^{2} - t^{2}}) \end{aligned} $$

13P4B3S2

… with looking at its asymptotic behavior for $x^2 \gg t^2$ (well outside the light-cone), using the method of stationary phase. The function $px - t \sqrt{p^{2} + m^{2}}$ has a stationary point at $p = imx/\sqrt{x^{2} - t^{2}}$.

Instead of doing the transformation of momentum space in polar coordinates (here), we use the idea of stationary phase approximation. For the given, $$ U(t) = \frac{1}{(2\pi)^3} \int d^{3}p ~ e^{i\left(\textcolor{green}{\vec{p}\cdot \vec{x} - t \sqrt{p^{2} + m^{2}}}\right)} ~e^{-i \vec{p}\cdot\vec{x}_{0}}, $$ the phase function of $U(t)$ is, $\phi(p) = \textcolor{green}{\vec{p}\cdot \vec{x} - t \sqrt{p^{2} + m^{2}}}$. The stationary point is given by $\frac{d\phi(p)}{dp} = 0$ so we proceed:

$$ \begin{aligned} & x - \frac{t}{\cancel{2}} (p^2 + m^2)^{\frac{1}{2} - 1} \cancel{2}p = 0 \\ \text{or, }& x - \frac{t p}{\sqrt{p^2 + m^2}} = 0 \\ \text{or, }& x\sqrt{p^2 + m^2} = tp \\ \text{or, }& x^{2} (p^2 + m^2) = t^2 p^2 \\ \text{or, }& p^{2}(x^2 - t^2) = -x^2 m^2 \\ \therefore ~&~ p = i m x / \sqrt{x^2 - t^2} \end{aligned} $$

The above result holds only when $x^2 \gg t^2$.

Noether Theorem

17E(2.11)-(2.12)

Consider infinitesimal form $\mathcal{L}(x) \to \mathcal{L’}(x) = \mathcal{L}(x) + \alpha \Delta\mathcal{L}(x)$. Note that $\mathcal{L}(x)$ and $\mathcal{L}$ mean the same. So, $\alpha\Delta\mathcal{L} = \mathcal{L’} - \mathcal{L’} := \delta \mathcal{L}$. We can write:

$$ \begin{aligned} \alpha\Delta\mathcal{L} &= \delta\mathcal{L} \\ &= \text{Using the definition of the partial derivative.} \\ &= \frac{\partial \mathcal{L}}{\partial \phi} \delta\phi + \left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \right) \delta(\partial_{\mu}\phi) \\ &= \text{Using eqn (2.9) i.e. } \delta\phi = \alpha\Delta\phi \text{ and } \delta(\partial_{\mu}\phi) = \partial_{\mu} (\delta\phi) = \partial_{\mu}(\alpha\Delta\phi) \\ &= \frac{\partial \mathcal{L}}{\partial \phi} (\alpha\Delta\phi) + {\color{red} \left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \right) \partial_{\mu}(\alpha\Delta\phi)} \\ &= \text{Using chain rule in the red colored equation.} \\ &= \alpha \frac{\partial \mathcal{L}}{\partial \phi} (\Delta\phi) + {\color{red} \partial_{\mu}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} (\alpha\Delta\phi) \right) - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \right) (\alpha\Delta\phi)} \\ &= \alpha \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi \right) + \alpha \left[ {\color{green}\frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \right)} \right] \Delta\phi \quad (2.11) \\ &= \text{Green colored equation is EOM (2.3) equals to 0.} \\ &= \alpha \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi \right) \end{aligned} $$

Comparing above result with eqn (2.10). We get $\Delta\mathcal{L} = \partial_{\mu}\mathcal{J}^{\mu}(x)$. i.e.

$$ \begin{aligned} \partial_{\mu}\mathcal{J}^{\mu}(x) &= \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi \right) \\ \therefore \partial_{\mu} \left( {\color{blue}\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi - \mathcal{J}^{\mu}} \right) &:= \partial_{\mu}{\color{blue}j^{\mu}(x)} = 0. \quad (2.12) \end{aligned} $$

Blue colored equation is defined as current. The eqn (2.12) means the divergence of current vanishes.

If the symmetry involves more than one field (say $\phi_{1}, \phi_{2}, \ldots, \phi(n)$), the first term of the eqn (2.12) for $j^{\mu}(x)$ should be replaced by a sum of such terms, one for each field. i.e.

$$ \begin{aligned} \displaystyle\sum_{\mu=0}^3 \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi &\xrightarrow[\text{by}]{\text{Replace}} \displaystyle\sum_{i=1}^n \displaystyle\sum_{\mu=0}^3\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{i})} \Delta\phi_{i} \\ &\equiv \text{In Einstein's convention, } \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{i})} \Delta\phi_{i} \end{aligned} $$

Note: we will use Einstein’s convention most of the time unless it need explicit form.

18P2B0

… so we conclude that the current $j^{\mu} = \partial^{\mu}\phi$.

Given $\mathcal{L} = \frac{1}{2}(\partial_{\mu}\phi)^{2} \equiv \frac{1}{2} (\partial_{\mu}\phi)(\partial^{\mu}\phi)$ and $\phi \to \phi + \alpha$ so deformation of field $\Delta\phi = 1$ i.e. we only rescale our field $\phi$ by scale $\alpha$. Our Lagrangian is scale invariance (a symmetry) because $\partial_{\mu}\alpha = 0$. Thus, we demands one conservation law because of Noether’s theorem. Note we assume $\mathcal{J}^{\mu} = 0$. Hence, current is

$$ \begin{aligned} j^{\mu} &= \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi - \mathcal{J}^{\mu} \\ &= \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \\ &= \frac{1}{2}\frac{\partial \left( (\partial_{\mu}\phi)(\partial^{\mu}\phi) \right)}{\partial (\partial_{\mu}\phi)} \\ &= \frac{1}{2} \left[ \frac{\partial (\partial_{\mu}\phi)}{\partial (\partial_{\mu}\phi)} (\partial^{\mu}\phi) + (\partial_{\mu}\phi) \frac{\partial ({\color{green}\partial^{\mu}}\phi)}{\partial (\partial_{\mu}\phi)} \right] \\ &= \text{Using } {\color{green}\partial^{\mu} = g^{\mu\nu} \partial_{\nu}}. \\ &= \frac{1}{2} \left[ \partial^{\mu}\phi + (\partial_{\mu}\phi) {\color{green}g^{\mu\nu}} \frac{\partial ({\color{green}\partial_{\nu}}\phi)}{\partial (\partial_{\mu}\phi)}\right] \\ &= \frac{1}{2} \left[ \partial^{\mu}\phi + (\partial_{\mu}\phi) g^{\mu\nu} \delta^{\nu}_{\mu} \right] \\ &\quad \text{where } \frac{\partial (\partial_{\nu}\phi)}{\partial (\partial_{\mu}\phi)} = \delta^{\nu}_{\mu} := \text{ Kronecker delta}. \\ &= \frac{1}{2} \left[ \partial^{\mu}\phi + (\partial_{\mu}\phi) g^{\mu\nu} \delta^{\nu}_{\mu} \right] \\ &= \frac{1}{2} \left[ \partial^{\mu}\phi + {\color{red}(\partial^{\nu}\phi) \delta^{\nu}_{\mu}} \right] \\ &= \frac{1}{2} \left[ \partial^{\mu}\phi + {\color{red}\partial^{\mu}\phi} \right] \\ \therefore j^{\mu} &= \partial^{\mu} \phi. \end{aligned} $$

18P2E(2.15)-(2.16)

Given the Lagrangian for complex scalar field, $$\mathcal{L} = |\partial_{\mu}\phi |^{2} - m^{2}|\phi|^{2} \equiv (\partial_{\mu}\phi)(\partial^{\mu}\phi^{*}) - m^{2} \phi\phi^{*} \quad (2.14)$$ … the transformation $\phi \to e^{i\alpha}\phi$; for an infinitesimal transformation with $\mathcal{O}(\alpha^{2})$ we have

$$ \begin{aligned} \phi \to \phi' &= e^{i\alpha} \phi \\ &= \text{Under infinitesimal transformation}\\ &= (1 + i\alpha)\phi \\ &= \phi + i\alpha\phi \end{aligned} $$

Similarly,

$$ \begin{aligned} \phi^{*} \to {\phi^{*}}' &\equiv (\phi')^{*} = (e^{i\alpha} \phi)^{*} = e^{-i\alpha} \phi^{*} \\ &= \text{Under infinitesimal transformation}\\ &= (1 - i\alpha)\phi^{*} \\ &= \phi^{*} - i\alpha\phi^{*} \end{aligned} $$

Comparing above results with $\phi \to \phi' = \phi + \alpha\Delta\phi$ and $\phi \to {\phi^{*}}' = \phi^{*} + \alpha\Delta\phi^{*}$ respectively. We get $\Delta\phi = i\phi$ and $\Delta\phi^{*} = -i\phi^{*}\quad (2.15)$.

… conserved Noether current is

$$ \begin{aligned} j^{\mu} &= \displaystyle\sum_{i=1}^n \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_{i})} \Delta\phi_{i} - \mathcal{J}^{\mu} \\ &\quad \text{Since, } \mathcal{J}^{\mu} = 0 \\ &= {\color{red}\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \Delta\phi} + {\color{green}\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^{*})} \Delta\phi^{*}} \\ &= {\color{red} \frac{\partial [(\partial_{\mu}\phi)(\partial^{\mu}\phi^{*}) - m^{2} \phi\phi^{*}]}{\partial (\partial_{\mu}\phi)}} (i\phi) + {\color{green} \frac{\partial (\partial_{\mu}\phi)(\partial^{\mu}\phi^{*}) - m^{2} \phi\phi^{*}}{\partial (\partial_{\mu}\phi^{*})}} (-i\phi^{*}) \\ &= \text{Note that } \frac{\partial (\phi\phi^{*})}{\partial (\partial_{\mu}\phi)} = 0. \text{ Also, $\phi$ and $\phi^{*}$ are independent fields.} \\ &= {\color{red} (\partial^{\mu}\phi^{*}) (i\phi)} - {\color{green} (\partial_{\mu}\phi) g^{\mu\nu}\frac{\partial (\partial_{\nu}\phi^{*})}{\partial (\partial_{\mu}\phi^{*})} (i\phi^{*})} \\ &= i[(\partial^{\mu}\phi^{*})\phi - \phi^{*}(\partial^{\mu}\phi)] \quad (2.16) \end{aligned} $$

18P2B3

… the divergence of this current vanishes by using the Klein-Gordon equation. i.e.

$$ \begin{aligned} \partial_{\mu} j^{\mu} &= i[\partial_{\mu}\{(\partial^{\mu}\phi^{*})\phi\} - \partial_{\mu}\{\phi^{*}(\partial^{\mu}\phi)\}] \\ &= \text{Using product rule.} \\ &= i [(\partial_{\mu}\partial^{\mu} \phi^{*})\phi + \cancel{(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)} - \cancel{(\partial_{\mu}\phi^{*})(\partial^{\mu}\phi)} - \phi^{*}(\partial_{\mu}\partial^{\mu}\phi)] \\ &= \partial_{\mu}\partial^{\mu} \equiv \partial^{\mu}\partial_{\mu} := \square^{2} \text{ called as d'Alembertian, and} \\ &\quad (\partial^{\mu}\phi^{*})(\partial_{\mu}\phi) \equiv (\partial_{\mu}\phi^{*})(\partial^{\mu}\phi) \equiv |\partial_{\mu}\phi|^{2}. \\ &= i [(\square^{2}\phi^{*})\phi - \phi^{*}(\square^{2}\phi)] \\ &= \text{Using Euler-Lagrange EOM: } \\ &\quad\quad \frac{\partial \mathcal{L}}{\partial \phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)} \right) = 0 \\ &\quad\quad \text{or, } -m^{2}\phi^{*} - \partial_{\mu}\partial^{\mu}\phi^{*} = 0 \\ &\quad\quad \therefore (\square^{2} + m^{2})\phi^{*} = 0 \text{ is Klein-Gordon eqn.}\quad (2.7) \\ &\quad\quad\text{or, } \square^{2}\phi^{*} = -m^{2}\phi^{*}\quad \times \phi \text{ both sides} \\ &\quad\quad \therefore (\square^{2}\phi^{*})\phi = -m^{2}\phi^{*}\phi \equiv -m^{2} |\phi|^{2}\\ &\quad \text{Similarly, } \\ &\quad\quad \frac{\partial \mathcal{L}}{\partial \phi^{*}} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^{*})} \right) = 0 \\ &\quad\quad \text{or, } -m^{2}\phi - \partial_{\mu} \left[ \partial_{\mu}\phi \frac{\partial(\partial^{\mu}\phi^{*}) }{\partial (\partial_{\mu}\phi^{*})} \right] = 0 \\ &\quad\quad \text{or, } -m^{2}\phi - \partial_{\mu} \left[ \partial_{\mu}\phi ~g^{\mu\nu}~ \frac{\partial (\partial_{\nu}\phi^{*})}{\partial (\partial_{\mu}\phi^{*})}\right] = 0 \\ &\quad\quad \text{or, } -m^{2}\phi - \partial_{\mu}[\partial^{\nu}\phi~ \delta^{\nu}_{\mu}] = 0 \\ &\quad\quad \text{or, } -m^{2}\phi - \partial_{\mu}\partial^{\mu}\phi = 0 \\ &\quad\quad \text{or, } \square^{2}\phi = -m^{2}\phi \quad \times \phi^{*} \text{ from left} \\ &\quad\quad \therefore \phi^{*}\square^{2}\phi = -m^{2}|\phi|^{2}\\ &\quad \text{Substituting the acquired identities.} \\ &= i[-\cancel{m^{2}|\phi|^{2}} + \cancel{m^{2}|\phi|^{2}}] \\ &= 0. \end{aligned} $$

Note: we’re using d’Alembertian operator (or simply d’Alembertian) as $\square^{2}$ but other books might write just $\square$.