# 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 completeness theorem, } \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} \\ &= \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}

## Noether’s 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 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.

### 18P2

… 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}

### 18P2

… 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 $\square$.

Published on Aug 24, 2021

Last revised on Mar 3, 2022