場の変分は$\delta\phi_{r}(x) \equiv \phi^{\prime}_{r}(x) - \phi_{r}(x)$と定められる。
加えて，以下の座標変換を考える。
\[
x^{\prime}_{\mu} = x_{\mu} + \varepsilon_{\mu\nu}x^{\nu} + a_{\mu}
\]
次に，座標変換後の場ともともとの場の変分を$\delta_{T}\phi_{r}(x)$とすると，
\begin{align*}
\delta_{T}\phi_{r}(x) &= \phi^{\prime}_{r}(x^{\prime}) - \phi_{r}(x) \\
&= \phi^{\prime}_{r}(x^{\prime}) - \phi_{r}(x^{\prime}) + \phi_{r}(x^{\prime}) - \phi_{r}(x) \\
&= \delta \phi_{r}(x^{\prime}) + \partial_{\mu} \phi_{r}(x) \delta x^{\mu}
\end{align*}
となる。ただし，無限小変換を考えている限り，$\delta\phi_{r}(x^{\prime})=\delta\phi_{r}(x)$
となるので，
\[
\delta_{T}\phi_{r}(x) = \delta \phi_{r}(x) + \partial_{\mu} \phi_{r}(x) \delta x^{\mu}
\]
が得られる。ここで，座標変換まで含んだ作用変分
\[
\delta_{T}S = \int d^{4}x^{\prime} \mathcal{L}(\phi_{r}^{\prime}(x^{\prime}),\partial_{\mu}^{\prime}\phi_{r}^{\prime}(x^{\prime}))
- \int d^{4}x \mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))
\]
を考える。積分変数を揃えるために第1項目を書き換える。
\[
d^{4}x^{\prime}
= \dfrac{\partial(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})}
{\partial(x^{0},x^{1},x^{2},x^{3})}d^{4}x
= \begin{vmatrix}
\partial x^{\prime 0}/\partial x^{0} & \partial x^{\prime 0}/\partial x^{1} & \partial x^{\prime 0}/\partial x^{2} & \partial x^{\prime 0}/\partial x^{3} \\
\partial x^{\prime 1}/\partial x^{0} & \partial x^{\prime 1}/\partial x^{1} & \partial x^{\prime 1}/\partial x^{2} & \partial x^{\prime 1}/\partial x^{3} \\
\partial x^{\prime 2}/\partial x^{0} & \partial x^{\prime 2}/\partial x^{1} & \partial x^{\prime 2}/\partial x^{2} & \partial x^{\prime 2}/\partial x^{3} \\
\partial x^{\prime 3}/\partial x^{0} & \partial x^{\prime 3}/\partial x^{1} & \partial x^{\prime 3}/\partial x^{2} & \partial x^{\prime 3}/\partial x^{3}
\end{vmatrix}
d^{4}x
\]
無限小変換の場合
\[
\dfrac{\partial(x^{\prime 0},x^{\prime 1},x^{\prime 2},x^{\prime 3})}
{\partial(x^{0},x^{1},x^{2},x^{3})}
= 1 + \partial_{\nu}\delta x^{\nu}
\]
となるから，
\begin{align*}
\delta_{T} S&=\int d^{4}x\bigl[ \mathcal{L}(\phi^{\prime}(x), \partial_{\mu}^{\prime}\phi_{r}^{\prime}(x^{\prime})) (1 + \partial_{\nu}\delta x^{\nu})
- \mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))\bigr] \\
&= \int d^{4}x \biggl[\biggl\{
\mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x)) + \biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr)\delta_{\rm T}\phi_{r} + \biggl(\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\partial_{\mu}\phi_{r}\biggr\}
(1 + \partial_{\nu}\delta x^{\nu})
- \mathcal{L}(\phi{r}(x),\partial_{\mu}\phi_{r}(x)) \biggr] \\
&=
\int d^{4}x \biggl[\biggl\{\mathcal{L}(\phi_{r}(x),\partial_{\mu}\phi_{r}(x))\partial_{\nu}\delta x^{\nu}
+ \biggl\{\partial_{\mu}\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\phi_{r} + \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{\rm T}\partial_{\mu}\phi_{r} \biggr\}
\biggr]
\end{align*}
となる。ここで，作用積分$S$が不変であれば
\begin{align}
\biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr)\delta_{T}\phi_{r}(x)
+ \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{T}\partial_{\mu}\phi_{r}(x)
+ \mathcal{L}\partial_{\nu}\delta x^{\nu}=0
\end{align}
となる。また，
\begin{align}
\delta\phi_{r}(x) = \delta_{\rm T}\phi_{r}(x) - \partial_{\mu}\phi_{r}(x)\delta x^{\mu}
\end{align}
であり，変分$\delta$は微分と交換可能なことから
\begin{align*}
\delta_{T}\partial_{\mu}\phi_{r}(x) &=
\delta\partial_{\mu}\phi_{r}(x) + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\\
&= \partial_{\mu} \delta\phi_{r}(x) + \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} \\
&= \partial_{\mu}[\delta_{T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\delta x^{\nu}]
+ \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu} \\
&= \partial_{\mu}\delta_{T}\phi_{r}(x) - \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}
- \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}
+ \partial_{\mu}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\\
&= \partial_{\mu}\delta_{\rm T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}
\end{align*}
となることを用いると，作用変分は
\begin{align*}
\delta_{\rm T}S &= \int d^{4}x \biggl[
\mathcal{L}\partial_{\nu}\delta x^{\nu}
+ \biggl(\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\biggr) \delta_{T} \phi_{r}(x)
+\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)
\bigl\{\partial_{\mu}\delta_{\rm T}\phi_{r}(x) - \partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}\}\biggr] \\
&= \int d^{4}x \biggl[
\mathcal{L}\partial_{\nu}\delta x^{\nu}
+ \partial_{\mu}\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr) \delta_{T} \phi_{r}(x)
+\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\mu}\delta_{\rm T}\phi_{r}(x)
- \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\partial_{\mu}\delta x^{\nu}\biggr] \\
&= \int d^{4}x \biggl[
\partial_{\mu}\{\mathcal{L}\delta x^{\mu}\} - \partial_{\mu}\mathcal{L}\delta x^{\mu}
+ \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\delta_{T}\phi_{r}(x)\biggr\}
- \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\delta x^{\nu}\biggr\}
+ \partial_{\mu}\biggl\{\biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\biggr)\partial_{\nu}\phi_{r}(x)\biggr\}\delta x^{\nu} \biggr]\\
&= \int d^{4}x \biggl[
\partial_{\mu}\biggl\{\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} \delta_{T}\phi_{r}(x)
- \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\nu}\phi_{r}(x)\delta x^{\nu}
+ \mathcal{L}(\phi_{r},\partial_{\nu}\phi_{r})\delta x^{\mu}\biggr\} \\
&\qquad\qquad
+ \biggl\{\dfrac{\partial\mathcal{L}}{\partial\phi_{r}}\partial_{\nu}\phi_{r}(x)
+ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\mu}\partial_{\nu}\phi_{r}(x)
- \partial_{\mu}\mathcal{L} \biggr\}\delta x^{\mu} \biggr] \\
&= \int d^{4}x \biggl[
\partial_{\mu}\biggl\{ \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi{r})}\delta_{T}\phi_{r}(x)
- \biggl(\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial_{\nu}\phi_{r}(x)
- g^{\mu\nu}\mathcal{L}\biggr)\delta x^{\nu} \biggr\} \biggr]
\end{align*}
が得られる。これにより，最終的に
\begin{align}
\delta_{T}S &= \int d^{4}x
\partial_{\mu}\biggl[
\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\delta_{T}\phi_{r}(x)
- \mathcal{T}^{\mu\nu}\delta x_{\nu}\biggr]
\end{align}
を得る。ここで，
\begin{align}
\mathcal{T}^{\mu\nu}
&= \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\partial^{\nu}\phi_{r}
- g^{\mu\nu}\mathcal{L}
\end{align}
はエネルギー運動量テンソルであり，保存則の式として
\begin{align}
\partial_{\mu}\biggl[
\dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}\delta_{T}\phi_{r}(x)
- \mathcal{T}^{\mu\nu}\delta x_{\nu}\biggr] = 0
\end{align}
が得られ，場の理論におけるNoether カレントの保存則を表す。
\subsection{時間と空間の平行移動に対する保存量}
時間と空間の並進について考えると，$x^{\prime\mu} = x^{\mu} + \varepsilon^{\mu}$
であり，$\delta_{T}\phi_{r} = 0$となっているので，保存カレントは$\mathcal{T}^{\mu\nu}$
となり，$\partial_{\mu}\mathcal{T}^{\mu\nu} = 0$が成り立つ。これより，保存量は
\begin{align}
P^{\mu} \equiv \int d^{3}\bm{x} \mathcal{T}^{\mu0}
= \it d^{3}\bm{x} \bigl\{ \pi_{r}(x)\partial^{\nu}\phi_{r}(x) - \mathcal{L}g^{\mu0}\bigr\}
\end{align}
となる。
時間成分と空間成分をそれぞれ書くと，
\begin{align}
P^{0} &= \int d^{3}\bm{x}\{ \pi_{r}(x)\dot{\phi}_{r}(x) - \mathcal{L}(\phi_{r},\partial_{\mu}\phi)\}
= \int d^{3}\bm{x} \mathcal{H} = H \\
P^{j} &= \int d^{3}\bm{x} \{\pi_{r}(x) \partial^{j}\phi_{r}(x)\}
\end{align}
となり，ハミルトニアンと運動量であることがわかる。
\subsection{ローレンツ変換に対する保存量}
無限小ローレンツ変換は$\varepsilon_{\mu\nu} = -\varepsilon_{\nu\mu}$を用いて
\begin{align}
\delta x_{\mu} = \varepsilon_{\mu\nu} x^{\nu},\quad
\delta_{T}\phi_{r}(x) = \dfrac{1}{2}\varepsilon_{\mu\nu}(S^{\mu\nu})_{rs} \phi_{s}(x)
\end{align}
$S^{\mu\nu}$はローレンツ変換の生成子の一つの表現で
\begin{align}
\begin{cases}
\text{スピン0(スカラー場)なら} S^{\mu\nu}= 0\\
\text{スピン$\dfrac{1}{2}$(フェルミオン場)なら} S^{\mu\nu}= \dfrac{1}{4}[\gamma^{\mu},\gamma^{\nu}]
\text{スピン1(ベクトル場)なら} S^{\mu\nu}= S^{\mu\nu} = L^{\mu\nu}
\end{cases}
\end{align}
ベクトル場の場合，
\begin{align*}
\delta x_{\mu} = \varepsilon_{\mu\nu}x^{\nu} = \dfrac{1}{2}\varepsilon_{\alpha\beta} (S^{\alpha\beta})_{\mu\nu} x^{\nu}
\end{align*}
より，
\begin{align*}
\varepsilon_{\mu\nu} x^{\nu}
= \varepsilon_{\alpha\beta}\delta^{\alpha}_{\;\mu}
\delta^{\beta}_{\;\nu} x^{\nu}
= \dfrac{1}{2}\varepsilon_{\alpha\beta}(S^{\alpha\beta})_{\mu\nu}x^{\nu}
\end{align*}
また，
\begin{align*}
\varepsilon_{\mu\nu}x^{\nu} = \dfrac{1}{2}\varepsilon_{\beta\alpha}(-S^{\alpha\beta})_{\mu\nu} x^{\nu}
\end{align*}
より，
\begin{align*}
\varepsilon_{\mu\nu}x^{\nu} = -\varepsilon_{\beta\alpha} \delta^{\beta}_{\;\alpha}
\delta^{\alpha}_{\;\nu} x^{\nu}
= \dfrac{1}{2} \varepsilon_{\beta\alpha} (-S^{\alpha\beta})_{\mu\nu} x^{\nu}
\end{align*}
よって，
\begin{align*}
\dfrac{1}{2}(S^{\alpha\beta})_{\mu\nu} &= \delta^{\alpha}_{\;\mu} \delta^{\beta}_{\;\nu} \\
-\dfrac{1}{2}(S^{\alpha\beta})_{\mu\nu} &= \delta^{\beta}_{\;\mu} \delta^{\alpha}_{\;\nu}
\end{align*}
以上より，
\begin{align}
(S^{\alpha\beta})_{\mu\nu} = \delta^{\alpha}_{\;\mu} \delta^{\beta}_{\;\nu}
- \delta^{\beta}_{\;\mu} \delta^{\alpha}_{\;\nu}
\end{align}
保存則の式に代入すると，
\begin{align*}
&\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})}
\delta_{T}\phi_{r}(x) - \mathcal{T}^{\mu\nu}\delta x_{\nu} \\
&= \dfrac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} \dfrac{1}{2}\varepsilon_{\nu\lambda} (S^{\nu\lambda})_{rs} \phi_{s}
- \mathcal{T}^{\mu\nu} \varepsilon_{\nu\lambda} x^{\lambda} \\
&= \dfrac{1}{2}\varepsilon_{\nu\lambda} \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs}\phi_{s}
- \dfrac{1}{2}\varepsilon_{\nu\lambda} \mathcal{T}^{\mu\nu} x^{\lambda}
+ \dfrac{1}{2}\varepsilon_{\lambda\nu} \mathcal{T}^{\mu\lambda} x^{\nu} \\
&= \dfrac{1}{2} \varepsilon_{\nu\lambda} \left[
\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs} \phi_{s}
- \mathcal{T}^{\mu\nu} x^{\lambda} + \mathcal{T}^{\mu\lambda} x^{\nu} \right]
\end{align*}
これより，Noetherカレントは
\begin{align}
\mathcal{M}^{\mu\nu\lambda} =
\dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{r})} (S^{\nu\lambda})_{rs} \phi_{s}
- \mathcal{T}^{\mu\nu} x^{\lambda} + \mathcal{T}^{\mu\lambda} x^{\nu}
\end{align}
となる。この保存則は$\partial_{\mu}\mathcal{M}^{\mu\nu\lambda}$と書かれるので，
保存量は
\begin{align*}
M^{\mu\nu} &= \int d^{3}\bm{x} \mathcal{M}^{0\mu\nu} \\
&= \int d^{3}\bm{x} \left[
\dfrac{\partial \mathcal{L}}{\partial(\partial_{0}\phi_{r})}(S^{\mu\nu})_{rs}\phi_{s}(x)
- \mathcal{T}^{0\mu}x^{\nu} + \mathcal{T}^{0\nu} x^{\mu} \right] \\
&= \int d^{3}\bm{x}\left[\pi_{r}(x)(S^{\mu\nu})_{rs}\phi_{s}(x)
- \mathcal{T}^{0\mu}x^{\nu} + \mathcal{T}^{0\nu} x^{\mu} \right]
\end{align*}
特に，空間成分については
\begin{align*}
M^{ij} &= \int d^{3}\bm{x} \left[
\pi_{r}(x) (S^{ij})_{rs} \phi_{s}(x) - p^{i}x^{j} + p^{j}x^{i}\right]
\end{align*}