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joshphysics
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One obtains those expressions by considering a particular action of the Poincare group on fields.

Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$. Let $\mathcal F$ denote the space of such fields. Define an action $\rho_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows \begin{align} \rho_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a)) \end{align} Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra). We can use this generator to define a corresponding infinitesimal generator for group action $\rho_\mathcal F$ as follows: \begin{align} G_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|_{\epsilon = 0} \end{align}

Example - translations. Consider the translation generators $P^\mu$ which have the property \begin{align} e^{-ia_\mu P^\mu}x = x+a \end{align} The generator of $\rho_\mathcal F$ corresponding to $P^0$, for instance, is \begin{align} (P^0)_\mathcal F(\phi)(x) &= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\ &= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\ &= i\partial_0\phi(x) \end{align}\begin{align} (P^0)_\mathcal F(\phi)(x) &= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\ &= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\ &= i\partial^0\phi(x) \end{align} where $e_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives \begin{align} (P^\mu)_\mathcal F = i\partial_\mu. \end{align}\begin{align} (P^\mu)_\mathcal F = i\partial^\mu. \end{align}

Example - Lorentz boosts.

If you use this same procedure for Lorentz boost generators, you will find that \begin{align} (M^{\mu\nu})_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu \end{align}

Disclaimer about signs etc. There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.

One obtains those expressions by considering a particular action of the Poincare group on fields.

Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$. Let $\mathcal F$ denote the space of such fields. Define an action $\rho_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows \begin{align} \rho_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a)) \end{align} Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra). We can use this generator to define a corresponding infinitesimal generator for group action $\rho_\mathcal F$ as follows: \begin{align} G_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|_{\epsilon = 0} \end{align}

Example - translations. Consider the translation generators $P^\mu$ which have the property \begin{align} e^{-ia_\mu P^\mu}x = x+a \end{align} The generator of $\rho_\mathcal F$ corresponding to $P^0$, for instance, is \begin{align} (P^0)_\mathcal F(\phi)(x) &= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\ &= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\ &= i\partial_0\phi(x) \end{align} where $e_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives \begin{align} (P^\mu)_\mathcal F = i\partial_\mu. \end{align}

Example - Lorentz boosts.

If you use this same procedure for Lorentz boost generators, you will find that \begin{align} (M^{\mu\nu})_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu \end{align}

Disclaimer about signs etc. There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.

One obtains those expressions by considering a particular action of the Poincare group on fields.

Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$. Let $\mathcal F$ denote the space of such fields. Define an action $\rho_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows \begin{align} \rho_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a)) \end{align} Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra). We can use this generator to define a corresponding infinitesimal generator for group action $\rho_\mathcal F$ as follows: \begin{align} G_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|_{\epsilon = 0} \end{align}

Example - translations. Consider the translation generators $P^\mu$ which have the property \begin{align} e^{-ia_\mu P^\mu}x = x+a \end{align} The generator of $\rho_\mathcal F$ corresponding to $P^0$, for instance, is \begin{align} (P^0)_\mathcal F(\phi)(x) &= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\ &= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\ &= i\partial^0\phi(x) \end{align} where $e_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives \begin{align} (P^\mu)_\mathcal F = i\partial^\mu. \end{align}

Example - Lorentz boosts.

If you use this same procedure for Lorentz boost generators, you will find that \begin{align} (M^{\mu\nu})_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu \end{align}

Disclaimer about signs etc. There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.

Source Link
joshphysics
  • 58.3k
  • 5
  • 144
  • 205

One obtains those expressions by considering a particular action of the Poincare group on fields.

Consider, for example, a single real scalar field $\phi:\mathbb R^{3,1}\to\mathbb R$. Let $\mathcal F$ denote the space of such fields. Define an action $\rho_\mathcal F$ of $P(3,1)$ acting on $\mathcal F$ as follows \begin{align} \rho_\mathcal F(\Lambda,a)(\phi)(x) = \phi(\Lambda^{-1} (x-a)) \end{align} Sometimes people will write this as $\phi'(x) = \phi(\Lambda^{-1} x)$ for brevity. Now let $G$ denote a generator of the Lie algebra of the Poincare group (namely an element of a chosen basis for this Lie algebra). We can use this generator to define a corresponding infinitesimal generator for group action $\rho_\mathcal F$ as follows: \begin{align} G_\mathcal F(\phi) = i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon G})(\phi)\bigg|_{\epsilon = 0} \end{align}

Example - translations. Consider the translation generators $P^\mu$ which have the property \begin{align} e^{-ia_\mu P^\mu}x = x+a \end{align} The generator of $\rho_\mathcal F$ corresponding to $P^0$, for instance, is \begin{align} (P^0)_\mathcal F(\phi)(x) &= i\frac{\partial}{\partial\epsilon}\rho_\mathcal F(e^{-i\epsilon P^0})(\phi)\bigg|_{\epsilon = 0} \\ &= i\frac{\partial}{\partial\epsilon}\phi(x + \epsilon e_0)\bigg|_{\epsilon = 0} \\ &= i\partial_0\phi(x) \end{align} where $e_0 = (1,0,0,0)$, and similarly for the other $P^\mu$, which gives \begin{align} (P^\mu)_\mathcal F = i\partial_\mu. \end{align}

Example - Lorentz boosts.

If you use this same procedure for Lorentz boost generators, you will find that \begin{align} (M^{\mu\nu})_\mathcal F = i(x^\mu\partial^\nu - x^\nu\partial^\mu) = x^\mu p^\nu - x^\nu p^\mu \end{align}

Disclaimer about signs etc. There are a lot of conventional factors of $i$ and negative signs floating around which I wasn't super careful to keep track of, if you notice an error in this regard, please let me know and I'll fix it.