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• Write the Lagrangian for free Schrodinger field and obtain an expression for the Hamiltonian.
• Using the Poisson bracket form of equations of motion show that the Galilean boost \[\vec{G}[\psi]=\int d^3 x\, \psi^\dagger (m\,\vec{x}+ it \hbar \nabla)\psi,\] is a conserved quantity. How do you interpret this conservation law?

 We use the notation \(x=(x_1,x_2,x_3)\).

• The Lagrangian density is given by \[\Lsc = i\hbar\psi^\dagger (x,t) \dot{\psi}(x,t) - \frac{\hbar^2}{2m}|\nabla\psi(x)|^2 \] The momentum canonically conjugate to the field \(\psi(x,t)\) is \begin{equation*} \pi(x,t) = \pp[\Lsc]{\dot{\psi}}= i\hbar \dot{\psi}^\dagger \end{equation*} Therefore the Hamiltonian density is \begin{eqnarray}\nonumber H &=& \int d^3x \pi(x,t) \dot{\psi}(x,t) -\int d^3x \Lsc\\ &=& \frac{\hbar^2}{2m}\int d^3x |\nabla\psi(x,t)|^2. \\ &=& \frac{-i\hbar}{2m}\int d^3x (\nabla\pi)(\nabla\psi) \end{eqnarray}
• The Galilean boost is given to be \begin{eqnarray} {G_k}[\psi,\pi] &=& \int d^3x \frac{1}{i\hbar} \pi(x,t)(m x_k +it\hbar\partial_k ) \psi(x,t) \end{eqnarray} Hence the time derivative of the boost \(G_k\) is given by \begin{eqnarray}\nonumber \dd[G_k]{t} &=& \int d^3x\, \pp[G_k]{t} +\big\{G_k, H\}_\text{PB}\\\label{EQ04} &=& =\big\{G_k,H\big\} + \int d^3x\, [\pi(x)\partial_k \psi(x)]. \end{eqnarray} We now compute the Poisson bracket \(\big\{G_k, H\}_\text{PB}\). \begin{eqnarray}\nonumber \big\{G_k, H\}_\text{PB} &=& -\frac{i\hbar}{2m}\int d^3x\, \big\{\nabla\pi(x,t), H\big\}_\text{PB}(mx_k +it\hbar\partial_k) (\partial_k\psi(x,t)) \\\label{EQ05} && -\frac{i\hbar}{2m}\int d^3x\, (\nabla\pi(x,t)) (mx_k +it\hbar\partial_k )\big\{\nabla\psi, H\big\}_\text{PB} \end{eqnarray} Now we use the Poisson brackets \begin{eqnarray}\nonumber \big\{\pi(x), H\big\}_\text{PB} &=& -\frac{i\hbar}{2m}\int d^3y\big\{\pi(x), (\nabla\pi)(\nabla \psi)\}_\text{PB}\\\nonumber & =& -\frac{i\hbar}{2m}\int d^3y (\nabla \pi(y))(-1)\nabla_y \delta^{(3)}(x-y)\\ &=& \frac{-i\hbar}{2m}\nabla^2 \pi(x)\\\nonumber \big\{\psi(x), H\big\}_\text{PB} &=&-\frac{i\hbar}{2m}\int d^3y\big\{\psi(x), (\nabla\pi)(\nabla \psi)\}_\text{PB} \\\nonumber &=& -\frac{i\hbar}{2m}\int d^3y \nabla_y\delta^{(3)}(x-y)(\nabla \psi(y)) \\ &=&\frac{i\hbar}{2m}\nabla^2 \psi(x) \end{eqnarray} Therefore the Poisson bracket \(\big\{G, H\big\}_\text{PB}\), \eqRef{EQ05}, is given by \begin{eqnarray}\nonumber \big\{G_k, H\big\}_\text{PB} &=& -\frac{1}{2m} \int d^3x\Big\{ (\nabla^2\pi)(mx_k+ it\hbar \partial_k) \psi + \big(mx_k +it\hbar\partial_k\big)\nabla^2\psi\Big\}\\\label{EQ12} &=& \frac{it\hbar}{2m}\int d^3x (\nabla^2\pi(x))(\partial_k\psi(x))-(\pi(x)) \nabla^2\partial_k\psi(x))\\\nonumber && \qquad - \frac{1}{2}\int d^3x \Big((\nabla^2\pi(x))(x_k \psi(x)) - (x_k \pi(x)) (\nabla^2\psi(x))\Big) \\ \label{EQ13} \end{eqnarray} Integrating by parts and, as we shall show below, discarding surface terms we get \begin{eqnarray}\label{EQ15} \big\{G_k, H\big\}_\text{PB} = - \int d^3x\, [\pi(x)\partial_k \psi(x)]. \end{eqnarray} Using \eqRef{EQ04} and \eqRef{EQ15} we get \(\dd[G]{t}=0.\) Therefore \(G\) is constant of motion.

Proof of \eqRef{EQ15
It is easy to see that the expression\eqRef{EQ12} is a total divergence. In fact \begin{eqnarray} \frac{it\hbar}{2m}\lefteqn{\int d^3x (\nabla^2\pi(x))(\partial_k\psi(x))-(\pi(x)) \nabla^2\partial_k\psi(x))}\\ &=&\frac{it\hbar}{2m} \int d^3x \nabla\Big((\nabla\pi(x))(\partial_k\psi)-(\pi(x))(\nabla\partial_k \psi(x))\Big) \end{eqnarray} which becomes surface term on using divergence theorem and hence is zero. % Therefore the Poisson bracket, \eqRef{EQ13}, becomes \begin{eqnarray} \big\{G,H\big\}_\text{PB}= - \frac{1}{2}\int d^3x \Big((\nabla^2\pi(x))(x_k \psi(x)) - (x_k \pi(x)) (\nabla^2\psi(x))\Big)\label{EQ17} \end{eqnarray} Next we use divergence theorem to integrate both the terms by parts and discard the surface term, we get \begin{eqnarray}\nonumber \lefteqn{ \int d^3 x (\nabla^2\pi(x))(x_k\psi(x))}\\\nonumber &=&- \int d^3x (\nabla\pi(x)) \nabla(x_k\psi(x))\\ &=&- \int d^3x (\nabla_k\pi(x)) \psi(x)) + (\nabla\pi(x))x_k(\nabla\psi(x))\label{EQ18} \end{eqnarray} and \begin{eqnarray}\nonumber \lefteqn{\int d^3 x (x_k\pi(x))(\nabla^2\psi(x))} \\\nonumber &=& -\int d^3x \nabla(x_k\pi(x))(\nabla\psi(x))\\ &=& -\int d^3x \pi(x) \nabla_k \psi(x) + (\nabla\pi(x)) x_k(\nabla\psi(x)) \label{EQ19} \end{eqnarray} Therefore from \eqRef{EQ17}, we get \begin{eqnarray}\nonumber \big\{G,H\big\} &=& \frac{1}{2}\int d^3x \big\{(\nabla_k\pi(x))(\psi(x)) - \pi(x) \nabla_k\psi(x)\big\}\\\label{EQ20} &=& \frac{1}{2}\int d^3x\nabla_k[\pi(x)\psi(x)] -\int d^3x [\pi(x) \nabla_k \psi(x)] \\ &=& -\int d^3x [\pi(x) \nabla_k \psi(x)] \end{eqnarray} where the last term in \eqRef{EQ20} becomes zero on integration by parts.

Write an infinitesimal Lorentz transformation as
\begin{equation} \widetilde{X{'}} = \widetilde{\Lambda} \widetilde{X} \end{equation} where \(\widetilde{\Lambda}\) is a four by four matrix. Taking \(Lambda\) to differ infinitesimal transformation of the form \[\Lambda = I + \Omega \] and find restrictions on the matrix \(\Omega\) using the fact that under Lorentz transformations \(x^\mu x_\mu \) remains invariant. Count independent number of parameters in \(\Omega\) and write its most general form. 

 The most general form of \(\Omega\) is \begin{equation*} \Omega = \begin{pmatrix} a & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_1 & b & \alpha_4 & \alpha_5 \\ \alpha_2 & -\alpha_4 & c & \alpha_6\\ \alpha_3 & -\alpha_5 & -\alpha_6 & d \end{pmatrix} \end{equation*} }\end{Answer} \begin{Solution} The Lorentz invariance of \(x^\mu x_\mu= x^\mu g_{\mu\nu} x^\nu\) implies \[ \widetilde{X}^{{'} \,T} \tilde{g} \widetilde{X}{'} = {X}^{{'} \,T} \tilde{g} {X}{'}\] where \(\tilde{g}\) is the diagonal matrix with diagonal entries (1,-1,-1,-1). The requirement of Lorentz invariance of the four scalar product gives the following condition on \(\Lambda\). \begin{equation} \widetilde{\Lambda} \tilde{g} \widetilde{\Lambda} = \tilde{g} \end{equation} Using \(\widetilde{X}^{{'} \,T} g \widetilde{X}{'}\) We rewrite the above equation in the form \begin{equation} (I + \Omega^T) \tilde{g} (1+ \Omega) = \tilde{g} \end{equation} Simplifying and keeping only first order terms in \(\Omega\), gives \begin{equation}\label{EQ04} \Omega^T \tilde{g} + \tilde{g} \Omega = 0. \end{equation}
Writing and multiplying out the matrices gives
\begin{eqnarray} \Omega \tilde{g} &=& \begin{pmatrix} a & -\alpha_1 & -\alpha_2 & -\alpha_3 \\ \alpha_1 & b & \alpha_4 & \alpha_5 \\ \alpha_2 & -\alpha_4 & c & \alpha_6\\ \alpha_3 & -\alpha_5 & -\alpha_6 & d \end{pmatrix}\\ 
\tilde{g}\Omega &=& \begin{pmatrix} a & \alpha_1 & \alpha_2 & \alpha_3 \\ \alpha_1{'} & b & \alpha_4 & \alpha_5 \\ \alpha_2{'} & -\alpha_4{'} & c & \alpha_6\\ \alpha_3{'} & -\alpha_5{'} & -\alpha_6{'} & d \end{pmatrix} \end{eqnarray}