3
$\begingroup$

I'm having an issue with obtaining the Dirac bracket in the Madelung (polar) representation of the Schrödinger field: \begin{equation} \Psi=\sqrt{\rho}e^{i\theta/\hbar}. \label{eq:WavefunctionPolarForm} \end{equation}

Background:

It has been shown (for instance by Nonnenmacher https://link.springer.com/article/10.1007%2FBF02817982 and Guerra https://journals.aps.org/prd/abstract/10.1103/PhysRevD.28.1916) that in this representation, $\theta$ and $\rho$ play the role of conjugate variables in phase space $\Gamma=\left( \rho,\theta \right)$ with a Poisson bracket given by \begin{equation} \left\{ f,g \right\}=\int d\vec{r}\left(\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta}-\frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}\right)=\left\{ f,g \right\}_{\rho,\theta}. \label{} \end{equation} Basically I would like to derive this result by applying the Dirac-Bergmann algorithm for constrained Hamiltonian systems. However, there is an additional factor of $2$ which pops up in the resulting Dirac bracket, so that \begin{equation} \left\{ f,g \right\}_D=2\left\{ f,g \right\}_{\rho,\theta} \label{} \end{equation} as shown below. To begin with, note that a Hermitian Lagrangian density for the free Schrodinger field, may be written as (see for instance Henley & Thirring's 'Elementary QFT' or Peter Holland's 'The quantum theory of motion') \begin{equation} \mathcal{L}=\frac{i\hbar}{2}\left( \Psi^*\dot{\Psi}-\dot{\Psi}^*\Psi\right)-\frac{\hbar^2}{2m}\nabla\Psi\nabla\Psi^*. \label{} \end{equation} Variation of the action $I=\int dt d^3x\mathcal{L}$ with respect to $\Psi^*$ yields the Schroedinger equation and variation with respect to $\Psi$ yields its complex conjugate. Substituting the polar form for $\Psi$ into this expression for $\mathcal{L}$, we obtain the following form for $\mathcal{L}$: \begin{equation} \mathcal{L}=-\rho\left(\dot{\theta}+\frac{(\nabla\theta)^2}{2m}\right)-\frac{\hbar^2}{8m\rho}\left( \nabla\rho \right)^2. \label{eq:LagrangianPolarForm} \end{equation} Variation with respect to the field $\theta$ yields an equation of continuity: \begin{equation} \dot{\rho}+\vec{\nabla}\cdot\vec{J}=0 \label{} \end{equation} where $\vec{J}=\rho\vec{\nabla}\theta/m$, while variation with respect to $\rho$ yields the quantum Hamilton-Jacobi equation: \begin{equation} \dot{\theta}+\frac{(\nabla\theta)^2}{2m}-\frac{\hbar^2}{2m\sqrt{\rho}}\nabla^2\sqrt{\rho}=0. \label{} \end{equation} It is well known that these $2$ wave equations map onto the Schroedinger equation. Now, the canonical momenta, are then $\pi_{\rho}=0$ and $\pi_{\theta}=-\rho$, leading to the constraint equations $C_1=\pi_{\rho}\approx 0$ and $C_2=\pi_{\theta}+\rho\approx 0$ in the full phase space $(\rho,\theta,\pi_{\rho},\pi_{\theta})$, where following Dirac the symbol '$\approx$' denotes weak equality on hypersurface defined by the constraints. The canonical Hamiltonian density is given by \begin{equation} \mathcal{H}_c=\pi_{\theta}\dot{\theta}+\pi_{\rho}\dot{\rho}-\mathcal{L}\approx \rho\left( \frac{(\nabla\theta)^2}{2m}\right)+\frac{\hbar^2}{8m\rho}\left( \nabla\rho \right)^2. \label{eq:canonicalHamiltonianDensity} \end{equation} The Poisson bracket of the constraints, shows that they are second class: $\left\{ C_1\left( \vec{r} \right), C_2 \left( \vec{r}' \right) \right\}=-\delta \left( \vec{r}-\vec{r}' \right)$.

The matrix of constraint Poisson brackets with elements $Q_{ij}\left( \vec{r},\vec{r}' \right)=\left\{ C_i\left( \vec{r} \right),C_j\left( \vec{r}' \right) \right\}$, is then \begin{equation} Q\left( \vec{r},\vec{r}' \right)= \begin{pmatrix} 0 & -1\\ 1 & 0 \\ \end{pmatrix} \delta \left( \vec{r}-\vec{r}' \right), \label{eq:ConstraintPoissonMatrix} \end{equation} whose inverse is \begin{equation} Q^{-1}\left( \vec{r},\vec{r}' \right)= \begin{pmatrix} 0 & 1\\ -1 & 0 \\ \end{pmatrix} \delta \left( \vec{r}-\vec{r}' \right). \label{eq:ConstraintPoissonMatrixInverse} \end{equation}

The Dirac bracket may be constructed, as \begin{equation} \left\{ f\left(\vec{x} \right),g\left( \vec{y} \right) \right\}_D=\left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}-\sum_{i,j=1,2}\iint d\vec{r}d\vec{r}'\left\{ f\left( \vec{x} \right),C_i\left( \vec{r} \right) \right\}Q^{-1}_{ij}\left( \vec{r},\vec{r}' \right)\left\{ C_j\left( \vec{r}' \right),g\left( \vec{y} \right) \right\}=\left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}-R_{12}-R_{21}. \label{} \end{equation} Now for $R_{12}$ one finds \begin{equation} R_{12}=\int d\vec{r}\left( \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}-\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} \right), \label{} \end{equation} and for $R_{21}$: \begin{equation} R_{21}=\int d\vec{r}\left( \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}-\frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho} \right). \label{} \end{equation} Hence, we have \begin{equation} \left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}_D=\int d\vec{r}\left( \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}- \frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho} + \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\pi_{\theta}}- \frac{\delta f}{\delta\pi_{\theta}}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}+\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}+\frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho}\right)= \int d\vec{r}\left( \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\pi_{\theta}}- \frac{\delta f}{\delta\pi_{\theta}}\frac{\delta g}{\delta\theta} +\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}\right). \label{} \end{equation} Now if we make use of the constraint equation $\pi_{\theta}=-\rho$, we get that the Dirac bracket reduces to \begin{equation} \left\{ f,g \right\}_D=2 \left\{ f,g \right\}_{\rho,\theta}. \label{} \end{equation} So the phase space is reduced to the variables $\rho$ and $\theta$ but the factor of $2$ really shouldn't be there as it leads to inconsistent wave equations for the $\rho$ and $\theta$ variables under e.g. $\dot{\theta}=\left\{ \rho,H_c \right\}_D$. I have tried to add a total time derivative to the Lagrangian density to start with.. For instance \begin{equation} \mathcal{L}\rightarrow\mathcal{L}'=\mathcal{L}+\frac{d}{dt}\left( \rho\theta/2 \right). \label{} \end{equation} But this ends up giving a factor of $4$ instead of $2$.. I have noticed that if the canonical momenta lead to the constraints $C_1=\pi_{\rho}-2\theta\approx 0$ and $C_2=\pi_{\theta}+2\rho\approx 0$, then the Dirac bracket reduces to the Poisson bracket $\left\{ f,g \right\}_{\rho,\theta}$ without any prefactor.. But it doesn't seem possible to add a total time derivative to $\mathcal{L}$ which achieves this. Any thoughts at all?

Thanks!

$\endgroup$
3
$\begingroup$

Your equation only contains first-order time derivatives and so is already of Hamiltonian action integral form: $$ S= \int (p_i\dot q_i -H(p,q)) dt $$ with $$ p_i\mapsto \rho(x),\\ q_i \mapsto \theta(x),\\ i\mapsto x $$ Dirac brackets are therefore unnecessary. So, from the continuum version of $\{p_i,q_j\}=\delta_{ij}$ we read off that $\{\rho(x),\theta(x')\}= \delta(x-x')$.

$\endgroup$
  • $\begingroup$ Thank you for your response. It did seem to me that might be the case. From the $4$ phase space variables $\rho,\theta,\pi_{\rho},\pi_{\theta}$, one of them is zero: $\pi_{\rho}=0$ while the other momentum is simply another field variable: $\pi_{\theta}=-\rho$. It was clear to me that $-\rho$ and $\theta$ are conjugate variables for the reason you have just stated - computing the Poisson bracket ${\rho(x),\theta(x')}=\delta(x-x')$ over the full phase space variables when supplementing the constraint $\pi_{\theta}=-\rho$. $\endgroup$ – muscaria Jul 31 '18 at 13:29
  • $\begingroup$ But I still wanted to check that the Dirac bracket reduced to the Poisson bracket on the reduced phase space, yet this factor of $2$ pops up which shouldn't! It does seem that it has to do with double counting in some way from the $\pi_{\theta}=-\rho$, but not sure how to get rid of it. $\endgroup$ – muscaria Jul 31 '18 at 13:29
  • $\begingroup$ I up-voted your answer but as I do not have 15 points yet it is not displayed.. $\endgroup$ – muscaria Jul 31 '18 at 13:30
2
$\begingroup$

User mike stone is right. No need to go through the full Dirac-Bergmann analysis of constraints, which is done in this Phys.SE post. The Faddeev-Jackiw method suffice: ${\cal L}$ is already on Hamiltonian first-order form, and $\rho$ and $\theta$ are canonical variables with canonical Poisson brackets $\{\rho({\bf x}),\theta({\bf y})\}=\delta^3({\bf x}-{\bf y})$.

$\endgroup$
  • $\begingroup$ Thanks for reminding me of Jackiw's paper. It has the neatest proof of Darboux that I know. I've been arguing that Dirac brackets are not always necessary for perhaps even longer than Roman J. See for example my ancient Phys. Rev. Lett. 63, 731 (1989)! $\endgroup$ – mike stone Jul 31 '18 at 14:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.