# Dirac bracket for the Madelung (polar) form of the Schrodinger field

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

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 $$\left\{ f,g \right\}=\int d\vec{r}\left(\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta}-\frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}\right)=\left\{ f,g \right\}_{\rho,\theta}. \label{}$$ 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 $$\left\{ f,g \right\}_D=2\left\{ f,g \right\}_{\rho,\theta} \label{}$$ 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') $$\mathcal{L}=\frac{i\hbar}{2}\left( \Psi^*\dot{\Psi}-\dot{\Psi}^*\Psi\right)-\frac{\hbar^2}{2m}\nabla\Psi\nabla\Psi^*. \label{}$$ 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 $\mathcal{L}$: $$\mathcal{L}=-\rho\left(\dot{\theta}+\frac{(\nabla\theta)^2}{2m}\right)-\frac{\hbar^2}{8m\rho}\left( \nabla\rho \right)^2. \label{eq:LagrangianPolarForm}$$ Variation with respect to the field $\theta$ yields an equation of continuity: $$\dot{\rho}+\vec{\nabla}\cdot\vec{J}=0 \label{}$$ where $\vec{J}=\rho\vec{\nabla}\theta/m$, while variation with respect to $\rho$ yields the quantum Hamilton-Jacobi equation: $$\dot{\theta}+\frac{(\nabla\theta)^2}{2m}-\frac{\hbar^2}{2m\sqrt{\rho}}\nabla^2\sqrt{\rho}=0. \label{}$$ 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 $$\mathcal{H}_c=\pi_{\theta}\dot{\theta}+\pi_{\rho}\dot{\rho}-\mathcal{L}\approx \rho\left( \frac{(\nabla\theta)^2}{2m}\right)+\frac{\hbar^2}{8m\rho}\left( \nabla\rho \right)^2. \label{eq:canonicalHamiltonianDensity}$$ 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 $$Q\left( \vec{r},\vec{r}' \right)= \begin{pmatrix} 0 & -1\\ 1 & 0 \\ \end{pmatrix} \delta \left( \vec{r}-\vec{r}' \right), \label{eq:ConstraintPoissonMatrix}$$ whose inverse is $$Q^{-1}\left( \vec{r},\vec{r}' \right)= \begin{pmatrix} 0 & 1\\ -1 & 0 \\ \end{pmatrix} \delta \left( \vec{r}-\vec{r}' \right). \label{eq:ConstraintPoissonMatrixInverse}$$

The Dirac bracket may be constructed, as $$\left\{ f\left(\vec{x} \right),g\left( \vec{y} \right) \right\}_D=\left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}-\sum_{i,j=1,2}\iint d\vec{r}d\vec{r}'\left\{ f\left( \vec{x} \right),C_i\left( \vec{r} \right) \right\}Q^{-1}_{ij}\left( \vec{r},\vec{r}' \right)\left\{ C_j\left( \vec{r}' \right),g\left( \vec{y} \right) \right\}=\left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}-R_{12}-R_{21}. \label{}$$ Now for $R_{12}$ one finds $$R_{12}=\int d\vec{r}\left( \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}-\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} \right), \label{}$$ and for $R_{21}$: $$R_{21}=\int d\vec{r}\left( \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}-\frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho} \right). \label{}$$ Hence, we have $$\left\{ f\left( \vec{x} \right),g\left( \vec{y} \right) \right\}_D=\int d\vec{r}\left( \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}- \frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho} + \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\pi_{\theta}}- \frac{\delta f}{\delta\pi_{\theta}}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\pi_{\rho}}+\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}+\frac{\delta f}{\delta\pi_{\rho}}\frac{\delta g}{\delta\rho}\right)= \int d\vec{r}\left( \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\pi_{\theta}}- \frac{\delta f}{\delta\pi_{\theta}}\frac{\delta g}{\delta\theta} +\frac{\delta f}{\delta\rho}\frac{\delta g}{\delta\theta} - \frac{\delta f}{\delta\theta}\frac{\delta g}{\delta\rho}\right). \label{}$$ Now if we make use of the constraint equation $\pi_{\theta}=-\rho$, we get that the Dirac bracket reduces to $$\left\{ f,g \right\}_D=2 \left\{ f,g \right\}_{\rho,\theta}. \label{}$$ 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 $$\mathcal{L}\rightarrow\mathcal{L}'=\mathcal{L}+\frac{d}{dt}\left( \rho\theta/2 \right). \label{}$$ 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!

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 $\{\rho(x),\theta(x')\}= \delta(x-x')$.
• 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 $\pi_{\theta}=-\rho$. – muscaria Jul 31 '18 at 13:29
• 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. – muscaria Jul 31 '18 at 13:29
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 $\{\rho({\bf x}),\theta({\bf y})\}=\delta^3({\bf x}-{\bf y})$.