The usual Schrodinger Lagrangian is $$ \tag 1 i(\psi^{*}\partial_{t}\psi ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi, $$ which gives the correct equations of motion, with conjugate momentum for $\psi^{*}$ vanishing. This Lagrangian density is not real but differs from a real Lagrangian density $$ \tag 2 \frac{i}{2}(\psi^{*}\partial_{t}\psi -\psi \partial_{t}\psi^{*} ) + \frac{1}{2m} \psi^{*}(\nabla^2)\psi $$ by a total derivative.

My trouble is that these two lagrangian densities lead to different conjugate momenta and hence when setting equal time commutation relations, I am getting different results (a factor of 2 is causing the problem). I can rescale the fields but then my Hamiltonian also changes. Then applying Heisenberg equation of motion, I don't get the operator Schrodinger equation.

Is it possible to work with the real Lagrangian density and somehow get the correct commutation relations? I would have expected two Lagrangians differing by total derivative terms to give identical commutation relations (since canonical transformations preserve them). But perhaps I am making some very simple error. Unless all conjugate momenta are equivalent for two Lagrangians differing by total derivatives, how does one choose the correct one?

I guess the same thing happens for other first order systems like Dirac Lagrangian also.

  • 3
    $\begingroup$ I don't have time for a detailed reply to your question, but it may help to have a look at the end of Sec. 7.2 in Weinberg's textbook (vol. 1). He discusses the effect of adding a total time derivative to the Lagrangian and shows that while modifying the canonical momentum, it does not affect the commutation relations of the theory. $\endgroup$ Sep 30, 2011 at 10:17
  • $\begingroup$ This may help: arxiv.org/abs/hep-th/0301052 $\endgroup$
    – Quillo
    Apr 7, 2021 at 19:43

1 Answer 1


Here we will for simplicity only consider the Schrödinger system. We will assume that

$$\phi~=~(\phi^1+i\phi^2)/\sqrt{2} \tag{A}$$

is a bosonic complex field, and that

$$\phi^*~=~(\phi^1-i\phi^2)/\sqrt{2} \tag{B} $$

is the complex conjugate, where $\phi^a$ are the two real component fields, $a=1,2$. [Note the change in notation $\psi\longrightarrow\phi$ as compared with the OP's question (v1).]

  1. The Lagrangian density

$${\cal L}~:=~ i\phi^{*}\dot{\phi} + \frac{1}{2m} \phi^* \nabla^2\phi \tag{C} $$

for the Schrödinger field $\phi$ is already on the Hamiltonian form

$${\cal L}~=~ \pi\dot{\phi} - {\cal H}. \tag{D} $$

Simply define complex momentum

$$\pi~:=~i \phi^{\ast}, \tag{E} $$

and Hamiltonian density

$${\cal H}~:=~-\frac{1}{2m} \phi^{\ast} \nabla^2\phi. \tag{F} $$

More generally, this identification is a simple example of the Faddeev-Jackiw method.

  1. Recall that the Euler-Lagrange equations do not change by adding a $4$-divergence $d_{\mu}\Lambda^{\mu}$ to the Lagrangian density

$${\cal L} ~~\longrightarrow~~ {\cal L}^{\prime}~:=~{\cal L} + d_{\mu}\Lambda^{\mu},\tag{G}$$

cf. e.g. this Phys.SE post. [We use the symbol $d_{\mu}$ (rather than $\partial_{\mu}$) to stress the fact that the derivative $d_{\mu}$ is a total derivative, which involves both implicit differentiation through the field variables $\phi^a(x)$, and explicit differentiation wrt. $x^{\mu}$.] Therefore, we can (via spatial integration by parts) choose an equivalent Hamiltonian density

$$\begin{align}{\cal H} ~~\longrightarrow~~ {\cal H}^{\prime}~:=~&\frac{1}{2m}|\nabla\phi|^2\cr ~=~&\frac{1}{4m}(\nabla\phi^1)^2 +\frac{1}{4m}(\nabla\phi^2)^2,\end{align}\tag{H} $$

and we can (via temporal integrations by part) choose an equivalent kinetic term

$$\begin{align} i\phi^*\dot{\phi}~=~ \pi\dot{\phi} ~~\longrightarrow~&~ \frac{1}{2}(\pi\dot{\phi}-\phi\dot{\pi})\cr ~=~& \frac{i}{2}(\phi^*\dot{\phi}-\phi\dot{\phi}^*)\cr ~=~&\frac{1}{2}(\phi^2\dot{\phi}^1-\phi^1\dot{\phi}^2)\cr ~~\longrightarrow~&~\phi^2\dot{\phi}^1. \end{align}\tag{I} $$

The last expression shows (in accordance with the Faddeev-Jackiw method) that

$$ \text{The second component }\phi^2 \\ \text{ is the momenta for the first component }\phi^1. \tag{J}$$

  1. Alternatively, we can perform a Dirac-Bergmann analysis$^1$ directly. Consider for instance the Lagrangian density

$${\cal L}^{\prime}~=~ (\alpha+\frac{1}{2})\phi^2\dot{\phi}^1+(\alpha-\frac{1}{2})\phi^1\dot{\phi}^2 - {\cal H}^{\prime},\tag{K} $$

where $\alpha$ is an arbitrary real number. [The term $d(\phi^1\phi^2)/ dt$, which is multiplied by $\alpha$ in ${\cal L}^{\prime}$, is a total time derivative.] Let us check that the quantization procedure does not depend on this parameter $\alpha$. We introduce canonical Poisson brackets

$$\begin{align} \{\phi^a({\bf x},t),\phi^b({\bf y},t)\}_{PB} ~=~&0, \cr \{\phi^a({\bf x},t),\pi_b({\bf y},t)\}_{PB} ~=~&\delta^a_b ~ \delta^3 ({\bf x}-{\bf y}), \cr \{\pi_a({\bf x},t),\pi_b({\bf y},t)\}_{PB} ~=~&0,\end{align} \tag{L}$$

in the standard way. The canonical momenta $\pi_a$ are defined as

$$\begin{align} \pi_1~:=~&\frac{\partial {\cal L}^{\prime}}{\partial \dot{\phi}^1} ~=~(\alpha+\frac{1}{2})\phi^2,\cr \pi_2~:=~&\frac{\partial {\cal L}^{\prime}}{\partial \dot{\phi}^2} ~=~(\alpha-\frac{1}{2})\phi^1.\end{align}\tag{M}$$

These two definitions produce two primary constraints

$$\begin{align}\chi_1~:=~&\pi_1-(\alpha+\frac{1}{2})\phi^2~\approx~0,\cr \chi_2~:=~&\pi_2-(\alpha-\frac{1}{2})\phi^1~\approx~0,\end{align}\tag{N}$$

where the $\approx$ sign means equal modulo constraints. The two constraints are of second-class, because

$$ \{\chi_2({\bf x},t),\chi_1({\bf y},t)\}_{PB}~=~\delta^3 ({\bf x}-{\bf y})~\neq~0. \tag{O} $$

Thus the Poisson bracket should be replaced by the Dirac bracket. [There are no secondary constraints, because

$$\begin{align} \dot{\chi}_a({\bf x},t) ~=~&\{\chi_a({\bf x},t), H^{\prime}(t)\}_{DB} ~=~ 0, \cr H^{\prime}(t)~:=~& \int d^3y \ {\cal H}^{\prime}({\bf y},t),\end{align} \tag{P} $$

are automatically satisfied.] The Dirac bracket between the two $\phi^a$'s is

$$\{\phi^1({\bf x},t),\phi^2({\bf y},t)\}_{DB}~=~\delta^3 ({\bf x}-{\bf y}), \tag{Q}$$

leading to the same conclusion (J) as the Faddeev-Jackiw method. Note that the eqs. (O) and (Q) are independent of the parameter $\alpha$.

  1. In all cases, the canonical equal-time commutator relations for the corresponding operators become

$$\begin{align} [\hat{\phi}^1({\bf x},t), \hat{\phi}^2({\bf y},t)] ~=~& i\hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}), \cr [\hat{\phi}({\bf x},t), \hat{\phi}^{\dagger}({\bf y},t)] ~=~& \hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}), \cr [\hat{\phi}({\bf x},t), \hat{\pi}({\bf y},t)] ~=~& i\hbar {\bf 1}~\delta^3 ({\bf x}-{\bf y}).\end{align} \tag{R}$$


$^1$ See, e.g., M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1992.

  • $\begingroup$ Thanks a lot to Qmechanic for very detailed answer. I really appreciate the help. $\endgroup$
    – user5468
    Oct 8, 2011 at 17:39
  • 1
    $\begingroup$ Notes for later: The Schrödinger action for multiple particles is $$I[\psi]~=~\int\! dt~\left[ \prod_{j=1}^N d^3{\bf x}_j \right] {\cal L}.$$ $\endgroup$
    – Qmechanic
    Jun 26, 2018 at 14:02
  • 1
    $\begingroup$ Notes for later: The Schrödinger Lagrangian density is $$ {\cal L} ~=~\frac{i\hbar}{2}(\psi^{\ast}\dot{\psi}-\dot{\psi}^{\ast}\psi) - \sum_{j=1}^N\frac{\hbar^2}{2m_j}|\nabla_j\psi|^2 -V |\psi|^2$$ $$~=~ -\rho\dot{S}-\sum_{j=1}^N\frac{\hbar^2}{2m_j}(\nabla_j\sqrt{\rho})^2-\sum_{j=1}^N\frac{\rho}{2m_j}(\nabla_j S)^2 -\rho V,$$ where we rewrote the wavefunction $\psi=\sqrt{\rho}\exp\left(\frac{i}{\hbar}S\right)$ in "polar" coordinates $\rho$ and $S$. $\endgroup$
    – Qmechanic
    Jun 26, 2018 at 14:12
  • 1
    $\begingroup$ Notes for later: Case $N=1$: Note that $\rho$ and $S$ are canonical variables $\{\rho({\bf x}),S({\bf y})\}=\delta^3({\bf x}-{\bf y})$, and that ${\cal L}$ is already on Hamiltonian first-order form. The Schrödinger picture suggests second quantization, cf. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 26, 2018 at 14:59
  • 1
    $\begingroup$ Notes for later: The EL eqs. wrt. $\rho$ and $S$ are $$\dot{S} - \sum_{j=1}^N\frac{\hbar^2}{2m_j\sqrt{\rho}}\nabla_j^2\sqrt{\rho} +\sum_{j=1}^N\frac{1}{m_j}(\nabla_jS)^2 + V ~\approx~ 0\qquad\text{and}\qquad \dot{\rho} + \sum_{j=1}^N\frac{1}{m_j}\nabla_j \cdot (\rho \nabla_jS) ~\approx~ 0,$$ respectively. The Madelung equations with ${\bf u}_j :=\frac{1}{m_j}\nabla_jS$ are straightforward consequences (and analogues of Navier–Stokes eqs). See also de Broglie–Bohm pilot-wave theory. $\endgroup$
    – Qmechanic
    Jun 29, 2018 at 16:55

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