In page 179 of Altland and Simon, Condensed Matter Field Theory, the author obtained the action

\begin{equation} S[\theta]=\frac{1}{2\pi}\int dx\,d\tau\,\left[(\partial_x\theta)^2+(\partial_\tau\theta)^2\right] \tag{4.48b} \end{equation}

The author then obtained the canonical momentum corresponding to $\theta$ as

$$\pi_\theta=\partial_{\partial_\tau \theta}\mathcal{L}=\partial_\tau\theta/\pi.\tag{4.48c}$$

According to Hamiltonian mechanics,

\begin{equation} \mathcal{H}=\dot{q}\frac{\partial\mathcal{L}}{\partial\dot{q}}-\mathcal{L}=\dot{q}p-\mathcal{L} \end{equation}

taking $\theta\leftrightarrow q$ and making use of $\partial_\tau\theta=\pi\pi_\theta$, we should have \begin{align} \mathcal{H}&=(\partial_\tau\theta)\pi_\theta-\frac{1}{2\pi}\left[(\partial_x\theta)^2+(\partial_\tau\theta)^2\right]\\ &=\frac{1}{2\pi}\left[\pi^2\pi_\theta^2-(\partial_x\theta)^2\right]. \end{align}

However, this expression is different from the Hamiltonian density given in the textbook $$\mathcal{H}=\frac{1}{2\pi}\left[(\partial_x\theta)^2+\pi^2\pi_\theta^2\right].\tag{4.48d}$$ What did I do wrong here? How to obtain the Hamiltonian given in the textbook? And also how to obtain the new action

\begin{equation} S[\theta,\pi_\theta]=\frac{1}{2}\int dx\,d\tau\,\left(\frac{1}{\pi}(\partial_x\theta)^2 +\pi\pi_\theta^2 +2i\partial_\tau\theta\pi_\theta\right)~?\tag{4.48e} \end{equation}

In particular, where is the last term in the parenthesis $2i\partial_\tau\theta\pi_\theta$ coming from?

  • 1
    $\begingroup$ Looks to me like the first action you wrote deals with imaginary time, while Hamiltonian formalism uses real-time, hence the minus sign. $\endgroup$
    – Dimitri
    Aug 30, 2016 at 9:49
  • 1
    $\begingroup$ Thanks Dimitri, I think it is that. The Hamiltonian above is in imaginary time hence the minus sign. However, I still don't understand how to obtain the last term in the action $2i\partial_\tau \theta\pi_\theta$, unless there is the typo in the text. $\endgroup$
    – JNL
    Sep 11, 2016 at 5:47

2 Answers 2


TL;DR: The trick is not to Wick-rotate the momentum field $$ \Pi_M~=~i\Pi_E, \tag{1}$$ because it would otherwise lead to a divergent Gaussian momentum integral in the Euclidean (E) path integral. So we will keep the Minkowski (M) momentum $\Pi_M\in\mathbb{R}$ even in the Euclidean formulation.

Further details: Standard conventions for the Wick rotation are $$ -S_E~=~iS_M, \qquad t_E~=~it_M, \qquad {\cal L}_E~=~-{\cal L}_M, \tag{2}$$ cf. p. 106 in Ref. 1.

The potential density is $${\cal V}~=~\frac{1}{2\pi}(\partial_x\Theta)^2.\tag{3}$$

The Minkowski & Euclidean Hamiltonian densities read

$${\cal H}_M~=~\frac{\pi}{2}\Pi_M^2+{\cal V},\tag{4M}$$ $${\cal H}_E~=~\frac{\pi}{2}\Pi_E^2-{\cal V}~=~-\frac{\pi}{2}\Pi_M^2-{\cal V}.\tag{4E}$$

The Minkowski & Euclidean Hamiltonian Lagrangian densities read

$$\begin{align} {\cal L}_H^M&~=~\Pi_M\frac{d\Theta}{dt_M} - {\cal H}_M ~\stackrel{(4M)}{=}~ \Pi_M\frac{d\Theta}{dt_M}-\frac{\pi}{2}\Pi_M^2-{\cal V} \cr &\stackrel{\text{int. out } \Pi_M}{\longrightarrow}\quad {\cal L}_M~=~\frac{1}{2\pi}\left(\frac{d\Theta}{dt_M}\right)^2 - {\cal V}. \tag{5M} \end{align}$$ $$\begin{align} {\cal L}_H^E&~=~\Pi_E\frac{d\Theta}{dt_E} - {\cal H}_E ~\stackrel{(4E)}{=}~ \color{Red}{-}i\Pi_M\frac{d\Theta}{dt_E} + \frac{\pi}{2}\Pi_M^2+{\cal V}\cr &\stackrel{\text{int. out } \Pi_M}{\longrightarrow} \quad {\cal L}_E~=~\frac{1}{2\pi}\left(\frac{d\Theta}{dt_E}\right)^2 + {\cal V}. \tag{5E}\end{align}$$ The last expression in eq. (5E) corresponds to OP's first eq. (4.48b). The second expression in eq. (5E) corresponds to OP's sought-for eq. (4.48e), although with an opposite sign marked in red. We have not investigated further the origin of this discrepancy.


  1. A. Altland & B. Simons, Condensed matter field theory, 2nd ed., 2010.
  • 1
    $\begingroup$ I believe the + sign in 4.48e is a mistake in Altland's book. Indeed, in the next paragraphs he writes the Noether current with a minus sign ($-i\partial_x\phi/\pi$). $\endgroup$
    – Patrick
    Jan 20, 2018 at 17:09
  • 1
    $\begingroup$ OK, then we are at least 2 that think there is a sign mistake in Ref. 1. $\endgroup$
    – Qmechanic
    Jan 21, 2018 at 21:12

Here $\tau$ is imagine time, which may lead to the difference with real time condition. You can first let $\tau = it$ and get a result, then let $t=-i\tau$ to the imagine time. Maybe you can get the result.


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