Let us consider a LC circuit containing an electric dipole moment, the quantum system (electric field $E$ coupled with a dipole moment) can be described by the path integral $$Z=\int DEDxe^{i\int dtL},$$ where the total Lagrangian $$L=\frac{1}{2g}(\dot{E}^2-\omega _{LC}^2E^2)+\frac{m}{2}\dot{x}^2-\frac{m\omega _{0}^2}{2}x^2+exE.$$ After integrating out the dipole $x$, we obtain an effective Lagrangian $L_{eff}$ for the electric field $$L_{eff}=\frac{1}{2g}(\dot{E}^2-\omega _{LC}^2E^2)+\frac{e^2}{2m}E(\partial_t^2+\omega _{0}^2)^{-1}E.$$

On the other hand, from the classical point of view, by solving the total Lagrangian $L$, we can obtain a 4th order equation of motion for the electric field $$[\partial_t^4+(\omega _{0}^2+\omega _{LC}^2)\partial_t^2+\omega _{0}^2\omega _{LC}^2-\frac{e^2g}{m}]E=0.\tag{a}$$

My questions are:

  1. Can the second term in $L_{eff}$ be written as a function of $\dot{E}$ and $E$?

  2. Can we derive an 'Euler-Lagrangian' equation from the effective Lagrangian $L_{eff}$? If yes, is this equation the same as the above 4th order equation of motion $(a)$ of the classical system?

  3. Can we construct another effective Lagrangian from the classical dynamics which gives rise to Eq.$(a)$? Is the concept of effective Lagrangian ONLY meaningful for the quantum system?

  • 1
    $\begingroup$ I think I can answer my Q3 now: Classically, without the path integral method, express $x$ in terms of $E$ by solving the Euler-Lagrangian equation for $x$ derived from the total Lagrangian $L$, then substitute $x$ back into $L$, and we will obtain the same effective Lagrangian $L_{eff}$ containing only $E$. $\endgroup$ – Kai Li Aug 6 '14 at 14:24
  • $\begingroup$ I remember that what one gets from the EOMs of $L$'s corresponding $H$ can be regarded as not only classical EOMs but also operator equations in the sense of canonical quantization. As shown by you, classical $L_{eff}$ is equal to quantum $L_{eff}$. Therefore, I guess this just means canonical quantization reproduces the same result as path integral does. Not sure anyway... How do think? $\endgroup$ – xiaohuamao Sep 20 '14 at 15:39

1 : I don't think so

2 : Note that $L_{eff}$, may be written, thanks to an integration by parts $(\partial_t E)^2 = \partial_t(E\partial_t E) - E \partial_t^2E$, and neglecting the surface term due to the the total derivative :

$$ L_{eff}=E\quad (\frac{1}{2g}(-\partial_t^2 -\omega _{LC}^2)+\frac{e^2}{2m}(\partial_t^2+\omega _{0}^2)^{-1})\quad E \tag{1}$$

The equation of movement is then :

$$(\frac{1}{2g}(-\partial_t^2 -\omega _{LC}^2)+\frac{e^2}{2m}(\partial_t^2+\omega _{0}^2)^{-1})\quad E = 0 \tag{2}$$

Multiplying $(2)$ by $(\partial_t^2+\omega _{0}^2)$ gives you the equation $(a)$

3 : Different lagrangians may give the same equation of movement.

  • $\begingroup$ Wow, your explanation to Q2 is very clear, thank you very much. BTW, I'm not clear how you get Eq(2) from Lagrangian (1)? I only know the so-called Euler-Lagrangian equation derived from an ordinary Lagrangian $L(E,\dot{E})$ which is a function of only coordinate and velocity variables. $\endgroup$ – Kai Li Aug 2 '14 at 13:53
  • 2
    $\begingroup$ @K-boy : I concede that it is more a formal derivation, than a standard derivation. I suppose here that a lagrangian of form $L_{eff}=E\, O \, E$, where $O$ is an operator, corresponds to equations of movement $O \,E=0$, because, in some sense, the lagrangian depends only on $E$ (and not on $\dot E$), so you may apply formally the Euler-Lagrange equations, here $\frac{\partial L_{eff}}{\partial E} = 0$ $\endgroup$ – Trimok Aug 3 '14 at 11:16
  • $\begingroup$ :Inspired by your comment, if there exists a least action principle for the effective action $S_{eff}=\int dtL_{eff}$, and further the operator $O$ is assumed to be positive, then $OE=0$ may be one possible solution. So what's your opinion? $\endgroup$ – Kai Li Aug 3 '14 at 13:58
  • $\begingroup$ I must confess I don't see why the positivity of $O$ should be important here. $\endgroup$ – Trimok Aug 4 '14 at 7:54
  • $\begingroup$ If $O$ is not positive, then there exists a function $f(t)$ such that $\int dtfOf<0=\int dtEOE$, where $OE=0$, which means that the solution $E$ does NOT correspond to the minimal action $S_{eff}$. $\endgroup$ – Kai Li Aug 4 '14 at 8:14

OP's system is two coupled harmonic oscillators

$$\tag{1} L~=~\frac{1}{2}(m\dot{x}^2 - k x^2) + \frac{1}{2}(M\dot{y}^2 - K y^2) - \kappa xy. $$

It seems a steep price to pay to create a non-local formulation by integration out one variable by brute force as OP does. Here we instead find the normal modes of the system of two coupled harmonic oscillators.

The equations of motion are

$$\tag{2} \begin{pmatrix}\ddot{x} \\ \ddot{y}\end{pmatrix} ~=~- \Lambda \begin{pmatrix}x \\ y\end{pmatrix}, \qquad \Lambda ~:=~ \begin{pmatrix}\frac{k}{m}&\frac{\kappa}{m} \\ \frac{\kappa}{M} &\frac{K}{M} \end{pmatrix}. $$

Interestingly, the real $2\times 2$ matrix $\Lambda$ is not symmetric if $m\neq M$. The two eigenvalues of $\Lambda$ are real

$$\tag{3} \lambda_{\pm} ~=~ \frac{{\rm tr}(M)}{2} \pm \sqrt{\Delta}, $$

$$\tag{4} \Delta~:=~\left(\frac{{\rm tr}(M)}{2}\right)^2-\det(M)~\geq~0. $$

If the matrix $T$ diagonalizes the matrix

$$\tag{5}\Lambda~=~TDT^{-1}, \qquad D ~:=~ \begin{pmatrix}\lambda_+&0\\ 0 &\lambda_- \end{pmatrix}, $$

then define new variables

$$ \tag{6} \begin{pmatrix}x_+ \\ x_-\end{pmatrix} ~=~T^{-1} \begin{pmatrix}x \\ y\end{pmatrix}. $$

Then the equations of motion are

$$ \tag{7} \ddot{x}_{\pm}~=~-\lambda_{\pm} x_{\pm}, $$

with Lagrangian

$$ \tag{8} \tilde{L}~=~\sum_{\pm} \frac{m_{\pm}}{2}\left(\dot{x}^2_{\pm} - \lambda_{\pm} x^2_{\pm} \right), $$

where the masses $m_{\pm}$ depends on the parameters of the theory.

  • $\begingroup$ ♦ Thanks for your answer. You have given the classical equations of motion using normal coordinates. But I'm not interested in how to solve the classical harmonic oscillators, instead, here what I concerned is the concept of effective Lagrangian, either quantum version or classical version. $\endgroup$ – Kai Li Aug 2 '14 at 13:59
  • $\begingroup$ The method of normal coordinates also works quantum mechanically. $\endgroup$ – Qmechanic Aug 2 '14 at 14:03
  • $\begingroup$ ♦ Yes. And I wonder whether we could define an effective Lagrangian containing only the electric field $E$ from the classical point of view? Thanks. $\endgroup$ – Kai Li Aug 2 '14 at 14:07

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.