# QFT-Style Lagrangian for a system of two symmetrized bosons

I'm wondering if anybody may have suggestions regarding the following problem. The Hamilton operator of the quantum harmonic oscillator (QHO) can be written as follows: $$\hat{\mathcal{H}}_{QHO} = \dfrac{1}{2m} \hat{P}^2_S + \dfrac{m \omega^2}{2} \hat{X}^2_S = \dfrac{1}{2m} \hat{P}^2_H + \dfrac{m \omega^2}{2} \hat{X}^2_H$$ where indices stand for the Heisenberg and Schrödinger pictures, correspondingly.

In the Heisenberg picture, the quantum-mechanical operators obey same differential equation as their classical counterparts: $$\left( \partial^2_t + \omega^2 \right) \hat{X}_H = 0$$ Their solution is $$\begin{cases} \hat{X}_H(t) &= \hat{X}_S \cos (\omega t) + \dfrac{1}{m \omega_0} \hat{P}_S \sin(\omega t) \\ \hat{P}_H(t) &= \hat{P}_S \cos (\omega t) + m \omega_0 \hat{X}_S \sin(\omega t) \end{cases}$$

Evidently, the Lagrange operator, from which such the equation above can be formally derived, mimics the classical one: $$\hat{\mathcal{L}}_{QHO} = \dfrac{1}{2} \hat{X}_H \left( - \partial^2_t - \omega^2 \right) \hat{X}_H$$

The latter equation is precisely the manifestation of the fact that the QHO can be ragarded as the $(0+1)$ Klein-Gordon equation. See detailed treatment e.g. in 1.

Now, here is my question. I would like to know how to write an analogous Lagrange operator for a system of, say, two non-interacting QHOs. It would be interesting to consider two cases: when the particles are distinguishable and when they are not. My guess is that in the first case the Lagrange operator will have exactly same form as for a single QHO ― due to the fact that the Lagrange operator of a quantum system has to have same form as of the classical system (which can be easily identified in this case). Second option seems to be more interesting.

Below you can find my attempt to start. First of all, we postulate that the Hamiltonian of a composite system is but the sum of two copies: $$\hat{\mathcal{H}}_{\text{two/symm}} = \sum \limits_{k=1,2} {}^{(k)}\hat{\mathcal{H}} = \dfrac{1}{2m} \hat{P}^2_S + \dfrac{m \omega^2}{2} \hat{X}^2_S = \dfrac{1}{2m} \hat{P}^2_H + \dfrac{m \omega^2}{2} \hat{X}^2_H$$

Next, we need to derive the differential equation for the operators $\hat{P}_H$ and $\hat{X}_H$, similar to what we have for a single QHO. But I am no longer sure how to write it. An obvious suggestion is that it's same as for two distinguishable particles, but this would require some justification or, at least, hand waving...

P.S. I refrain from using words like 'Hamiltonian' or 'Lagrangian' in order not to confuse the operators with the functionals which arise when QM is treated as a classical field theory: $H = \langle\psi|\hat{\mathcal{H}}|\psi\rangle$, and $L = \pi \dot{\psi} - H$ with $\pi = i \hbar \psi^*$.

P.P.S. Let me provide a motivation for this questions. From the numerous books on advanced QM and solid-state physics (see eg. 2 or 3) we know that the second quantization of the system of bosons can be performed by taking the Schrödinger Lagrangian (precisely the one which I have mentioned in the previous paragraph) and replacing the wave functions with the field operators (which are, btw, non-Hermitean). This may, one obtains the Lagrange operator which has the first time derivative. However, nothing stops us from considering a system of only $N=2$ identical non-relativistic particles. Which is precisely the case I consider in my question. So, the solid-state approach gives the first time derivative, while the above described approach - the second time derivative, just like for a single QHO (or not??).

-------------------------------- UPDATE --------------------------------

I am slowly coming to the conclusion that the order of the time derivative in the Langrangian is not something fundamental. Let me explain what I mean by that. First of all, the Lagrangian $L[\psi]$ can always be replaced by another $\xi(L[\psi])$, assuming the latter has same minima as the former. This is the trick which we are constantly employing when writing the Lagrangians of the relativistic equations. E.g., for a Klein-Gordon Lagrangian or, say, when replacing Nambu-Gotto string action with the Polyakov action.

The inverse procedure (reducing the second time derivative to the first one) is also well-known, since the order of a system of differential equations can always be decreased by adding extra variables and equations.

Let me provide few examples of switching between the first and second time derivatives in Lagrangian.

EXAMPLE 1 (2nd to 1st time derivative)

One of the most beautiful examples of going from the second to the first time derivative in the Lagrangian is related to the Klein-Gordon equation. Namely, it's the Feshbach-Villars representation, see e.g. 3, 1 or 4. The Klein-Gordon equation and the corresponding Lagrangian can be written in the two-component form as follows: \begin{gather} (\Box + m^2) \psi = 0 \\ \phi_+ = \dfrac{1}{\sqrt{2m}} \left( \phantom{-} i\dfrac{\partial}{\partial t} + m \right) \psi \\ \phi_- = \dfrac{1}{\sqrt{2m}} \left( - i\dfrac{\partial}{\partial t} + m \right) \psi \\ \phi = \begin{pmatrix} \phi_+ \\ \phi_- \end{pmatrix} \\ H = \begin{pmatrix} m +\dfrac{\bf{p}^2}{2m} & \dfrac{\bf{p}^{2}}{2m} \\ -\dfrac{\bf{p}^2}{2m} & -m-\dfrac{\bf{p}^2}{2m} \end{pmatrix} \\ i \partial_t \phi = H \phi \\ \bar{\phi} = \phi^\dagger \tau_3 \\ L = i \bar{\phi} \partial_t \phi - \dfrac{1}{2m} \nabla \bar{\phi} (\tau_3 + i \tau_2) \nabla \phi - m \bar{\phi} \tau_3 \phi = \bar{\phi} (i \partial_t - H) \phi \end{gather}

Here we notice a typical feature of the Schrödinger-like Lagrangians: the canonical momentum does not contain a time derivative of the wave function, instead it is obtained via some sort of conjugation (same in the Dirac case).

EXAMPLE 2 (same, applied to the Lagrange operator of the QHO)

What follows is closely related to my original question about the second-quantised form of the Lagrange operator for a system of non-interacting non-relativistic particles.

As I have already mentioned above, the Lagrange operator of the QHO can be written as a $(0+1)$-dimensional second-quantized Klein-Gordon Lagrangian. If we apply Feshbach-Villars machinery to the harmonic oscillator, we get:

\begin{gather} \left( \partial^2_t + \omega^2 \right) \hat{X}_H = 0 \\ \hat{\phi}_+ = \alpha \dfrac{1}{\sqrt{2\omega}} \left(\phantom{-} i \partial_t + \omega\right) \hat{X}_H \\ \hat{\phi}_- = \alpha \dfrac{1}{\sqrt{2\omega}} \left(- i \partial_t + \omega\right) \hat{X}_H \\ \alpha = \sqrt{m} \\ \hat{\phi}_+ = \sqrt{\dfrac{m \omega}{2}} (\hat{X} + \dfrac{i}{m \omega} \hat{P}) = \hat{a}\\ \hat{\phi}_- = \sqrt{\dfrac{m \omega}{2}} (\hat{X} - \dfrac{i}{m \omega} \hat{P}) = \hat{a}^\dagger\\ \hat{\phi} = \begin{pmatrix} \hat{a}^{\phantom{\dagger}} \\ \hat{a}^\dagger \end{pmatrix} \\ \mathcal{H} = \begin{pmatrix} \omega & 0 \\ 0& -\omega \end{pmatrix} \\ i \partial_t \hat{\phi} = \mathcal{H} \hat{\phi} \\ \hat{\mathcal{L}} = \bar{\hat{\phi}} (i \partial_t - \mathcal{H}) \hat{\phi} \end{gather}

Let me summarize now the intermediate result which I have achieved.

When treating QHO as a $(0+1)$-dimensional system, the equation of motion for the operator in the Heisenberg picture can be derived from:

1) The 'Lagrangian' with the second time derivative (formally looks like a 'second-quantized' Klein-Gordon equation). In this case the evolving operator is the usual position $\hat{X}_H$. $$\left( \partial^2_t + \omega^2 \right) \hat{X}_H = 0$$ $$\hat{\mathcal{L}}_{QHO} = \dfrac{1}{2} \hat{X}_H \left( - \partial^2_t - \omega^2 \right) \hat{X}_H$$ 2) The 'Lagrangian' with the first time derivative (formally looks like a 'second-quantized' Schrödinger equation). In this case the evolving operator is $\hat{\phi} = \begin{pmatrix} a & a^\dagger \end{pmatrix}^T$ \begin{gather} i \partial_t \hat{\phi} = \mathcal{H} \hat{\phi} \\ \hat{\mathcal{L}} = \bar{\hat{\phi}} (i \partial_t - \mathcal{H}) \hat{\phi} \end{gather}

Be mindful however, that the Schrödinger-like operator equation above is not actually the one which we would obtain by usual 'second quantization of the Schrödinger equation'. In that case (see 2, 6 or any book on solid state physics) after taking $N=1$ for the number of particles, we still would have $$\hat{\mathcal{H}}_{QHO} \propto \sum \limits_{k=0}^\infty \hat{a}_k^\dagger\hat{a}_k$$ I have discussed this issue in another question of mine.