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I would like to write a simple QFT simulation for a free scalar field with a cubic interaction term. However, I got stuck a bit. I will try to describe what I think I understand.

I want to have a look at a field with Lagrangian density

$$\mathcal{L} = \frac{1}{2} \dot{\phi}^2 - \frac{1}{2} \phi'^2 - K \phi^3.$$

Then, for a classical field, the equation of motion (from Euler-Lagrange equation) would be

$$\ddot{\phi} - \phi'' = -3K \phi^2.$$

However, I would like to work with a quantum field. Using the standard procedure of second quantisation, I should construct the Hamiltonian

$$H = \int \mathrm{d} x \left [ \frac{1}{2} \Pi^2 + \frac{1}{2} \phi'^2 + K \phi^3 \right ]$$

where $\Pi = \partial \mathcal{L} / \partial \dot{\phi}$.

Now I should make a transformation $\Pi(x,t) \to \hat{\Pi}(x,t)$ and $\phi(x,t) \to \hat{\phi}(x,t)$. So far so good. Defining commutation relation $[\hat{\phi}(x,t),\hat{\Pi}(x',t')] = i \delta(x-x',t-t')$, I should somehow make contact with the ladder operators $a$ and $a^{\dagger}$. However, I don't really see how. For a free field, I find the free classical wave-like solutions, and then promote the weights of different modes to operators. For a field with interaction, I don't really know what to do. This is my first problem.

Secondly, once I make contact with ladder operators, I can then express any state of the system as a concatenation of ladder operators on the vacuum state $|0\rangle$ such as $a^{\dagger}_{p_2} a^{\dagger}_{p_1} |0\rangle$. A general state should then be described as a superposition of all possible states of all particle numbers

$$|\psi\rangle = \left( \lambda_0 + \sum_{\mathrm{p_1}} \lambda_1(p_1) a^{\dagger}(p_1) + \sum_{\mathrm{p_1,p_2}} \lambda_2(p_1,p_2) a^{\dagger}(p_2) a^{\dagger}(p_1) + \begin{bmatrix}\mbox{higher particle}\\\mbox{number states}\end{bmatrix} \right) |0\rangle.$$

Such state has parameters functions $\lambda_0$, $\lambda_1(p_1)$, $\lambda_2(p_1,p_2)$ etc. giving amplitudes of different particular particle number states with different momenta.

Now, I am confused as to whether the evolution of the system should be entirely contained in the $\lambda$s as functions of time, or whether the ladder operators should evolve in time. In any case, how can I get the equation of motion of the QFT system? This is my second problem.

I would be grateful if anyone could help me with this. In particular, my goal is to see the functions $\lambda$ evolving in time on a computer.

Thanks a lot.

SSF

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    $\begingroup$ Your first commutation relations are wrong. They should be at equal time, i.e $[{\hat \phi} (t,x) , {\hat \Pi}(t,x') ] = i \delta(x-x')$. $\endgroup$ – Prahar Nov 3 '15 at 17:40
  • $\begingroup$ I get the $i$ bit - that was a mistake. But why should I write them manifestly at the same time. After all, the objects $\phi$ and $\Pi$ in classical field theory were functions of both space and time. $\endgroup$ – SSF Nov 3 '15 at 18:19
  • $\begingroup$ In canonical quantization, one quantizes the theory on an arbitrary time slice $t$. One then defines fields and their conjugate momenta on this time slice and finally, imposes "equal-time commutation relations" on this time slice. Everything is done on a single time slice. To move between slices, you can then use the time-evolution operator. $\endgroup$ – Prahar Nov 3 '15 at 18:33
  • $\begingroup$ Indeed, one of the issues people have with the Hamiltonian formulation is that it breaks manifest Lorentz invariance by picking out a special time as described above. $\endgroup$ – Okazaki Nov 3 '15 at 18:58
  • $\begingroup$ Note that the system you are simulating is unstable. It has no vacuum. You're going to get a mess if you try to simulate it on a computer. Try starting instead with the simple harmonic oscillator, considered as a 1d QFT. $\endgroup$ – user1504 Nov 3 '15 at 20:26
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Well, I think that first of all, you should understand, that the reason why you can decompose free field into plain waves is that equations of motions are linear. In case of interaction, equations are motions are nonlinear, so any linear combination of its solutions is no longer a solution.

In the second place, about your second question: it depends on representation: in Heisenberg's representation operators involve in time, in Schroedinger - they don't evolve, but the state evolves. In QFT in case of nonzero interaction the most usable way is to work in Dirac's representation.

The commutation relation that you wrote (be careful - it should be taken at equal times, without delta function like you wrote) holds for equal times in Heisenberg's representation and for any times in Schroedinger's one.

But tell me, I don't understand - what do you mean by 'simulation'? What do you want to get?

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    $\begingroup$ Thanks. I want to write a program that evolves the system in time and shows me how it behaves (i.e. what the probabilities of difference particle numbers are etc). $\endgroup$ – SSF Nov 9 '15 at 20:59
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    $\begingroup$ To find probabilities you should work with Green's functions. In case of free field such probability is just a Feynman's propagator. But if you have an interaction, those propagators are modified and this modifications can be obtained with perturbation theory. So I think you can do such simulations only within given order of perturbation. $\endgroup$ – newt Nov 10 '15 at 7:23
  • $\begingroup$ I see. I thought that there might be a general way - without the need for a perturbative series solution. The same way, when in ordinary QM, I can just use $e^{-i \hat{H} t} \approx 1 - i\hat{H}t$ and evolve the system step by step, even without knowing the actual solution to $\hat{H}$. $\endgroup$ – SSF Nov 10 '15 at 9:16
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    $\begingroup$ Have a look at lattice QCD and lattice gauge theory, it seems as if that is the popular way to simulate in a non-perturbative fashion. There it seems you don't really "timestep" anything, instead you generate 4D configurations stochastically, weigh them and add up their amplitudes. Would be cool if there was another (tractable) way to numerically do this though. $\endgroup$ – BjornW Jun 9 '17 at 14:02

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