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According to "Modern Quantum Chemistry" by Szabo and Ostlund, the Hamiltonian for a system of non-interacting electrons can be written as $$\mathcal{H}=\sum^N_i\mathcal{h(i)}$$ here $\mathcal{h(i)}$ is the one electron Hamiltonian. $\mathcal{h(i)}$ consists of a kinetic energy term for electron $i$ and the potential felt by electron $i$. Why can we add one electron Hamiltonians to get the Hamiltonian for the complete system? In addition, the page about Hamiltonians in wikipedia

the result is that the Hamiltonian of the system is the sum of the separate Hamiltonians for each particle

Intuitively, this makes sense. But is there an mathematical proof? Is there an "if and only if" proof?

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    $\begingroup$ This is a definition, not a theorem (or, more specifically, the QM recipe for quantizing systems with multiple particles). How would such a 'proof' look like? $\endgroup$ – Emilio Pisanty Dec 14 '16 at 17:35
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    $\begingroup$ Apart from @EmilioPisanty's comment, I would say that this is also true in classical mechanics, where the Hamiltonians are just the energies of each particle, and so naturally, the sum of the individual energies is the total energy and therefore the total Hamiltonian. $\endgroup$ – march Dec 14 '16 at 17:55
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I'll write about classical Hamiltonian systems for now. Maybe add later about quantum theory.

Knowing Hamiltonian equation of motion for some system means that you know derivatives of the Hamiltonian, \begin{equation} \frac{\partial H}{\partial P_I}=F_I(P,Q),\quad \frac{\partial H}{\partial Q_I}=G_I(P,Q) \end{equation} You can then recover the Hamiltonian up to the constant by path-independent integration, \begin{equation} H=\int\limits_{(P_0,Q_0)}^{(P,Q)} \sum_I\Big(dP_I\,F_I(P,Q) + dQ_I\,G_I(P,Q)\Big) \end{equation}

That two subsystems with canonical variables $(P_I,Q_I)=\begin{pmatrix}(p_i,q_i)\\(\tilde{p_j},\tilde{q_j})\end{pmatrix}$ don't interact means that $(\dot{p}_i,\dot{q}_i)$ don't depend on $(\tilde{p}_j,\tilde{q}_j)$ and vice versa. From Hamiltonian equation of motion that requires, \begin{align} \frac{\partial H}{\partial p_i}=F_i(P,Q)=f_i(p,q),\quad \frac{\partial H}{\partial q_i}=G_i(P,Q)=g_i(p,q),\\ \frac{\partial H}{\partial \tilde{p}_j}=F_j(P,Q)=\tilde{f}_j(\tilde{p},\tilde{q}),\quad \frac{\partial H}{\partial \tilde{q}_j}=G_j(P,Q)=\tilde{g}_j(\tilde{p},\tilde{q}) \end{align} The above integral then separates into two terms, \begin{equation} H=\left[\int\limits_{(p_0,q_0)}^{(p,q)}\sum_i\Big(dp_i\,f_i(p,q)+dq_i\,g(p,q)\Big)\right] + \left[\int\limits_{(\tilde{p}_0,\tilde{q}_0)}^{(\tilde{p},\tilde{q})}\sum_j\Big(d\tilde{p}_j\,f_j(\tilde{p},\tilde{q})+d\tilde{q}_j\,\tilde{g}_j(\tilde{p},\tilde{q})\Big)\right] \end{equation}

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  • $\begingroup$ Could you elaborate on the relation between $f$ and $F$ (and $g$ and $G$)? I am not how you move from the first Hamiltonian to the second. Is $P$ supposed to be a multivariable coordinate $P=(p,\tilde{p})$? $\endgroup$ – CoffeeIsLife Dec 14 '16 at 19:49
  • $\begingroup$ @QuantumAMERICCINO Exactly, hopefully now it's more clear $\endgroup$ – OON Dec 14 '16 at 20:45
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There can't be a "proof", since this just the hamiltonian choosen to give the right results: In QM, you START with the hamiltonian, which then is supposed to describe your system. It's always "Whoo, this is the Hamiltonian, let's see how my system behaves. Oh, it behaves the same way I can measure stuff. This must be the right Hamiltonian".

However, there are ways to make it plausible to use a certain Hamiltonian. For a single particle you want to reproduce classical mechanics in the limit of $\hbar$ being small. So you take the Hamilton-Function, you've been used to, and replace position and momentum with Operators.

In classical mechanics, for serveral noninteracting particles, you just add the hamilton-functions, because then every particle, now described by the multi-particle-hamilton-function, will behave according to the same equations of motion. And it's the same with Quantum-mechanics:

You want a Hamiltonian, according to which the particles behave independently of each other (that would be the non-interacting hamiltonian by definition), and in the same manner they would act when they were the only particle. The only way to achieve that is to add all the Hamiltonians of the single particles.

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  • $\begingroup$ An intuitive handle is to think that the Hamiltonian is the energy operator, and to start with energy is a conserved scalar additive quantity. So to a first order approximation adding the individual Hamiltonians should not be far wrong, unless diabolic potentials are involved, like the ones in QCD. Higher orders will come from variables mixing $\endgroup$ – anna v Dec 14 '16 at 18:49
  • $\begingroup$ The question whether any Hamiltonian describing independent evolution of two subsystems should be sum of two Hamiltonians for each subsystem is perfectly valid. Your and others opposition to it is ridiculous. $\endgroup$ – OON Dec 14 '16 at 18:51
  • $\begingroup$ I do not oppose the question, I answered it. $\endgroup$ – Quantumwhisp Dec 14 '16 at 19:04
  • $\begingroup$ Well, no. You just stated that "The only way to achieve that is to add all the Hamiltonians of the single particles." $\endgroup$ – OON Dec 14 '16 at 19:06
  • $\begingroup$ As OP said intuitively it makes sense but can we prove that this is the only way. You just state it's the only way and shrug off the question of proof. $\endgroup$ – OON Dec 14 '16 at 19:08

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