# Independent systems and Lagrangians

Definition 1:

The notion of independent systems has a precise meaning in probabilities. It states that the (joint) probability or finding the system ($S_1S_2$) in the configuration ($C_1C_2$) is equal to the probability of finding the system ($S_1$) in the configuration ($C_1$) times the probability of finding the system ($S_2$) in the configuration ($C_2$).

Definition 2:

However, in we consider fields systems, the practical tool is Lagrangians. So I should say that 2 systems are independent if :

$$Lagrangian (S_1S_2) = Lagrangian (S_1) + Lagrangian (S_2)$$

The question:

Now, what is the relationship between these 2 definitions? They could be only compatible, or they could be equivalent in the field domain. Is there a way to "demonstrate" the latter from the former ?

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You might be interested in this question and its answer by David Bar Moshe. (It doesn't answer your question exactly, but seems related.) – Nathaniel May 4 '13 at 9:59
Yes, It is very interesting, of corse, but my question is a different question. I am OK with the $x,p$ parametrization for probabilities, so the question is about considering two independent systems (in this case a $x_1,p_1,x_2,p_2$ parametrization for the probability law) and find a way to express the global Lagrangian as the sum of individual Lagrangians. I cannot see an obvious link between probabilities and Lagrangians. – Trimok May 4 '13 at 15:59
The question I linked to is indeed quite different from yours, but nevertheless I think it's relevant because both questions are about the relationship between probabilities and Lagrangians. In the classical world this relationship occurs because of statistical mechanics, which is the subject of the question I linked to. In the quantum world it's more straightforward because of the path integral formalism, as levitopher points out. – Nathaniel May 4 '13 at 17:22
Can you please state explicitly, are you interested in statistical mechanics, or in quantum mechanics only? – Peter Kravchuk May 4 '13 at 17:49
The problem is the relationship between a probabilistic point of view, and a Lagrangian point of view. So, each theory which uses Lagrangians is OK for me. – Trimok May 4 '13 at 17:56

How about path integrals? The probability that a system evolves between state $|\phi_1\rangle$ and $|\phi_2\rangle$ is

$$\langle \phi_2|\phi_1 \rangle =\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)$$

where the measure $\mathcal{D}\phi$ is suitably defined and the action $S(\phi)$ is the integral of the Lagrangian (over whatever the physical coordinates are).

Consider two systems, described by states $|\phi\rangle$ and $|\psi\rangle$, which are independent. Then the action is

$$S(\phi,\psi)=S(\phi)+S(\psi)$$

And the probability of evolving between two configurations is

\begin{align} \langle\phi_2,\psi_2|\phi_1,\psi_1\rangle&=\int_{(\phi_1,\psi_1)}^{(\phi_2,\psi_2)}\mathcal{D}\phi\mathcal{D}\psi \exp \left(\frac{i}{\hbar}(S(\phi)+S(\psi))\right)\\ &=\int_{\phi_1}^{\phi_2}\mathcal{D}\phi \exp \left(\frac{i}{\hbar}S(\phi)\right)\int_{\psi_1}^{\psi_2}\mathcal{D}\psi \exp \left(\frac{i}{\hbar}S(\psi)\right)\\ &=\langle \phi_2|\phi_1\rangle \langle \psi_2|\psi_2\rangle\\ \end{align}

So the probability is a product if the systems are independent. I picked specific states here but take $|\phi_i\rangle$ describing a system $S_i$ in configuration $C_{i1}$ and I think this gets what you want.

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+1 for you because your idea is very interesting. But there is a problem. What you demonstrate, is, that, if S(ϕ,ψ)=S(ϕ)+S(ψ), then ⟨ϕ2,ψ2|ϕ1,ψ1⟩ = ⟨ϕ2|ϕ1⟩⟨ψ2|ψ2⟩. But the last quantities are transition probabilities, so they are not exactly probabilities to be in some configuration.But maybe, we can prove a relation between these 2 kind of probabilities – Trimok May 4 '13 at 16:07
And interestingly, it is a quantum argument, which is fine for me, because nature is quantum, and maybe your argument is deeper than it looks – Trimok May 4 '13 at 16:12
There is only a little problem of circular argument, when you say, that if |ϕ⟩ and |ψ⟩ are independent, then S(ϕ,ψ)=S(ϕ)+S(ψ). No, it is precisely what we want to demonstrate... But the following logic is correct. – Trimok May 4 '13 at 16:15
...'where the measure $\mathcal{D}\phi$ is suitably defined' so far the most popular way of pretending to bother about the definition=) – Peter Kravchuk May 4 '13 at 16:17
Well, I think that we can consider that the initial state is a kind of unique generic state, or generic configuration, and that the others configurations are calculated from transitions from this generic configuration. So it is maybe almost the answer. – Trimok May 4 '13 at 16:25

Well, you are not specifying the kind of probabilities you are talking about. Therefore, I refer to the tags and assume that we are talking about quantum (not statistical) probabilities. (See the edit at the end)

Then, I have to note that your probabilistic definition of independece does not make much sense.

Edit

Well, according to the comments, I should state that my point is that these definitions are different in their meaning. The first is the statistical independence (and should be refined for the case of QM) and the second is the absence of interaction. The following should be understood as a reference to quantum entanglement as a fact that non-interacting systems can be correlated.

End of edit

First of all, we usually believe that if we have the Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ of two systems, then the Hilbert space for the compound system is $\mathcal{H}=\mathcal{H}_1\otimes\mathcal{H}_2$. This is regardless of the interaction between the systems.

Next, when we talk about the probability of finding the system in state $|a\rangle$ while it is in state $|b\rangle$ (all normalized) we usually mean $|\langle a|b \rangle|^2$. Now, consider the following state in $\mathcal{H}$: $$|a\rangle=\frac{1}{\sqrt{2}}|1\rangle\otimes|\alpha\rangle+\frac{1}{\sqrt{2}}|2\rangle\otimes|\beta\rangle$$ There is $0$ probability to find such system in the state $|1\rangle\otimes|\beta\rangle$ while there is $0.5$ prabability to find the first system in state $|1\rangle$ and $0.5$ probability to find the second system in the state $|\beta\rangle$. And such a state exists regardless of the interactions between the systems. See quantum entanglement.

Also, we can consider the state $$|a\rangle=|1\rangle\otimes|\alpha\rangle$$ for which your rule more or less works.

Finally, we see that there are states of non-interacting systems such that your probabilistic equality does not hold, as well as there are states of interacting systems such that your equality holds. So, there is no connection between this equality and independence of the systems.

What is true, however, that the transition amplitudes (that is, matrix elements of $\exp(-iHt/\hbar)$) are indeed multiplicative for independent systems. That is because $$\exp(-iHt)=\exp(-it(H_1\otimes 1+1\otimes H_2))=\exp(-it H_1\otimes 1)\exp(-it1\otimes H_2)\\ =\exp(-iH_1t)\otimes\exp(-iH_2t)$$ acts on the components of tensor product independently (also see levitopher's answer for path integral pespective).

What about the converse? We cannot say that any matrix element for the whole system is the product of the matrix elements for the separate systems because a priori we dont have the latter. We have to say something different, but I am now not sure how to state it conveniently.

Edit The tags have changed a bit, but still we can construct density matrices (and pretend that they correspond to some non-equilibrium situations) with similar properties. I think that the pure-state density matrix is ok.

For equilibrium density matrices, namely $\rho=\exp(-\beta H)$ it is true that probablilities mulptily for independent systems by the same argument as for the evolution operator. The same can be done via Euclidian path integrals.

As for the converse, I do not know the answer at the moment.

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I don't understand your point.My probabilistic definition of independence is correct. When you exhibit your entangled state, you cannot consider it as a separable state, and so it is not a system which will be the tensorial product of 2 sub-systems 1 and 2. For the entangled state, there is no notion of independent sub-systems, the density matrix is not a tensorial product, and so on. And, after, you make a circular argument with the hamiltonians. It is not a demonstration because you admit that for independent systems, H = H1⊗1+1⊗H2. But this is precisely what we want to demonstrate ! – Trimok May 4 '13 at 17:32
@Trimok: First, about the Hamiltonians. Sorry, I thought that you are interested in the consequence is either direction. Second, about your first point. Systems are never 'tensorial products', but states can be. Indeed, my state is indecomposable, but there is the notion of independent sub-systems. I can entangle two photons, send the first to the mars and measure the second in my lab. I can do this many times and estimate the probability distribution for some characteristic the second proton, and I can ask my friend on mars to do the same with the first. – Peter Kravchuk May 4 '13 at 17:39
@Trimok, in mathematical language, probability to find the first system in a state $|a\rangle$ is the squared norm of the projection of the state of the full system onto the subspace $|a\rangle\otimes\mathcal{H}_2$. – Peter Kravchuk May 4 '13 at 17:42
No, if your state is entangled (and in your example it is a maximally entangled state), there is no notion of independent sub-systems. These are correlated sub-systems. And the spatial separation has nothing to do with this. – Trimok May 4 '13 at 17:51
@Trimok. Ok, so we are using different definitions of independence. Then my claim is 'Definition 2 does not imply Definition 1, negation of Definition 1 does not imply negation of Definition 2' – Peter Kravchuk May 4 '13 at 17:54