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A quote from Landau & Lifshitz (Quantum Mechanics - Non-relativistic Theory, §2):

"Let us consider a system composed of two parts, and suppose that the state of this system is given in such a way that each of its parts is completely described. Then we can say that the probabilities of the coordinates q_1 of the first part are independent of the probabilities of the coordinates q_2 of the second part, and therefore the probability distribution for the whole system should be equal to the product of the probabilities of its parts."

How does independence of the two systems follow from the fact that they are both completely described? What does it mean exactly? Does independence mean no interaction?

Update: No interaction is not assumed, as Martin pointed out in his answer. I added a screenshot added below, yellow text corresponds to the above quote. Also, what is meant by complete description is of course the following:

"Completely described states occur as a result of the simultaneous measurement of a complete set of physical quantities."

enter image description here enter image description here

Update 2: Ján Lalinský mentioned the density matrix description. I think this another quotation might be crucial for understanding of the above problem:

enter image description here enter image description here

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    $\begingroup$ Could you please add which book and section they wrote this in? $\endgroup$ Feb 21 '15 at 11:01
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    $\begingroup$ That quote is very strange indeed. $\endgroup$ Feb 21 '15 at 11:02
  • $\begingroup$ @JánLalinský updated $\endgroup$
    – wondering
    Feb 21 '15 at 11:21
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How does independence of the two systems follow from the fact that they are both completely described? What does it mean exactly?

I think what they really mean by "system is completely described" is

The state of the system is best described with complex function of coordinates of the system only.

There are other modes of description, via density matrix, via function $\Psi_{whole}(q_1,q_2)$, via point in phase space etc.. L&L assume we have a special case when $\psi_1(q_1)$ is best of these and call it "completely described".

The idea that I can decipher from their quoted text is this:

If both subsystems are described by separate function of their respective coordinates - $\psi_1(q_1)$, $\psi_2(q_2)$ - in the Born sense, then both have probability distributions $\rho_1(q_1), \rho_2(q_2)$ that are functions of their respective coordinates only, i.e. $\rho_1(q_1)=|\psi_1(q_1)|^2$, $\rho_2(q_2)=|\psi_2(q_2)|^2$.

In words, the probability for a region of configurations $q_1$ is independent of the configuration $q_2$ and vice versa.

Such probability distributions are called independent (also systems are then called "statistically independent" although this seems like a misnomer) and if both subsystems have such, the probability distribution for the whole system is always their product:

$$ \rho_{whole}(q_1,q_2) = \rho_1(q_1)\rho_2(q_2). $$

If this is also the result of using the Born rule on $\Psi_{whole}(q_1,q_2)$ as well, we need to assume there is a restriction on $\Psi_{whole}$:

$$ \Psi_{whole} (q_1,q_2) = e^{i\alpha}\psi_1(q_1)\psi_2(q_2), $$

where $\alpha$ is some real number from the interval $\langle 0,2\pi)$. Since the value of $\alpha$ has no consequence on probabilities, by simplicity we choose $\alpha=0$.

Does independence mean no interaction?

It does not mean no interaction; it means there are separate probability distributions for the parts and the probability distribution for the whole system is their product.

However, if there is interaction and it is strong enough, it is not reliable to assume that the systems are independent.

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  • $\begingroup$ Ok, so if we take the hydrogen atom, as an example, which is a system of bounded proton and electron (hence a system with interaction), than this simply means that we can separate the wave equations of the two constituents, what we usually do in the treatment of the hydrogen atom - we separate the equation for the centre of mass and that for the reduced mass. $\endgroup$
    – wondering
    Feb 21 '15 at 12:25
  • $\begingroup$ No, L&L talk about two systems being already described separately and how to combine two such descriptions into one. For example, two hydrogen atoms far away from each can be assigned separate $\psi$ function and these can be multiplied to obtain $\psi$ function of the pair. $\endgroup$ Feb 21 '15 at 16:54
  • $\begingroup$ In one hydrogen atom, however, the situation is the opposite: the whole system is assumed to be described by $\psi$ function $\Psi(q_e,q_p)$, but the constituents, the electron and the proton, are not assigned separate $\psi$ functions. They could be, but it would give much less accurate description than $\Psi(q_e,q_p)$ because the two particles positions and momenta are correlated due to strong mutual EM interaction and separate $\psi$ functions cannot describe that. $\endgroup$ Feb 21 '15 at 16:55
  • $\begingroup$ You presented it as if the interaction was the problem, which is not, as seen from the quote. I think there is some gap in you comment, but I will have to think it through. $\endgroup$
    – wondering
    Feb 21 '15 at 17:17
  • $\begingroup$ The presence of interaction does not prevent us from writing and using product of two functions as function for the whole system, but it strongly suggests it is not an accurate description. $\endgroup$ Feb 21 '15 at 17:48
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It doesn't - at least I would say that it doesn't given only the words there.

I believe what they meant to say is that the system is completely described by describing each part individually. In this case, since describing one part of the system does not require knowledge of the other, the probabilities are independent.

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  • $\begingroup$ You are right that no interaction is assumed, it actually follows from what follows, see the update. $\endgroup$
    – wondering
    Feb 21 '15 at 11:23
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We can start by understanding what an incomplete description involves. A quantum mechanical system, isolated from the rest of the world, can be described by a density matrix (or state matrix) $\rho$, rather than a wavefunction. If a quantum mechanical system is described by a wavefunction $|\Psi\rangle$ then that simply means that it is in the unique state for which the observable $|\Psi\rangle\langle\Psi|$ has expected value $1$; in general the expected value of an observable operator $\hat O$ in state $\rho$ is given by the trace operator:

$ \langle O \rangle ~=~ \operatorname{Tr} ~\rho ~\hat O$

We could therefore say that when you do have a wavefunction, your state matrix is $\rho ~=~ |\Psi\rangle\langle\Psi|$.

Suppose you've got a system of two qubits; there are states like

$|\Psi\rangle ~=~ \sqrt{\frac{1}{2}}~\left(|01\rangle ~+~ |10\rangle\right)$

where the individual parts cannot be "completely described" by themselves.But they can be described incompletely by a state matrix. Any one-qubit observable $\hat O_1$ becomes the two-qubit observable $\hat O_1 \otimes \hat I$, and this can be viewed equivalently as "tracing out" the second qubit, $\rho_1 ~=~ \langle 0_2 | \rho | 0_2 \rangle ~+~ \langle 1_2 | \rho | 1_2 \rangle$, then doing $\operatorname {Tr} ~\rho_1 ~\hat O_1$.

So if you trace out the second qubit in the above state you get the state matrix:

$\rho_1 ~=~ \frac{1}{2} ~ |0\rangle\langle 0| ~+~ \frac{1}{2} ~ |1\rangle\langle 1| ~=~ \frac{1}{2}$

And here you can see that there is no operator $|\Psi\rangle\langle\Psi|$ such that the trace produces anything other than $1/2$. This first qubit is "classically" half in $|0\rangle$ and half in $|1\rangle$ if you throw away the second qubit's information.

On the flip side, there are two-qubit states like:

$|\Psi\rangle ~=~ \frac{1}{2}\left(|00\rangle - |01\rangle - |10\rangle + |11\rangle \right)$

which can be separated into two complete descriptions--in this case $\sqrt{1/2}(|0_1\rangle - |1_1\rangle) : \sqrt{1/2} (|0_2\rangle - |1_2\rangle)$, where $|a_1\rangle : |b_2\rangle = |ab\rangle$. Using a normal tensor product, this says that the state matrix for that above state can be produced as a Kronecker product of two 2x2 matrices:

$\rho = \frac{1}{4}~\left(\begin{array}{cccc}~1 & -1 & -1 & ~1 \\ -1 & ~1 & ~1 & -1 \\ -1 & ~1 & ~1 & -1 \\ ~1 & -1 & -1 & ~1 \\ \end{array}\right) = \frac{1}{2} \left(\begin{array}{cc}~1 & -1 \\ -1 & ~1\end{array}\right) \otimes \frac{1}{2} \left(\begin{array}{cc}~1 & -1 \\ -1 & ~1\end{array}\right) = \rho_1 \otimes \rho_2$.

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  • $\begingroup$ Thank you for the example, it helps +1. Actually I think the solution must be connected to density matrix (see update 2). $\endgroup$
    – wondering
    Feb 26 '15 at 8:34
  • $\begingroup$ I'll update my answer with a little more detail from this added context. $\endgroup$
    – CR Drost
    Feb 26 '15 at 18:51
  • $\begingroup$ Ok, looking forward! $\endgroup$
    – wondering
    Feb 26 '15 at 19:06
  • $\begingroup$ @wondering done. $\endgroup$
    – CR Drost
    Feb 26 '15 at 23:52

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