# Why is it valid to take the partial trace to describe a subsystem?

In derivations of decoherence, there eventually comes a point when we are asked to take the partial trace over the environment. Why should this be valid for an entangled system? Why should taking the partial trace give the correct description of the system? Taking the partial trace turns a pure density matrix into a mixed density matrix, but a pure state is always a pure state. In what sense can the reduced mixed density matrix be assigned probabilities?

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The density matrix given by the partial trace predicts the results of any experiment done an half of an entangled system as well as any method of prediction that does not look at the other half of the entangled system. So if you throw the other half of the system away, the behavior of the state you are in is given by the density matrix obtained by the partial trace. –  Peter Shor May 22 '11 at 18:30

Dear Gunther, a partial trace contains all the information about possible predictions made for the subsystem that you haven't traced over. It's trivial to see why.

First, take the whole system A+B. They may be entangled but we're only interested in future measurements of A. If the description for A+B is a pure state $|\psi\rangle$, then define $$\rho = |\psi\rangle \langle \psi |$$ The expectation values of the operator $P$ in $|\psi\rangle$ are trivially equal to ${\rm Tr}\,(\rho P)$, do you see why?

Now, $\rho$, the density matrix, is an operator acting on the total Hilbert space which is the tensor product $H_A\otimes H_B$. And operators $L$ that only act on the system $A$ have the form $$L = L_A \otimes {\bf 1}$$ They're just tensor product of an operator acting on $H_A$ only and the identity operator acting on the Hilbert space $H_B$. Do you understand why? Then it's obvious that $${\rm Tr} \,( \rho L) = {\rm Tr}_{ab} \,( \rho (L_A\otimes {\bf 1}))$$ where $ab$ indicates that the tracing goes over indices corresponding to the basis of $H_A$ as well as those of $H_B$. You can see that the tracing over both types of the indices may be made one by one. You may first trace over the $b$-type indices. However, only $\rho$ depends on $b$-type indices while $L_A$ doesn't and ${\bf 1}$ doesn't make a difference. So $${\rm Tr} \,( \rho L) = {\rm Tr}_{a} {\rm Tr}_{b} \,( \rho (L_A\otimes {\bf 1}))$$ But $L_A$ doesn't contain any $b$-type indices, so it's a constant that may be pulled in front of the trace over $b$-type indices: $${\rm Tr} \,( \rho L) = {\rm Tr}_{a} L_A {\rm Tr}_{b} \,( \rho {\bf 1}) = {\rm Tr}_a L_A \rho_{\rm traced\,over\,b}$$ So the trace over $b$ may be made first and the reduced density matrix may then be traced to find properties of the observables linked to the subsystem A.

The fact that this reduction must be possible if you only want to predict A should be intuitive. In classical physics, you also don't have to know another subsystem to write down probabilities of different properties of A - despite the fact that A,B may be correlated. You just integrate over all possible positions and momenta of B. The tracing is the exact quantum analogy of the integration over $x_B,p_B$.

For example, if you have an EPR experiment, A and B may be 100% correlated. But if you only measure A, you won't see the correlation because you don't have any results from B. So you will see that 50% of the photons on the A side are left-handed and 50% are right-handed: the density matrix will be proportional to the unit matrix (times one half) even though the original state for both A,B was pure - but entangled.

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