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I have been studying this lecture from Scott Aaronson: https://www.scottaaronson.com/democritus/lec9.html

In the section "Mixed states", he says you can find the combined state by using the density matrix, and summing. I understand that part.

Further down, in the section "Real vs Complex Numbers", he says if you know the state of two quantum systems, you can find their "combined" state by using the tensor product.

I don't understand when to use the density matrix combination method or the tensor product method. What is the difference between what these two methods calculate?

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Starting with a pure state $$ \rho_n = |\psi_n\rangle\langle\psi_n| , $$ one can form a mixed state by adding several such pure states $$ \rho = \sum_n P_n \rho_n , $$ where $P_n$ represents probabilities, such that $$ \sum_n P_n = 1 . $$ What this means is that there is some ignorance so that the best information is represented by having different pure states that can be observed with different probabilities. If the pure state represent a single particle, then the mixed state is still that of a single particle.

A tensor product represents multiple particles. I $\rho_1$ and $\rho_2$ are respectively single particle states, then $$ \rho=\rho_1\otimes\rho_2 $$ represents a two-particle state. In this case, the respective one particle states can either be mixed or pure.

Hope this explanation clarifies the difference.

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  • $\begingroup$ Thanks for your answer. However, I am still a little confused: if the mixed state is for a single particle, and the tensor product represents multiple particles... What if we considered each individual state of a "single" particle as different particles? For example: an election with |0> we call it "zectron", |1> we call it "onetron". Then it's "two" particles, and shouldn't their combination be a tensor product? $\endgroup$
    – l3utterfly
    Nov 5 '20 at 9:55
  • $\begingroup$ A single particle can have different states. Think for instance of the different states of polarization. One can form superpositions of such different states, but the result is still a single particle. $\endgroup$ Nov 5 '20 at 11:53

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