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I have read this and I want to know why is that the case, and why not simply write two (or more) particles with their 'pure' states, because the latter is much more straightforward?

Qmechanic
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draconianzoo
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1 Answers1

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There are two different kinds of uncertainty in quantum mechanics.

The first is the normal, intrinsic, quantum mechanical uncertainty. Knowing the state of the system completely, still means you only can make probabilistic statements about the outcomes of some experiments. For example, if we have two particles in a 2-state system (if it helps, say the two states are "spin up" and "spin down" for concreteness), then the system may be in an entangled state like \begin{equation} | \Psi \rangle = \frac{1}{\sqrt{2}} \left(|0 \rangle | 1 \rangle + | 1 \rangle | 0 \rangle \right) \end{equation} in which case it is not possible to say with certainty if a measurement of particle $1$ will yield the result $0$ or $1$.

The second is an uncertainty in what the state of the system is. For example, perhaps we think there's a 50% chance that the system is in the state $|\Psi\rangle$ given above, and a 50% chance that the system is in a different state $|\Phi\rangle$, given by \begin{equation} | \Phi \rangle = \frac{1}{\sqrt{2}} \left(|0 \rangle | 1 \rangle - | 1 \rangle | 0 \rangle \right) \end{equation} This kind of uncertainty cannot be represented nicely using only state vectors in Hilbert space. One option is to say in words "there's a 50% chance that the state is $|\Psi\rangle$ and a 50% chance that the state is $|\Phi\rangle$, but this becomes cumbersome. And, in situations like thermodynamics, you often need to specify the probability for an infinite number of possible states of the system.

The density matrix can encode both kinds of uncertainty. In the first case, where we know the state is $|\Psi\rangle$, the density matrix is \begin{equation} \rho = |\Psi \rangle \langle \Psi | \end{equation} In the second case, with a 50% chance that the state is $|\Psi\rangle$ or $|\Phi\rangle$, the density matrix is \begin{equation} \rho = \frac{1}{2} \left( |\Psi \rangle \langle \Psi | + |\Phi \rangle \langle \Phi | \right) \end{equation} Because the density matrix can handle both kinds of uncertainty, the density matrix generalizes the normal quantum state vector and lets us handle a wider variety of cases.

This flexibility also lets us formulate statements about entanglement in a precise way. For example, in the first case where the state is $|\Psi\rangle$, we can consider the state of knowledge we would have if we only have access to the first particle, by tracing over the second particle. This leads to a density matrix \begin{eqnarray} \rho' &=& {\rm Tr}_2 \rho \\ &=& \frac{1}{2} \left(|0\rangle \langle 0| + | 1 \rangle \langle 1 | \right) \end{eqnarray} where $\rm Tr_2$ denotes a trace over the Hilbert space of the second particle. This tells us that if we don't know anything about particle 2, then we only know that particle 1 has a 50% chance of being in state $|0\rangle$ or state $|1\rangle$. Of course, this uncertainty is different from particle 1 being in a superposition of states $|0\rangle$ and $|1\rangle$.

Andrew
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