# Separable state vs part of entangled state

1. May 1, 2012

### Fightfish

There is something that has been bothering me recently: that is, the distinction between a separable state and being part of an entangled state.
To make my query concrete, consider:
$\left|\psi\right\rangle = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$ and $\left|\Psi\right\rangle = \alpha \left|0\right\rangle_{1}\left|0\right\rangle_{2} + \beta \left|1\right\rangle_{1}\left|1\right\rangle_{2}$​
For the entangled state $\left|\Psi\right\rangle$, suppose that I am only interested in the 1st particle (thus in effect I discard the second particle).
Now, the density matrices are obviously different:
$\rho_{\psi} = \left( \begin{array}{cc}\alpha^{2} & \alpha \beta^{*} \\ \alpha^{*} \beta & \beta^{2} \end{array} \right)$ while $\tilde\rho_{\Psi, 1} = \left( \begin{array}{cc}\alpha^{2} & 0 \\ 0 & \beta^{2} \end{array} \right)$​
When measured in the basis $\left|0\right\rangle\langle 0|, \left|1\right\rangle\langle 1| \& \left|\psi\right\rangle\langle \psi|$, they give the same expectation values, but not for other basis. It would thus appear that $\tilde\rho_{\Psi, 1}$ is a mixture, not a pure state. In fact, we can express $\tilde\rho_{\Psi, 1} = \alpha^{2}|0 \rangle\langle 0|+ \beta^{2} |1 \rangle\langle 1|$.

Mathematically this is all nice and good, but physically the question is, how come I get a mixed state when I consider a particle that belongs to a pure entangled state, assuming that I have not done any measurements on the second particle. If my conjecture is correct though, would this be a property of the subsystem ie all is nice and good and pure with the whole system, but if I consider the subsystems then they are mixed?

2. May 2, 2012

### kith

Yes. This is a remarkable property of quantum mechanics and there is no consensus on how to interpret these mixed states.

First of all, you could interpret it as the failure of reductionism. It just wouldn't be meaningful to speak about the subsystem alone. Such an idea is the basis for Everett's Many Worlds interpretation, where both |0> and |1> are realized but in different "worlds".

Also, you can interpret quantum mechanics as a theory about statistical ensembles. If already the pure entangled states describe ensembles, there's nothing weird in the reduction from pure to mixed states.

In standard textbooks, kets are often introduced as states of a system and density matrices as ensembles of these states. I agree that from such a viewpoint, the reduction of a pure entangled state to two mixed states does seem weird. But I think there is nothing wrong in interpreting both kets and density matrices as (information) states of a system. This seems to be in the original Copenhagen spirit.

3. May 3, 2012

### Dali

This is an interesting example! I think one could view it as an extreme example of the principle of decoherence mimicking a measurement. Just think of particle 1 as "the system" and particle 2 as "the environment". This makes sense since you consider precisely the case when we don't (or can't) measure what happens to particle 2 - we just let it "fly away" unmeasured. Now, to predict the behavior of particle 1 alone, we average over the two possible outcomes of particle 2, which gives the diagonal density matrix $\tilde\rho_{\Psi, 1}$, which really is a classical mixture as you point out.

We can interpret this as if the vanishing particle 2 has the effect of decohering particle 1 from a superposition to a classical mixture of state $| 0 \rangle$ or $|1\rangle$ (with probability α2 and β2 respectively). This is exacly as if we had let a measurement apparatus measure particle 1, collapsing its wavefunction, but had not yet read the outcome of the measurement.

It is also true that decoherence in general picks out a special basis that it "collapses wavefunctions onto". I.e. it is the details of the interaction between "the system" and its environment that dictates a specific basis in which the density matrix will tend to become diagonal. (If the coupling would have been weaker than in this example of full entaglement, the off diagonal terms would decay exponentially with some very short time-scale).