Mixed states and pure states: what's the difference?

[jeebs]

Hi,
this is probably a straightforward question over something simple but it's confusing me. I don't get what the difference is between a mixed state, a superposition and a pure state. I'm looking through my notes about density operators and it's talking about a qubit system where $$|0> = \left(\begin{array}{c}1&0\end{array}\right)$$ and $$|1> = \left(\begin{array}{c}0&1\end{array}\right)$$.

It then goes on to talk about the system being in state $$|+> = \frac{1}{\sqrt{2}}(|0> + |1>)$$, but it calls it a "pure state". This does not look pure to me, I would have called |0> and |1> the pure states, and |+> superposition of the two.

Using the basis { |+>, |-> }, where $$|-> = \frac{1}{\sqrt{2}}(|0> - |1>)$$, the density operator can be found: $$\rho = \left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$.
The notes say that "the statistical mixture of pure states giving rise to the density operator is called a mixed state". |+> certainly looked like a statistical mixture of |0> and |1> to me. Is there a difference between a mixed state and a superposition?

 

[Hurkyl]

You're confusing "pure" with "basis". All states that can be named with a ket from that Hilbert space are pure.

In the (0,1) basis, that ket has density matrix
$$\left( \begin{matrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{matrix} \right)$$

An equally weighted statistical mixture of |0> and |1> would have density matrix
$$\left( \begin{matrix} 1/2 & 0 \\ 0 & 1/2 \end{matrix} \right)$$

 

[phyzguy]

Note that a state is a pure state if and only if the density matrix is such that rho^2 = rho. In Hurkyl's examples, note that the first density matrix of the superposition has the property that rho^2 = rho, but for the second one, rho^2 is not equal to rho. Alternatively, we can say that it is a pure state if and only if Trace(rho^2) = 1.