# Difference between mixed states and pure states.

1. Apr 11, 2012

### Casco

I am doing my thesis in quantum information and I am learning the background of it, I met this two things mixed and pure states. I have read a few books and a few articles too. But I still can not get the idea. So If anyone can give some help explaining this, would be such a good help. Either if you just want to give me a good reference it is welcomed, not forgetting that I am an undergraduated. Thanks

2. Apr 11, 2012

### Khashishi

A pure state is the basic way of representing a known quantum state. Basically, it means you know everything there is to know about the system or subsystem that the state refers to. That doesn't mean you can predict the results of a measurement, because the pure state might not be in an eigenstate of the measured quantity. But it is in an eigenstate in some other basis. On the other hand, a mixed state represents an unknown quantum state. Since you don't know the state, you don't know the state in any other basis either.

An example should make things clearer. A spin 1/2 particle can be thought of as being pointed in any direction. But when it is measured in some direction, it will instantly align to up or down in that direction. A spin right particle is in a pure state. You can't predict if it will become spin up or spin down if you measure it in that direction, but you know everything about the spin that there is to know. And if you choose to measure the spin in the left/right direction, you'll get spin right 100% of the time, because you know the state perfectly. On the other hand, if you don't know the spin, or you only have partial information about the spin, then you can treat it as a mixed state, which is a combination of some pure states which probabilities of being in that pure state. You could have a mixed state of 50% spin up, 50% spin right. But you can't have a pure state like that.

A superposition between pure states is still a pure state. It's not a mixed state. If you superimpose a spin up state with a spin down state, you will get a spin state in some other direction, depending on the relative phases of the input states. You can't throw away the phase information, or you lose information and end up with a mixed state.

3. Apr 12, 2012

### tom.stoer

A calculation might help.

Suppose you have an observable A and its orthonormalized eigenstates |a>; then you can introduce a density operator ρ w.r.t. to the A-basis

$$\rho = \sum_a p_a\,|a\rangle\langle a|$$

and calculate an expectation value (B)ρ of a second observable B in this generalized state ρ as

$$(B)_\rho = \text{tr}(\rho B) = \sum_{a\prime} \langle a^\prime| \sum_a p_a\,|a\rangle\langle a|B|a^\prime\rangle = \sum_a p_a \langle a|B|a\rangle = \sum_a p_a \langle B\rangle_a$$

Let's look at the equation

$$(B)_\rho = \sum_a p_a \langle B\rangle_a$$

It contains two expectation values, namely the expectation value <B>a for every pure state |a> and the expectation value (B)ρ in the generalized state ρ.

The big difference between the pure states |a> and the generalized state ρ is that in the latter one you introduce a classical probability pa for every |a>. Suppose you have B=A; then the last equation reduces to

$$(A)_\rho = \sum_a p_a \langle A\rangle_a = \sum_a p_a\,a$$

so for every eigenvalue you get a weight (a probability) pa

If the system is not in an eigenstate |a°> i.e. if you have non-zero pa for several different eigenvalues the system is in a mixed state. Then in addition to the expectation value <B>a for every pure state |a> you have to average over different states |a>.

If the system is in an eigenstate |a°> you have p=1 and pa'=0 for all a' ≠ a°. So the equation reduces to

$$(B)_\rho = \langle B\rangle_{a^0}$$

which means that in this special case where ρ corresponds to a pure state |a°>, the two expectation values are identical. You can see that the pure state is a special case of the generalized state ρ.

4. Apr 12, 2012

### Casco

So for example, I have the following system, the 0 represents spin down and the 1 represents spin up.

$\left| \psi \right \rangle = (a\left| 0 \right \rangle+b\left| 1\right \rangle)$

As you said, we can think a 1/2 spin particle like above. Pointying in any direction

$\left| \psi \right \rangle = \left| 0 \right \rangle$

And this would be a pure state of a 1/2 spin particle system, right?

well if I am not wrong, would the following be a mixed state?

$\left| \psi \right \rangle = \frac{1}{\sqrt{2}}(\left| 0 \right \rangle+\left| 1\right \rangle)$,

5. Apr 12, 2012

### tom.stoer

no, in order to describe mixed states you need density operators ρ as described above

6. Apr 12, 2012

### Khashishi

No, that would be a pure state, because it is a superposition of pure states, and the phase between the $\left| 0 \right \rangle$ and the $\left| 1\right \rangle$ is known. In fact, if $\left| 0 \right \rangle$ and $\left| 1\right \rangle$ are measured along the z direction, then $\frac{1}{\sqrt{2}}(\left| 0 \right \rangle+\left| 1\right \rangle)$ is pointing in the x direction. Remember that any choice of direction for z gives you a complete basis representation for the spin, which means you can express any (pure) spin state in the z basis, even if the spin isn't along the z direction. (Otherwise, how would you even consider a spin right particle? The spin can't change just because you changed the coordinates. You need to make a measurement for that.)

A mixed state cannot be represented as a ket vector. Any sum of ket vectors is a pure state. Instead, a density matrix is used to represent the state.

7. Apr 12, 2012

### Khashishi

There's nothing magical about the mixed state. It's really a simple, mundane concept that represents classical probability. If you have 3 red balls and 2 white balls in a bag, and you grab one without looking, then your ball is in a mixed state of 60% red, 40% white. That's just a lack of knowledge-nothing quantum going on.

A ball in a superposition state of red and white is a totally different beast. This thing actually is in a red and white state, and you can have 100% knowledge about this ball. I suppose you could, in theory, prepare such a ball using some kind of Shrodinger's box style device.

8. Apr 12, 2012

### Einstein Mcfly

If you want a good reference, try Cohen-Tanudji.

9. Apr 13, 2012

### Casco

Thanks I got it. Finally I understood

Good analogy, it helped me a lot.