# Preparation of Single qubit mixed states

Hi,

I would like to ask some questions in regards to preparing mixed states.

Assuming I am allowed to prepary any pure quantum state, perform measurements and apply unitaries, I am asked to prepare a single qubit quantum state of maximum entropy.

As I understand, a 1-qubit quantum state of maximal entropy is one that gives 50/50 chance of either measurement, whatever the two orthonormal states I pick. So basically, randomly firing a pure state would do and I don't unitaries/measurements for this? Would also firing just two chosen orthonormal pure states also work? Or should I prepare a pure state 1/sqrt(2)|0>+1/sqrt(2)|1> and measure it with respect to basis |0> and |1>, then there's a 50/50 chance of the qubit being either |0> or |1>, i.e. exhibiting the same probabilities as the mixed state of maximal entropy would.

Exactly I didn't understand your question. But I can say those. in Pure states be zero entropy, in Mixed states be " LnN" ; N: dimensionality;

then if N increase, entropy increase too . . .

Hi zygi, thanks for the answer. It's not quite what I was asking though I admit my phrasing isn't clear. Let me try ask a different question: Would measuring a state in equal superposition of two quantum states |0> and |1> give me the totally mixed state?

Hi macduy, I guess, I understand this time. For example, |S_x;+>= 1/Sqrt(2)(|+>+|->)
is |S_x;+> mixed state? This is your question.
I think, |S_x;+> still is pure state.
Now, I asked you. How do we know whether pure or mixed of any state.
As far as I know, it related to density matrix.
g= sum_i w_i*|u_i><u_i|; |u_i>: eigenvectors, w_i; probabilities. Here if g^2=g , pure state if not mixed state, or semi mixed state.

I hope I could help.
Now, I am going to sleep. See you tomorrow. . .

Right, |S_x;+> would still be a pure state. However, let's say now that you measure it using (|+><+| + |-><-|). Then, the state will either become |+> or |->, with equal probabilities. Either way, the state will still be pure *but* because we have no knowledge which one it would be, we refer to it as a mixed state.

According to wiki, an example of a quantum system with mixed state is normal unpolarized light. Each photon emitted has a certain polarization, therefore it is pure it is sense, but we just have no idea which kind of polarization.

So, the way I understand it, mixed states (and hence density matrices) are used when we lack knowledge about the preparation of a quantum state.

Let me know if this sounds right to you ;)

|+><+|+|-><-| state that you said is mixed state. Because, You can constitue infinite in different style. So we don't know anything about previous case.

For example, "Poincare-Block sphere " (we saw it in the lecture)
There is a sphere in our hand. İts outside(in other words its shell) represent the pure states. İts inside correspond the mixed states. Now, its inside imagine a point. there is infinite girders passing from this point. These girders be made up of pure states. we only know mixed state. We dont know previous case of the system.

See you...

There's been a misunderstanding, |+><+|+|-><-| is not a state (it can be but it isn't here), it's a measurement, with which I'm measuring the qubit. It's not the same thing because if I measure the qubit |+> with it, then I get |+> 100 percent of the time as the outcome, which clearly is not mixed. I come from a mathematical background and this is how we were taught to do so-called "measurements" :) not sure, is it the same for you?

To be honest, yes :) ,
Usually we deal with maths.
What do you mean "|+><+|+|-><-| is not a state (it can be but it isn't here) "
I not quite understand.

I think we are getting confused now. Let's keep it simple. I have a single qubit system, with a basis. I prepare a qubit which is a superposition of the basis vectors. Then I measure it in the basis. Do I get a mixed state?

right. you can look "Computation and Quantum ınformation " Authors: Micheal A. Nielsen and Isaac L. Chuang ...