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$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$Given a mixed state $\rho = \sum p_k \rho_k$ that is an statistical emsemble, where each "state" $\rho_k$ on the upper half of the Bloch sphere

$\hskip3in$enter image description here

appears with equal probability. The states don't lie on the surface of the Bloch sphere but at a radius of $1/{2\pi}$, such that the sum, which, in the continous case turns out to be an integral, properly works out. So the $\rho_k$ are not pure states!

If I now measure the system in the computational basis, I'll get $\ket0$ in 100% of the cases. So I would assume the state is a pure one, but is it?

We might need infinitely many states $\rho_k$, but maybe a big number is enough to get a good aprroximation. Or did I miss something else?

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  • $\begingroup$ If a state is inside the Bloch sphere, it is mixed. Only a pure state is on surface of the spehere. $\endgroup$ May 1 '20 at 21:45
  • $\begingroup$ Yes but I sum up a lot of them, let's say infinitley many sch that the integral looks like a vector ending at the north pole, which is $|0\rangle$... $\endgroup$
    – draks ...
    May 1 '20 at 21:52
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    $\begingroup$ @draks... Assume we have only two states $\rho_1$ and $\rho_2$ with different $r_1$ and $r_2$ vector lengths in the Bloch sphere. The state that corresponds to their statistical sum $p_1 \rho_1 + p_2 \rho_2$ will never have $r$ greater then $max(r_1, r_2)$, right?. If I am right I don't see a reason why this will not be true for infinite number of $\rho$s in the sum. So we will not obtain the $|0\rangle$ state, which has $r=1$, with sum of many states that have $r<1$. $\endgroup$ May 2 '20 at 10:12
  • $\begingroup$ @DavitKhachatryan ok I see your point. So then let's drop the scaling factor. My state is then a sum of pure states on the northern hemisphere. How would you interpret the resulting state? To me it feels like a state that would give identical results as a pure $|0\rangle$ would do... $\endgroup$
    – draks ...
    May 2 '20 at 20:56
  • $\begingroup$ @draks... I guess we should take very specific $p(\vec{r})$ distribution in order to obtain something like $\rho_0 = 0.999 | 0 \rangle \langle 0 | + 0.001 \tilde{\rho}$. $\endgroup$ May 3 '20 at 5:31
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You're forgetting the requirement that the probability weights must sum to 1.

You can't sum up all the mixed state 3-vectors corresponding to the $\rho_k$ with unit weight to get the 3-vector corresponding to $\rho = \iint_k \rho_k$ - that sum isn't properly normalized. You need to take a convex combination of the $\rho_k$, i.e. a weighted sum $\sum_k \rho_k$ in which the $p_k$ are nonnegative and sum to 1, which yours don't.

You are correct that a convex combination of qubit states maps to the same convex combination of the initial state's 3-vectors in the Bloch ball. But geometrically, a convex combination of vectors in $\mathbb{R}^n$ always yields a vector inside their convex hull, which (loosely) consists of "the space in between" the original vectors. So you can't take a convex combination of 3-vectors and get a 3-vector that "reaches outside" the original set, as you propose. In your case of a uniformly-weighted mixture, you'd end with a mixed state whose 3-vector on the Bloch ball lies at the geometric center of mass of the original vectors, which would still be inside the ball.

In particular, a nontrivial convex combination of several qubit states (by which I mean that multiple coefficients are positive) always has a purity that is strictly lower than the highest purity of the constituent qubit states.

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A pure state is any state that can be written in the form $|\alpha \rangle \langle \alpha|$, but you are free to write it any way you like, including as a sum.

A density $\rho$ is any Hermitian trace-1 positive semi-definite matrix. It can be written as $$\rho = \sum_i p_i |\psi_i\rangle \langle \psi_i|$$ with $\{p_i\}$ a probability distribution (i.e. all between 0 and 1 and they sum to 1) and each $|\psi_i \rangle$ a pure state. But, yes, you could write one density as a sum of other mixed densities, provided again your $p_i$ form a probability distribution.

The only trace-1 positive semi-definite matrix with $\langle 0 |\rho |0\rangle = 1$ is $$\rho = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}$$

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  • $\begingroup$ Thanks for your answer $\endgroup$
    – draks ...
    May 3 '20 at 17:06

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