$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>}$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?

  • $\begingroup$ If a state is inside the Bloch sphere, it is mixed. Only a pure state is on surface of the spehere. $\endgroup$ May 1, 2020 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, 2020 at 21:52
  • 2
    $\begingroup$ @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, 2020 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, 2020 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, 2020 at 5:31

2 Answers 2


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.


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 $$\rho = \begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}$$

  • $\begingroup$ Thanks for your answer $\endgroup$
    – draks ...
    May 3, 2020 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.