# Is my $|0\rangle$ state mixed or pure?


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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?

• If a state is inside the Bloch sphere, it is mixed. Only a pure state is on surface of the spehere. May 1 '20 at 21:45
• 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 $|0\rangle$... May 1 '20 at 21:52
• @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 $r<1$. May 2 '20 at 10:12
• @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... May 2 '20 at 20:56
• @draks... I guess we should take very specific $p(\vec{r})$ distribution in order to obtain something like $\rho_0 = 0.999 | 0 \rangle \langle 0 | + 0.001 \tilde{\rho}$. May 3 '20 at 5:31

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

• Thanks for your answer May 3 '20 at 17:06