Does normalizing a mixed state give a pure state?

Question

According to bloch sphere interpretation, any point on the surface of the sphere corresponds to a pure state and any point inside the sphere corresponds to a mixed state. Suppose you have a point inside the bloch sphere C corresponding to a mixed state. Draw a ray connecting C and origin O of the bloch sphere. Now extend this ray OC such that we get some point C' on the surface of the sphere. Now C' = cC for which c is a real number.

So can we say that the mixed state corresponding to point C when normalized becomes a pure state? Or is there a problem with this logic?

Answer 1

Bloch vector $\vec{v}$ of a density matrix $$ \rho_C = \frac{1}{2}(I + \vec{v}\cdot\vec{\sigma}) $$ can be shorter than 1 i.e. $|\vec{v}| \le 1$, when the the density matrix represents a mixed state.

If you normalize it $\vec{n} = \vec{v}/|\vec{v}|$ then you will get the density matrix of the C' pure state $$ \rho_{C'} = \frac{1}{2}(I + \vec{n}\cdot\vec{\sigma}). $$

When it comes to interpretations, I don't know of any simple relation between C and C' like e.g. measurements of C won't bring you to C.

Answer 2

An arbitrary one qubit density matrix $\rho$:

$$\rho = \begin{pmatrix} \rho_{00} & \rho_{01} \\ \rho_{10} & \rho_{11}\end{pmatrix} = \frac{I + \vec{r} \cdot \vec{\sigma}}{2}$$

where $\vec{r} \cdot \vec{\sigma} = r_x \sigma_x + r_y \sigma_y + r_z \sigma_z$, $\sigma$s are Pauli matrices, $r$ is the corresponding vector for the density matrix in the Bloch sphere.

The probability of measuring $|0\rangle$ state is equal to $\rho_{00}$ and the probability of measuring $|1\rangle$ state is equal to $\rho_{11}$. Thus, the normalization of the probabilities of measuring $|0\rangle$ or $|1\rangle$ states for the density matrix corresponds to the statement $Tr(\rho) = \rho_{00} + \rho_{11} = 1$. In this sense it is already a normalized state if $Tr(\rho) = 1$, but still the vector $\vec{r}$ in the Bloch sphere formalizm will be inside the sphere for a mixed state.

I am not sure how the action described in the question can be related to the normalization, because both mixed (inside the sphere) and pure (on the sphere) states should be normalized.