# Does normalizing a mixed state give a pure state?

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?

• rescaling the density matrix does not correspond to rescaling the corresponding Bloch vector
– glS
Commented May 24, 2020 at 10:50

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.

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.