# Can I understand mixed states using the Bloch sphere? [duplicate]

I'm a bit confused with the representation of mixed states in a Bloch sphere. Are they represented as points or vectors? For pure states, they're vectors on the surface of the Bloch sphere and have a norm of 1. If I have a mixed state $$\rho=\sum_sp_s|\psi_s\rangle\langle\psi_s|$$, is there a way I can see $$\sum_sp_s=1$$ on the Bloch sphere?

• They are the points inside the sphere. – KAJ226 Apr 15 at 1:43
• @KAJ226 Thanks! Why they're not vectors? – IGY Apr 15 at 2:00
• Note that any 2 by 2 unitary matrix $U$ can be written as $U = n_I I + n_X X + n_Y Y + n_Z Z$ where $I,X,Y,Z$ are Pauli matrices. Now, because $Tr(U) = 1$ because of preservation of probability, we must have $n_I = 1/2$. And if we let $\vec{r} = \langle n_X, n_Y, n_Z \rangle$ then $| \vec{r} | \leq 1$. If $| \vec{r} | = 1$ then this corresponds to the state on the sphere... these are the pure states. The mixed states are the points correspond to points where $|\vec{r}| < 1$. – KAJ226 Apr 15 at 3:06
• @KAJ226 This should be an answer! – Adam Zalcman Apr 15 at 5:46
• @KFBJN In addition to what KAJ226 wrote, note that a point $P$ (inside the sphere or not) corresponds to a vector from the origin to $P$, so the answer to the question whether mixed states are represented as points or as vectors is: "both" or "whichever you prefer". – Adam Zalcman Apr 15 at 5:48

An arbitrary $$2 \times 2$$ Hermitian matrix $$U$$ can be decomposed into

$$U = n_I I + n_X X + n_Y Y + n_Z Z$$

with $$n_X, n_Y, n_Z \in \mathbb{R}$$ and $$X, Y, Z$$ are Pauli matrices and $$I$$ is the $$2 \times 2$$ Identiy matrix.

$$I = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}, X= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}, Y = \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}, Z = \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}$$

These matrices, $$\{I,X,Y,Z\}$$, formed an orthorgonal basis for the real vector space for all $$2 \times 2$$ Hermitian matrices. But furthermore, they are an orthorgonal basis complex vector space for all $$2 \times 2$$ matrices.

Now, suppose that $$U$$ represents a quantum state (either pure or mixed), then it must be a Hermitian matrix with unit trace, $$Tr(U) = 1$$. This is to make sure we have preservation of probability. Now, if we calculate $$U$$ out explicitly, we have that

\begin{align} U &= n_I \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} + n_X \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}+ n_Y \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix} + n_Z \begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix} \\ &= \begin{bmatrix}n_I + n_Z & n_X - in_Y \\ n_X + in_Y & n_I - n_Z\end{bmatrix} \end{align}

and this implies that

$$Tr(U) = Tr\bigg( \begin{bmatrix}n_I + n_Z & n_X - in_Y \\ n_X + in_Y & n_I - n_Z\end{bmatrix} \bigg) = 2n_I = 1 \ \ \Rightarrow \ \ n_I = \dfrac{1}{2}$$

Thus, given an arbitrary single qubit state (pure or mixed), we can write it as:

$$U = \dfrac{1}{2} I + n_X X + n_Y Y + n_Z Z = \dfrac{I + \vec{r}\cdot \vec{\sigma} }{2}$$

where we use $$\vec{r}$$ to denote the vector $$\langle2 n_X, 2 n_Y, 2 n_Z \rangle$$ and $$\vec{\sigma}$$ to denote the vector $$\langle X, Y, Z \rangle$$.

Furthermore, we must have that $$|\vec{r}| \leq 1$$ because $$U$$ must be Positive semi-definite (Its eigenvalues must be non-negative). To see why $$|\vec{r}| \leq 1$$ will ensure the Positive semi-definite condition, you need to calculate the eigenvalues of $$U$$. We can do this in the usual way, by considering the equation:

\begin{align} det(U - \lambda I) &= det\bigg( \begin{bmatrix}n_I + n_Z - \lambda & n_X - in_Y \\ n_X + in_Y & n_I - n_Z - \lambda \end{bmatrix} \bigg) \\ &= \lambda^2 - 2n_I \lambda + n_I^2 - (n_X^2 + n_Y^2 + n_Z^2 ) = 0 \\ \end{align}

This implies that (use quadratic equation)

\begin{align} \lambda &= \dfrac{-(-2n_I) \pm \sqrt{(-2N_I)^2 - 4(n_I^2 - (n_X^2 + n_Y^2 + n_Z^2 ) ) } }{2} \\ &= \dfrac{ 2n_I \pm \sqrt{4 (n_X^2 + n_Y^2 + n_Z^2 ) } }{2}\\ &= n_I \pm \dfrac{|\vec{r}|}{2} \\ &= \dfrac{1}{2} \pm \dfrac{|\vec{r}|}{2} \hspace{1 cm} \textrm{since we know that} \ \ n_I = \dfrac{1}{2} \ \ \textrm{because} \ \ Tr(U) = 1 \end{align}

To make sure that $$\lambda \geq 0$$ we must have that $$| \vec{r} | \leq 1$$.

Now, note that if $$| \vec{r} | = 1$$ then there is only one non-zero eigenvalue, this implies that we have a pure state.

If $$| \vec{r} | < 1$$ then we will have two eigenvalues, says $$\lambda_1$$ and $$\lambda_2$$. This implies that this is a mixed state. Furthermore, also note that here we have $$0 < \lambda_1, \lambda_2 < 1$$ and so $$Tr(U^2) = tr( V^{-1} \Sigma^2 V ) = tr(\Sigma^2) = \lambda_1^2 + \lambda_2^2 < 1$$ here $$V$$ is composed of eigenvectors of $$U$$ and $$\Sigma$$ is diagonal matrix with eigenvalues $$( \lambda_1, \lambda_2 )$$ of $$U$$. If you look at the definition of denisty matrix you can see that this is a condition to determine whether a state is pure or mixed. Only pure states $$(\rho)$$ have the condition $$Tr( \rho^2) = 1$$.

• not unitary, Hermitian. – kludg Apr 15 at 7:41
• Note that $\{I, X, Y, Z\}$ is a basis of two vector spaces: the real vector space of $2\times 2$ Hermitian matrices and the complex vector space of all $2\times 2$ matrices. Here we implicitly mean the former and so the coefficients $n_I, n_X, n_Y, n_Z$ are real numbers. – Adam Zalcman Apr 15 at 14:51
• @kludg I didn't mean to indicate that density matrix is unitary. I started the answer by saying any 2 by 2 unitary matrix.. but I change that to just any 2 by 2 matrix as $I,X,Y,Z$ is the span of 2 by 2 matrices. I guess the way I started out the answer make its seems like I was trying to indicate that density matrix is unitary operator... . Sorry for the misunderstanding. – KAJ226 Apr 15 at 14:51

Since density matrix is Hermitian and has unit trace, for a single qubit it always can be written as $$\rho = \frac{1}{2}(I+n_x\sigma_x+n_y\sigma_y+n_z\sigma_z)= \frac{1}{2} \begin{pmatrix} 1+n_z & n_x-in_y \\ n_x+in_y & 1-n_z \end{pmatrix}$$ where $$n_i$$ are some real numbers.

Since density matrix is positive semidefinite, it should have non-negative eigenvalues and, consequently, non-negative determinant, $$det(\rho)=\frac{1}{2}(1-n_x^2-n_y^2-n_z^2)\geq 0$$ or $$n_x^2+n_y^2+n_z^2 \leq 1$$

This equation defines Bloch Sphere (actually ball, not sphere). The pure states have eigenvalues $$0$$ and $$1$$ which corresponds to $$n_x^2+n_y^2+n_z^2 = 1$$ (points on actual sphere). The points inside the sphere, $$n_x^2+n_y^2+n_z^2 < 1$$

are mixed states