2
$\begingroup$

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?

$\endgroup$
10
  • $\begingroup$ They are the points inside the sphere. $\endgroup$ – KAJ226 Apr 15 at 1:43
  • $\begingroup$ @KAJ226 Thanks! Why they're not vectors? $\endgroup$ – IGY Apr 15 at 2:00
  • 2
    $\begingroup$ 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 $|\vec{r}| < 1$. $\endgroup$ – KAJ226 Apr 15 at 3:06
  • $\begingroup$ @KAJ226 This should be an answer! $\endgroup$ – Adam Zalcman Apr 15 at 5:46
  • 2
    $\begingroup$ @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". $\endgroup$ – Adam Zalcman Apr 15 at 5:48
5
$\begingroup$

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 $.

$\endgroup$
3
  • $\begingroup$ not unitary, Hermitian. $\endgroup$ – kludg Apr 15 at 7:41
  • $\begingroup$ 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. $\endgroup$ – Adam Zalcman Apr 15 at 14:51
  • $\begingroup$ @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. $\endgroup$ – KAJ226 Apr 15 at 14:51
2
$\begingroup$

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

$\endgroup$

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