@AHusain mentions here that the purity of a qubit state can be expressed as a function of the radius from the center of a Bloch sphere. The state corresponding to the origin is maximally mixed whereas the states corresponding to the boundary points are pure. So far, I haven't seen this fact written anywhere explicitly. What would be the mathematical proof for this claim?


3 Answers 3


A density matrix $\rho$ has the properties of being Hermitian, non-negative and has trace 1.

Any $2\times 2$ matrix can be written in the form $$ \rho=\frac{n_0\mathbb{I}+\vec{n}\cdot\vec{\sigma}}{2}. $$ The trace being 1 fixes that $n_0=1$, while the Hermitian property imposes that $\vec{n}\in\mathbb{R}^3$, where $\vec{\sigma}$ is the vector of the 3 Pauli matrices. The eigenvalues are $$ \frac12(1\pm\sqrt{\vec{n}\cdot\vec{n}}), $$ so it's non-negative iff $|\vec{n}|\leq 1$. In the case where the length is 1, the eigenvalues are 0,1 so $\rho$ is a projector onto a pure state.

Now let's consider the purity, $\text{Tr}(\rho^2)$. In this way of writing, we have $$ \text{Tr}(\rho^2)=\frac12(1+\vec{n}\cdot\vec{n}) $$ Clearly, the largest value of the purity is 1, for pure states, and the smallest value corresponds to $\vec{n}\cdot\vec{n}=0$, i.e. the maximally mixed state.

Now, how do we depict a state on the Bloch Sphere? We simply plot $\vec{n}$, so the distance from the centre of the sphere is given by $|\vec{n}|=\sqrt{2\text{Tr}(\rho^2)-1}$.

Proof of the basic properties

Let's say we have a density matrix of the form $$ \rho=\sum_ip_i|\psi_i\rangle\langle\psi_i|, $$ where $\sum_ip_i=1$. Trivially, $\rho$ has trace 1 and is Hermitian. Non-negative matrices satisfy the property $$ \langle\phi|\rho|\phi\rangle\geq 0\qquad\forall|\phi\rangle. $$ So, consider a state $|\phi\rangle$: $$ \langle\phi|\rho|\phi\rangle=\sum_ip_i|\langle\phi|\psi_i\rangle|^2. $$ Every term on the right-hand side must be non-negative, and hence $\rho$ is non-negative.

  • $\begingroup$ How does trace being 1 fix n_0 to be 1? $\endgroup$ Commented May 23, 2020 at 19:12
  • $\begingroup$ @EeshStarryn Pauli matrices have trace 0. Identity has trace 2. $\endgroup$
    – DaftWullie
    Commented May 24, 2020 at 5:18

Let me supplement the other answer by also showing what happens in the general case of the Bloch representation of generic qudits of dimension $d$.

Let $\rho$ be an arbitrary state over $d$ modes, that is, a $d\times d$ positive semi-definite Hermitian matrix with unit trace.

As any other Hermitian matrix, we can decompose it in terms of a basis of Hermitian matrices, writing it as something like $\rho=\sum_k c_k\sigma_k$ with $\sigma_k$ an orthonormal basis of Hermitians and $c_k\in\mathbb R$. There are many ways to choose such a basis for the Hermitian matrices, but a fairly natural choice is to include the identity matrix in it. If we do, the constraint of orthogonality automatically imposes that all the other elements of the basis must be traceless (because we must have $\text{Tr}(\mathbb I^\dagger\sigma_k)=\operatorname{Tr}(\sigma_k)=0$). Moreover, we remember that we must have $\text{Tr}(\rho)=1$, and thus we can write the decomposition of $\rho$ as: $$\rho=\frac{1}{d}\left(\mathbb I+\sum_k c_k\sigma_k\right),$$ where $\text{Tr}(\sigma_j\sigma_k)=\delta_{jk}d$. Note that this choice of normalisation is conventional (we could have equally written $\rho=\mathbb I/d+\sum_k c_k\sigma_k$ with the $\sigma_k$ normalised to one), but is somewhat "natural" in that it puts $\sigma_k$ to a more even footing with $\mathbb I$, whose normalisation is instead "forced" by the normalisation constraint on $\rho$ (of course, this is conventional as well, but is a "very established" convention).

It is now not hard to compute $\text{Tr}(\rho^2)$, and find it equal to $$\text{Tr}(\rho^2)=\frac{1}{d}\left(1+\sum_k c_k^2\right)=\frac{1}{d}(1+\lvert\vec c\rvert^2).$$

The state $\rho$ being pure means that it is a projector over some pure state, $\rho=\lvert\psi\rangle\!\langle\psi\rvert$, and thus satisfies $\text{Tr}(\rho^2)=1$. It then follows that $\rho$ is pure if and only if $|\vec c|=\sqrt{d-1}$.

It should be noted that while this result could seem to suggest that the state space for higher-dimensional systems is not significantly more complex than the Bloch sphere we have for qubits, this is far from the case. Indeed, the state space is in general not a (hyper)sphere, nor an (hyper)ellipse or other simple geometrical figure. It stretches by different amounts at different angles, and its boundary contains non-pure states. Nevertheless, at least the length of pure states is easy to compute.


A simple way of proving that the purity (or any other property which only depends on the eigenvalues of $\rho$) can only depend on the distance from the center of the Bloch sphere is rotational invariance: Rotating a density matrix $\rho\mapsto U\rho U^\dagger$ corresponds to a rotation on the Bloch sphere. Thus, any property which only depends on the eigenvalues (which do not change under $\rho\mapsto U\rho U^\dagger$) can only depend on the distance from the center.

To get an explicit formula, you can then e.g. restrict to diagonal density matrices.


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