Consider a one-qubit system with Hilbert space $\mathscr H\simeq \mathbb C^2$.
Define the hermitian operator
$$\rho := \alpha\, \sigma_0 + \sum\limits_{i=1}^3 \beta_i\, \sigma_i \quad , \tag{1}$$
where $\alpha,\beta_i \in \mathbb R$, $\sigma_0 = \mathbb I_{\mathbb C^2}$ and $\sigma_i$ are the usual Pauli matrices. What are the necessary and sufficient conditions for $\rho$ to be a density operator, that is a positive semi-definite operator with unit trace? Under which conditions is $\rho$ pure? Can these conditions be derived without using the explicit matrix representation of the Pauli matrices?
Best Answer
Let us first derive necessary conditions on the coefficients, so assume $\rho$ is a density matrix. From $\mathrm{Tr} \rho =1$ it trivially follows that $\displaystyle \alpha=\frac{1}{2}$. To proceed, let $\lambda$ and $1-\lambda$ denote the eigenvalues of $\rho$. As shown e.g. here, we find $$\det \sum\limits_{i=1}^3 \beta_i \,\sigma_i = -\sum\limits_{i=1}^3 \beta_i^2$$ and thus
$$\det \left(\rho - \frac{\sigma_0}{2}\right) = -\sum\limits_{i=1}^3 \beta_i^2 \quad . $$
Further, since $[\rho,\sigma_0]=0$ trivially, we have that the eigenvalues of $\rho - \frac{\sigma_0}{2}$ are given by $\lambda-\frac{1}{2}$ and $1-\lambda - \frac{1}{2}$. Hence $$ \left(\lambda-\frac{1}{2}\right) \left(1-\lambda - \frac{1}{2}\right) = -\sum\limits_{i=1}^3\beta_i^2 \quad , $$
which eventually leads to $$\det \rho = \lambda \left(1-\lambda\right) = -\sum_{i=1}^3 \beta_i^2 +\frac{1}{4} \quad . $$
Because of $0 \leq \lambda\leq 1$, we require $\det \rho \geq 0$, so for $\rho$ in $(1)$ to be a density matrix the coefficients must fulfill: $$\alpha=\frac{1}{2} \quad \text{and} \quad \sum\limits_{i=1}^3 \beta_i^2 \leq \frac{1}{4} \quad . \tag{2} $$ Moreover, from $\det \rho = 0$ if and only if $\lambda=1$ or $\lambda=0$, we see that $\rho$ is pure if and only if the equality in $(2)$ holds.
Finally, note that these conditions are also sufficient: If an operator of the form $(1)$ obeys equation $(2)$, then $\mathrm{Tr} \rho=1$ and $\det \rho \geq 0$. It remains to show that both eigenvalues are non-negative. But since $\det \rho \geq 0$, we know that both eigenvalues have the same sign and from the trace condition it follows that both must be non-negative.