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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?

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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.

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    $\begingroup$ you can also prove this for more general bases of operators, see e.g physics.stackexchange.com/a/425101/58382. That gives you an iff condition for purity, replacing inequalities with identities, and using the correct length for pure states $\endgroup$
    – glS
    Commented Mar 16, 2022 at 9:04
  • $\begingroup$ @glS Thank you very much! Very interesting! $\endgroup$ Commented Mar 16, 2022 at 9:10
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A very simple derivation, without using the specific form of the Pauli matrices, can be obtained if one uses that the vector of Pauli matrices transforms as a $\mathrm{SO}(3)$ rotation under the adjoint action of $\mathrm{SU}(2)$ -- i.e, one has that $$ U (\vec r\cdot \vec\sigma) U^\dagger = (R_U\vec r)\cdot \vec\sigma $$ for any $U\in\mathrm{SU}(2)$, where $R_U$ is the $\mathrm{SO(3)}$ rotation corresponding to $U$ (modulo $\pm 1$).

Once you know this fact, $\rho = \alpha I + \sum \beta_i\sigma_i$ equals to $$ U\rho U^\dagger = \alpha I + |\vec\beta| \sigma_z\ . $$ Now you could use the explicit matrix form of $\sigma_z$ -- but you don't need to, all you need to know is that it has eigenvalues $\pm1$: Then it is immediate to see that $$ \mathrm{eig}(\rho) = \alpha\pm|\vec\beta| $$ and $$\mathrm{tr}(\rho) = 2\alpha\ . $$ This immediately answers all your questions:

  1. $\rho$ is a density operator iff $\ 2\alpha=1$ and $|\vec\beta|\le \alpha$.

  2. $\rho$ is pure iff $\ \alpha = |\vec\beta|$.

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  • $\begingroup$ Dear Norbert, thanks for your answer! A cute method indeed. $\endgroup$ Commented Nov 17, 2022 at 14:10
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    $\begingroup$ Thanks. I mostly wanted to write it down because I felt that the first equation is something worth knowing (and which can make life much simpler!). $\endgroup$ Commented Nov 17, 2022 at 14:12

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