# Necessary and sufficient conditions for operator on $\mathbb C^2$ to be a density matrix

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?

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.

• 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
– glS
Mar 16, 2022 at 9:04
• @glS Thank you very much! Very interesting! Mar 16, 2022 at 9:10

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 $$\ \alpha = |\vec\beta|$$.

• Dear Norbert, thanks for your answer! A cute method indeed. Nov 17, 2022 at 14:10
• 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!). Nov 17, 2022 at 14:12