# Can we write the density operator as a sum of mixed states?

In every resource I find (like Nielsen and Chuang or online courses), the density operator is defined as follows: we consider a sequence of pure states $$\left|\psi_i\right\rangle$$ with associated probabilities $$p_i$$. If the system is in state $$\left|\psi_i\right\rangle$$ with probability $$p_i$$, then its density operator is given by: $$\rho = \sum_ip_i\left|\psi_i\right\rangle\left\langle\psi_i\right|.$$ However, I do not understand the necessity of $$\left|\psi_i\right\rangle$$ being pure states for this. For instance, let us say that the system is in state (potentially mixed) $$\rho_i$$ with probability $$p_i$$ (that is, $$\rho_i$$ is a density operator). Then we can prove that its density matrix is given by: $$\rho=\sum_ip_i\rho_i.$$ For this, I use the Lemma defined in this answer, assuming it is correct. We consider an arbitrary unitary $$\mathbf{U}$$ and an arbitrary basis state $$|x\rangle$$. Applying $$\mathbf{U}$$ on the system, we obtain, with probability $$p_i$$: $$\mathbf{U}\rho_i\mathbf{U}^\dagger.$$ The probability of measuring $$|x\rangle$$ in this situation is thus given by: $$\mathrm{tr}\left(|x\rangle\langle x|\mathbf{U}\rho_i\mathbf{U}^\dagger\right).$$ Since this situation happens with probability $$p_i$$, the probability of measuring $$|x\rangle$$ is finally given by: $$\sum_ip_i\mathrm{tr}\left(|x\rangle\langle x|\mathbf{U}\rho_i\mathbf{U}^\dagger\right).$$ On the other hand, applying $$\mathbf{U}$$ on $$\rho$$ gives: $$\mathbf{U}\rho\mathbf{U}^\dagger=\sum_ip_i\mathbf{U}\rho_i\mathbf{U}^\dagger$$ which means that the probability of measuring $$|x\rangle$$ is given by, using the linearity property of the trace: $$\sum_ip_i\mathrm{tr}\left(|x\rangle\langle x|\mathbf{U}\rho_i\mathbf{U}^\dagger\right).$$ Hence, for any unitary $$\mathbf{U}$$ and any basis state $$|x\rangle$$, the probabilities of measuring $$|x\rangle$$ after having applied $$\mathbf{U}$$ on those systems are equal, hence their density matrices are identical.

I cannot see the error here, but on the other hand, I find it surprising that I did not find any mention to this (or I've missed it, which is also quite likely). Is this result correct? Otherwise, where's my mistake?

• Density operator is matrix, not vector, so it cannot be given by $\rho = \sum_ip_i\left|\psi_i\right\rangle$ Apr 15 '21 at 8:43
• Oops, forgot the $\left\langle\psi_i\right|$ indeed. Thanks! Apr 15 '21 at 8:44
• The formula $\rho = \sum_ip_i\left|\psi_i\right\rangle\left\langle\psi_i\right|$ shows that $\rho$ is a statistical mixture of pure states. The formula that you are proving, $\rho=\sum_ip_i\rho_i$ shows that $\rho$ can be written as a statistical mixture of statistical mixtures. Though probably correct (I did not check), I don't see much sense in it. Apr 15 '21 at 8:54
• @TristanNemoz oh, I see. You essentially want to prove that having "classical uncertainty" on which states you have amounts to describing your state as a statistical mixture of the individual states. That makes sense. I'd prove this more generally using a POVM then: you can show that the outcome probabilities resulting from having "one of the $\rho_i$ with probability $p_i$" are identical to those resulting from measuring $\rho=\sum_i p_i\rho_i$, and that this holds for any possible measurement. There is no need to assume purity in $\rho_i$ in this reasoning
– glS
Apr 15 '21 at 9:08
• I agree that it would be essentially equivalent. You can also simplify yours but dispensing with the unitary rotations by the way. You can simply show that for any $|\psi\rangle$ the expectation values $\langle\psi|\cdot|\psi\rangle$ match in the two situations ($\rho$ and mixture of $\rho_i$). This is enough to show that the two descriptions result in equal probabilities with respect to any projective measurement (btw, remember to tag people in comments, otherwise they might not get notified and not see your addressing them)
– glS
Apr 15 '21 at 10:23

All your working is correct. Of course, if you want to define a density operator/matrix, you cannot define it in terms of another density operator/matrix which is why you will not find an initial presentation of density matrices going in this way. That said, you may want to look up the definition of a mixed separable state. This is typically done in the form $$\rho_{\text{sep}}=\sum_ip_i\sigma^A_i\otimes \sigma^B_i$$ where $$\sigma^A_i$$ and $$\sigma^B_i$$ are density matrices. You could write them just as pure states but you will often see it in this form, consistent with the structure you're talking about.
Due to $$\rho$$ being a positive and adjoint, you can always spectrally decompose it into a convex combination of pure states, it's eigenvectors. However, since the measurement statistics, or the action of a unitary operator, as you observed, will not change if you do not represent it diagonally, a lot of papers or books will choose not to represent them as this decomposition.