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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?

glS
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Tristan Nemoz
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  • Density operator is matrix, not vector, so it cannot be given by $\rho = \sum_ip_i\left|\psi_i\right\rangle$ – kludg Apr 15 '21 at 08:43
  • Oops, forgot the $\left\langle\psi_i\right|$ indeed. Thanks! – Tristan Nemoz Apr 15 '21 at 08:44
  • 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. – kludg Apr 15 '21 at 08:54
  • Well, to be fair, I'd just like to use this result in a proof, not to describe an actual real-world case. This would be quite handy: knowing that the system is described by $\rho_i$ with probability $p_i$ would allow to compute the associated density matrix easily. – Tristan Nemoz Apr 15 '21 at 08:57
  • I don't quite understand what you are trying to prove with this reasoning. A convex combination of states is again a state, yes. A state is defined as a positive semidefinite operator with trace 1, so you only need to verify that these are verified for convex combinations of states to get the result – glS Apr 15 '21 at 08:58
  • @kludg, your point is that the result is (seemingly) correct, but since it does not make much sense to describe an actual system it is not described in resources, am I correct? – Tristan Nemoz Apr 15 '21 at 08:58
  • @glS, by doing so, I would show that this is a state, but not that this is necesarily the state associated with my system, am I correct? – Tristan Nemoz Apr 15 '21 at 09:00
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    @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 09:08
  • To be honest, I'm quite unfamiliar with POVM measurements (I've learnt about the density operator formalism fairly recently). From my understanding though, your proof is what I've tried to do in mine, since my goal is to show that after having applied an arbitrary unitary, the probability of a projective measurement is the same in both systems. Since according to NC "projective measurements augmented by unitary operations turn out to be completely equivalent to general measurements" and "POVMs are best viewed as a special case of the general measurement formalism". – Tristan Nemoz Apr 15 '21 at 09:21
  • Hence, I feel like I'm just reinventing the wheel here, a proof using POVM would be, I think, much shorter than mine. – Tristan Nemoz Apr 15 '21 at 09:22
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    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
  • @glS oh, that's much simpler indeed. Thank your very much! – Tristan Nemoz Apr 15 '21 at 10:56

2 Answers2

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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.

DaftWullie
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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.

GaussStrife
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