Is there a superposition of two density matrices?

I asked a similar question here.

You can generate superpositions of density matrices, but this is only super clear when everything is a pure state. This is a sort of trivial case, because the purpose of the density matrix is to work with mixed states, so if you're states are always pure it is usually pointless to try to generate the superposition of two mixed states.

My old question that I asked tried to work out the math for this, to figure out how to create superpositions in the density matrix picture without having to convert the pure state density matricies back to bras and kets and then convert back. I still don't know how to do this for general cases. So while it may be practically pointless to create superpositions in the density matrix framework when working exclusively with pure states, it's not so easy (to me) to work out the math for doing this in the general case.

But I am pretty sure that you are also able to create superpositions of mixed states.

A superposition of mixed states is a bit unusual conceptually, because a mixed state is created specifically to deal with states that are not in quantum superpositions - so I think there will be some that might insist such a thing doesn't exist, but I think you can do this.

Consider the following thought experiment, which I'm going to call Schrodinger's coin flip: I put my friend in a black box, and inside the box is a quantum coin, a regular coin, and a weighted coin. The "quantum coin" is in a superposition of heads and tails $|psirangle = |Hrangle + |Trangle$, the normal coin is a normal coin and would have to be represented as a density matrix: $begin{bmatrix} 1/2 & 0 0 & 1/2 end{bmatrix}$, and the weighted coin is some kind of coin used for cheating and has a 90% chance of being heads, and therefore has a density matrix of the form: $begin{bmatrix} 9/10 & 0 0 & 1/10 end{bmatrix}$.

So here's what I'm going to do. I'm going to ask my friend to make a measurement of the quantum coin, and he is going to flip the normal coin if he measures $|Hrangle$, and flip the weighted coin if he measures tails $|Trangle$.

Then in my perspective, he must be in a superposition of these two possibilities. But these two possibilities are mixed states! And therefore we have created a superposition of mixed states.

The state that I observe will have the form:

$$frac{1}{sqrt{2}}(|F_Hrangle otimes rho_{NC} + |F_Trangle otimesrho_{WC})$$

which is a superposition of two mixed states. (I should mention that writing an outer product of a density matrix and a pure state is very unconventional, but I am not sure that this can be written in a more conventional form since superpositions of mixed states aren't normally considered.) This is basically a copy of the Wigner's Friend thought experiment but where the friend flips classical coins when measuring a quantum state.

EDIT: Norbert criticizes that writing the above state is meaningless unless I've established a way of evaluating it, which is a good point. (Personally, I think it is conceptually obvious that this is a superposition of a mixed state and that developing a whole formalism to evaluate these states is overkill to answer that question, but I think its interesting nonetheless.)

if we considered two general coins with distributions:

$$rho_{wc} = begin{bmatrix} p & 0 0 & 1-p end{bmatrix}$$

$$rho_{nc} = begin{bmatrix} q & 0 0 & 1-q end{bmatrix}$$

So in our case, our mixed states are just classical probabilities, so we can work out each superposition as being generated from the combinatoric possibilities:

$$ begin{align*} p q%: & quad frac{1}{2} Big(|H_{nc}rangle otimes |F_{H,nc}rangle + |H_{wc}rangle otimes |F_{H,wc}rangle Big) p (1-q)%: & quad frac{1}{2} Big(|H_{nc}rangle otimes |F_{H,nc}rangle + |T_{wc}rangle otimes |F_{H,wc}rangleBig) (1-p) q %: & quad frac{1}{2} Big(|T_{nc}rangle otimes |F_{T,nc}rangle + |H_{wc}rangle otimes |F_{H,wc}rangle Big) (1-p) (1-q)%: & quad frac{1}{2} Big(|T_{nc}rangle otimes |F_{T,nc}rangle + |T_{wc}rangle otimes |F_{T,wc}rangleBig) end{align*} $$

So here we end up with a statistical mixture of multiple entangled states. If we want to keep track of the probabilities of each outcome, we can get these termms by collecting the elements in $rho_{wc}otimes rho_{nc}$

And if the normal coin and the weighted coin are the same coin (and maybe I'm just changing how I flip it), then this can be reduced to:

$$ begin{align*} p q%: & quad|Hrangle otimes |F_{H}rangle p (1-q)%: & quad frac{1}{2} Big(|Hrangle otimes |F_{H}rangle + |Trangle otimes |F_{H}rangleBig) (1-p) q %: & quad frac{1}{2} Big(|Trangle otimes |F_{T}rangle + |Hrangle otimes |F_{H}rangleBig) (1-p) (1-q)%: & quad |Trangle otimes |F_{T}rangle end{align*} $$

Overall, we see that there are meaningful ways of calculating and working with the state generated by the thought experiment. While it's true that such a state can be eventually reduced to a large density matrix, I think it's reasonable to conceptually think of this as a superposition of two statistical mixtures.

In the setting you describe, you cannot say much about the effect on the state on $lvert+ranglelangle+|$ (with $|+rangle=(|0rangle+|1rangle)/sqrt{2}$.

To see why, consider the two channels (acting on qubits) $$ S_1(rho) = tfrac12(rho+Zrho Z) , $$ which sets the off-diagonal entries to zero, and $S_2(rho)=rho$ (the identity channel).

Then, begin{align} S_1(|0ranglelangle0|)&=S_2(|0ranglelangle0|)=|0ranglelangle0| , S_1(|1ranglelangle1|)&=S_2(|1ranglelangle1|)=|1ranglelangle1| , end{align} that is, $S_1$ and $S_2$ act identically on $|0ranglelangle0|$ and $|1ranglelangle1|$.

On the other hand, $S_1(|+ranglelangle+|)=tfrac12 I$, while $S_2(|+ranglelangle+|)=|+ranglelangle+|$, so $S_1$ and $S_2$ have vastly different effects on $|+ranglelangle+|$.

Thus, knowing how a channel acts on $|0ranglelangle 0|$ and $|1ranglelangle 1|$ does not allow you to conclude how it acts on $|+ranglelangle+|$.

(As a side note, this is not that surprising, since the space of $2times 2$ matrices (to restrict to qubits) is four-dimensional, so you need to know the effect on at least 4 different states to uniquely fix the channel.)

Note that you can e.g. also consider $S_1'(rho)=tfrac12(rho+Xrho X)$ and $S_2'(rho)=tfrac12 I$: They map both $|0ranglelangle0|$ and $|1ranglelangle1$ to the same output $tfrac12 I$, so one might be tempted to think that $S'_bullet$ is constant, yet, they have very different effects on $|+ranglelangle+|$.

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It is an interesting question how much we can actually say about $S(|+ranglelangle+|)$ given $S(|kranglelangle k|)$ $k=0,1$. For sure, we can use that CPTP maps (or generally trace-nonincreasing CP maps) are contractive w.r.t. the trace norm, and thus, $|S(|+ranglelangle+|)-S(|kranglelangle k|)|_1le sqrt{2}$, which gives some non-trivial constraint (but as we saw from the example above, not enough to fix $S(|+ranglelangle+|)$, even if $|S(|0ranglelangle0|)-S(|1ranglelangle1|)|_1$ is maximal. Yet, there might some some other non-trivial things to say - but that is probably another question.