Mixed states and true randomness

Recently two nice papers appeared on the arXiv, the most recent by Galley and Masanes, and the oldest by López Grande et al.. They are both – although a bit indirectly – about the age old question of the equivalence between proper and improper mixtures.

A proper mixture is when you prepare the states $\ket{0}$ and $\ket{1}$ with probability $p$ and $1-p$, obtaining the density matrix
\[ \rho_\text{proper} = p\ket{0}\bra{0} + (1-p)\ket{1}\bra{1}.\] An improper mixture is when you prepare the entangled state $\sqrt{p}\ket{0}\ket{0} + \sqrt{1-p}\ket{1}\ket{1}$ and discard the second subsystem, obtaining the density matrix \[ \rho_\text{improper} = p\ket{0}\bra{0} + (1-p)\ket{1}\bra{1}.\] The question is then why do these different preparation procedures give rise to the same statistics (and therefore it is legitimate to represent them with the same density matrix).

Well, do they? I’m not so sure about that! The procedure to prepare the proper mixture is rather vague, so we can’t really answer whether is it appropriate to represent it via the density matrix $\rho_\text{proper}$. To remove the vagueness, I asked an experimentalist how she prepared the state $\frac12(\ket{0}\bra{0}+\ket{1}\bra{1})$ that was necessary for an experiment. “Easy”, she told me, “I prepared $n$ copies of $\ket{0}$, $n$ copies of $\ket{1}$, and then combined the statistics.

This sounds like preparing the state $\ket{0}^{\otimes n} \otimes \ket{1}^{\otimes n}$, not like preparing $\frac12(\ket{0}\bra{0}+\ket{1}\bra{1})$. Do they give the same statistics? Well, if I measure all states in the $Z$ basis, exactly $\frac12$ of the results will be $0$. But if I measure $\frac12(\ket{0}\bra{0}+\ket{1}\bra{1})$ in the $Z$ basis $2n$ times, the probability that $\frac12$ of the results are $0$ is
\[ \frac{1}{2^{2n}} {2n \choose n} \approx \frac{1}{\sqrt{n\pi}},\] so just by looking at this statistic I can guess with high probability which was the preparation. It is even easier to do that if I disregard her instructions and look at the order of the results: getting $n$ zeroes followed by $n$ ones is a dead giveaway.

Maybe one should prepare these states using a random number generator instead? If one uses the function rand() from MATLAB to decide whether to prepare $\ket{0}$ or $\ket{1}$ at each round one can easily pass the two randomness tests I mentioned above. Maybe it can even pass all common randomness tests available in the literature, I don’t know how good rand() is. But it cannot, however pass all randomness tests, as rand() is a deterministic algorithm using a finite seed, and is therefore restricted to outputting computable sequences of bits. One can, in fact, attack it, and this is the core of the paper of López Grande et al., showing how one can distinguish a sequence of bits that came from rand() from a truly random one. More generally, even the best pseudorandom number generators we have are designed to be indistinguishable from truly random sources only by polynomial-time tests, and fail against exponential-time algorithms.

Clearly pseudorandomness is not enough to generate proper mixtures; how about true randomness instead? Just use a quantum random number generator to prepare bits with probabilities $p$ and $1-p$, and use these bits to prepare $\ket{0}$ or $\ket{1}$. Indeed, this is what people do when they are serious about preparing mixed states, and the statistics really are indistinguishable from those of improper mixtures. But why? To answer that, we need to model the quantum random number generator physically. We start by preparing a “quantum coin” in the state
\[ \sqrt{p}\ket{H}+\sqrt{1-p}\ket{T},\] which we should measure in the $\{\ket{H},\ket{V}\}$ basis to generate the random bits. Going to the Church of the Larger Hilbert Space, we model the measurement as
\[ \sqrt{p}\ket{H}\ket{M_H}+\sqrt{1-p}\ket{T}\ket{M_T},\] and conditioned on the measurement we prepare $\ket{0}$ or $\ket{1}$, obtaining the state
\[ \sqrt{p}\ket{H}\ket{M_H}\ket{0}+\sqrt{1-p}\ket{T}\ket{M_T}\ket{1}.\] We then discard the quantum coin and the measurement result, obtaining finally
\[ p\ket{0}\bra{0} + (1-p)\ket{1}\bra{1},\] which is just the desired state, but now it is an improper mixture. So, at least in the Many-Worlds interpretation, there is no mystery about why proper and improper mixtures are equivalent: they are physically the same thing!

(A closely related question, which has a closely related answer, is why is it equivalent to prepare the states $\ket{0}$ or $\ket{1}$ with probability $\frac12$ each, or the states $\ket{+}$ or $\ket{-}$, again with probability $\frac12$? The equivalence fails for pseudorandomness, as shown by López Grande et al.; if we use true randomness instead, we are preparing the states
\[ \frac1{\sqrt{2}}(\ket{H}\ket{0}+\ket{T}\ket{1})\quad\text{or}\quad\frac1{\sqrt{2}}(\ket{H}\ket{+}+\ket{T}\ket{-})\] and discarding the coin. But note that if one applies a Hadamard to the coin of the first state one obtains the second, so the difference between then is just a unitary on a system that is discarded anyway; no wonder we can’t tell the difference! More generally, any two purifications of the same density matrix must be related by a unitary on the purifying system.)

Galley and Masanes want to invert the question, and ask for which quantum-like theories proper and improper mixtures are equivalent. To be able to tackle this question, we need to define what improper mixtures even are in a quantum-like theory. They proceed by analogy with quantum mechanics: if one has a bipartite state $\ket{\psi}$, and are doing measurements $E_i$ only on the first system, the probabilities one obtains are given by
\[ p(i) = \operatorname{tr}( (E_i \otimes \mathbb I) \ket{\psi}\bra{\psi} ),\] and the improper mixture is defined as the operator $\rho_\text{improper}$ for which
\[ p(i) = \operatorname{tr}( E_i \rho_\text{improper})\] for all measurements $E_i$.

In their case, they are considering a quantum-like theory that is still based on quantum states, but whose probabilities are not given by the Born rule $p(i) = \operatorname{tr}(E_i \ket{\phi}\bra{\phi})$, but by some more general function $p(i) = F_i (\ket{\phi})$. One can then define the probabilities obtained by local measurements on a bipartite state as
\[ p(i) = F_i \star \mathbb I (\ket{\psi}),\] for some composition rule $\star$ and trivial measurement $\mathbb I$, and from that an improper mixture as the operator $\omega_\text{improper}$ such that
\[ p(i) = F_i (\omega_\text{improper})\] for all measurements $F_i$.

Defining proper mixtures, on the other hand, is easy: if one can prepare the states $\ket{0}$ or $\ket{1}$ with probabilities $p$ and $1-p$, their proper mixture is the operator $\omega_\text{proper}$ such that for all measurements $F_i$
\[ p(i) = F_i(\omega_\text{proper}) = p F_i(\ket{0}) + (1-p) F_i(\ket{1}).\] That is, easy if one can generate true randomness that is not reducible to quantum-like randomness. I don’t think this makes sense, as one would have to consider a world where reductionism fails, or at least one where quantum-like mechanics is not the fundamental theory. Such non-reducible probabilities are uncritically assumed to exist anyway by people working on GPTs all the time1Not Galley and Masanes, I must add; they are explicitly addressing this question..

Now with both proper and improper mixtures properly defined, one can answer the question of whether they are equivalent: the answer is a surprising no, for any alternative probability rule that respects some basic consistency conditions. This has the intriguing consequence that if we were to modify the Born rule while keeping the rest of quantum mechanics intact, a wedge would be driven between the probabilities that come from the fundamental theory and some “external” probabilities coming from elsewhere. This would put the Many-Worlds interpretation under intolerable strain.

But such an abstract “no” result is not very interesting; I find it much more satisfactory to exhibit a concrete alternative to the Born rule where the equivalence fails. Galley and Masanes propose the function
\[ F_i(\ket{\psi}) = \operatorname{tr}(\hat F_i (\ket{\psi}\bra{\psi})^{\otimes 2})\] for some positive matrices $\hat F_i$ restricted by their consistency conditions. It is easy to see that the proper mixture of $\ket{0}$ and $\ket{1}$ described above is given by2Note that in this theory mixtures of vectors in $\mathbb C^d$ live in $\mathbb C^{d^4}$, whereas in quantum mechanics they live in $\mathbb C^{d^2}$
\[ \omega_\text{proper} = p \ket{00}\bra{00} + (1-p)\ket{11}\bra{11}.\] In quantum mechanics one would try to make it by discarding half of the state $\sqrt{p}\ket{0}\ket{0} + \sqrt{1-p}\ket{1}\ket{1}$. Here it doesn’t work, as nothing does, but I want to know what it gives us anyway. It is not easy to see that the improper mixture is given by the weirdo
\begin{multline} \omega_\text{improper} = (p^2 + \frac{p(1-p)}{3})\ket{00}\bra{00} + \\ \frac{2p(1-p)}{3} (\ket{01}+\ket{10})(\bra{01}+\bra{10}) + ((1-p)^2 + \frac{p(1-p)}{3})\ket{11}\bra{11}.\end{multline}