# Ensemble interpretation and density matrices - why is it impossible to distinguish two equivalent density matrices?

I read this in a paper : "... we cannot distinguish them by making measurements because they have the same density matrix". The authors are referring to two different decompositions of the same density matrix. Example:

\begin{aligned} &\rho=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}| 1\rangle\langle 1|\\ &\rho=\frac{1}{2}|+\rangle\langle+|+\frac{1}{2}|-\rangle\langle-| \end{aligned}

these cannot be experimentally distinguished.

My question is why that is the case? Because the ensemble is a classical superposition ie. the two (pure) states in each ensemble exist a priori to measurements. So if I measure the first one and get eigenvalue corresponding to $$|0\rangle$$ then I can say that the ensemble has to be the first one. Similarly, if I get value corresponding to the state $$|x\rangle$$ I can say the ensemble is the second one. This is true, then why the statement that we cannot experimentally distinguish?

Also, if ultimately the statement is true, then why to even write a density matrix in a concrete decomposed form, if that form and any other are physically indistinguishable?

Thanks

• In both cases, $\rho=\frac{1}{2}Id$. It's just a question of basis choice of decomposition. If you fix the basis, the matrix is uniquely determined by its coefficients. It's just a question of mathematical convenience. Sometimes, it is even better not to use a orthonormal basis (or even a basis).
– LPZ
Apr 21 at 14:58
• A classical potential is only unique up to a constant, all inertial systems are equivalent etc.. We have many situations in physics where the same physical reality is open to multiple equivalent mathematical descriptions. This does not look any different from that to me. Apr 21 at 14:59
• @FlatterMann Good point. But it seems there is some distinction here. The properties or the eigenvalues of these states must be objective, unlike potential (mathematical construct) and velocities (observer dependent or subjective upto a point). Apr 21 at 15:24
• @FlatterMann if I construct an end symbol of 1000 pure states st 500 each are states 0 and 1; this ensemble must be different from another ensemble where 500 each states are +x and -x. This is because the first ensemble should give me values 0 and 1 while the latter something else; even though the statistical average remains same Apr 21 at 15:26
• @Kobamschitzo. You are misinterpreting the meaning of the density matrix. The first one does not mean that you can only get 0 and 1 as the result of a measurement, because what if you were measuring $\sigma_x$ instead of $\sigma_z$? Then you should be able to get "+" and "-" (so to speak) as the result of the measurement, and again, both of those density matrices will yield the same results. Apr 21 at 15:30

It is not true that a measurement result of 0 distinguishes those two states. The probability of getting $$a$$ as a result of a measurement of the observable associated with the operator $$\hat{A}$$ is given by $$\Pr(m=a)=\operatorname{Tr}(\rho\hat{A})\,,$$ and so if we compute these probabilities with the density matrices above, we have to get the same result, since the density matrices are mathematically the same.

To see this explicitly for this case, we first note that $$\lvert{\pm}\rangle=\frac{\lvert{0}\rangle\pm\lvert{1}\rangle}{\sqrt{2}}\,.$$ Taking $$\hat{Z}=(1)\lvert{1}\rangle\langle1\rvert+(0)\lvert{0}\rangle\langle 0\rvert =\lvert{1}\rangle\langle1\rvert\,$$ as the operator we're measuring, we can then compute the probabilities. First, $$\Pr(m=0|\rho_1)= \operatorname{Tr}\left(\rho\hat{A}\right) =\operatorname{Tr}\left( \left( \frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}| 1\rangle\langle 1| \right) \lvert{1}\rangle\langle1\rvert\right) =\frac{1}{2} \operatorname{Tr}\left(\lvert{1}\rangle\langle1\rvert\right) = \frac{1}{2}\,.$$ A similar computation yields exactly the same result for the second state. (I leave it as an exercise. I didn't want to do it because as part of the computation, I would have just shown that the second density matrix can be written as the first.)

Any probability of an outcome for any measurement can be shown to yield the same result for these two density matrices (again, they are mathematically the same, so they have to yield the same results), and so these represent the same physical state.

I think perhaps the confusion comes in by interpreting $$\rho = \frac{1}{2}|0\rangle\langle0|+\frac{1}{2}|1\rangle\langle1|\,,$$ as being similar in nature to $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}\lvert0\rangle + \frac{1}{\sqrt{2}}\lvert1\rangle\,.$$ That is, if we consider the two states $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}\lvert0\rangle + \frac{1}{\sqrt{2}}\lvert1\rangle = \lvert+\rangle\,,$$ and $$\lvert\psi\rangle = \frac{1}{\sqrt{2}}\lvert+\rangle + \frac{1}{\sqrt{2}}\lvert-\rangle = \lvert0\rangle\,,$$ the probabilities of getting 0 as the result of a measurement of $$\hat{Z}$$ are clearly different! In one case the probability is 1/2 and in the other the probability is 1. Density matrices just don't work this way.

On the other hand, let's finally address the suggestion by the OP. Suppose we measured $$\hat{Z}$$ on only one of the systems in the ensemble. The OP suggests that upon getting the result 0, this would reveal that that particular system was one of the 50% of systems that was in state $$|0\rangle$$, thus revealing that the entire ensemble must consist of systems that are in either $$|0\rangle$$ or $$|1\rangle$$.

However, getting 0 as the result of the measurement doesn't tell us what the state before the measurement was. Rather, it only tells us that the system is in state $$|0\rangle$$ post-measurement*. Since there is a non-zero probability of getting 0 as the result of the measurement when half the systems are in state $$|+\rangle$$ and half are in state $$|-\rangle$$, this measurement cannot distinguish which statistical configuration the ensemble is in.

And, finally, in fact, in order to figure that out, you have to do a complete ensemble measurement, and the statistics of that sequence of measurements will be identical in both cases.

• Thx for the answer. I am clear in the latter part of your answer ie. I am not confusing a statistical mixture with a superposition. As for the first part of the answer, I am aware of finding the expected value using the Tr(rho.A). But it is an average value over the statistical mixure. When I said to measure 0, I meant take one particle from the ensemble - it will be either one of the two pure states; in this case lets say we get 0 state. Now measuring that will give us its eigenvalue, which will tell us that the ensemeble was the first one (of 0 and 1) and not of +x,-x. Apr 21 at 15:32
• Thus, if it is true, what I am suggesting is a way to distinguish these equivalent 'indistinguishable ' density matrices. Apr 21 at 15:33
• @Kobamschitzo. Well no, if you measure a single system in the ensemble and get the value 0, you actually have no idea what the state of that system was! All you know is that it is now in the pure state $|0\rangle$. Let's say that this particular system was in the state $|+\rangle$. Then you could get 0 as the result of the measurement, and if you did, you still wouldn't know that it used to be $|+\rangle$. Apr 21 at 15:39
• Agreed, I think that's what I was not seeing. In case of single measurements like that we don't know where it comes from , and in case of many measurements it converges to statistical averages. Thx! Apr 21 at 16:04
• @Kobamschitzo Since you agree, I'll add that to my post so that I can actually answer your question! Apr 21 at 17:55

Your two definitions of $$ρ$$ evaluate to the same matrix, so they can't be different. I think what you're really asking is why it's impossible to tell the difference between

• A state that is either $$|0\rangle$$ or $$|1\rangle$$, each with probability ½
• A state that is either $$|{+}\rangle$$ or $$|{-}\rangle$$, each with probability ½

The reason is that the only type of experiment you can do in quantum mechanics is measure a system in some orthogonal basis, and look at the probability of each outcome over many such experiments on identically prepared systems.

If the state of the system is $$|ψ\rangle$$, and you measure in a basis $$\{|x_i\rangle\}$$, the probability of getting outcome $$x_i$$ is $$\left|\langle ψ|x_i\rangle\right|^2$$, which could also be written $$\langle x_i|ψ\rangle\langle ψ|x_i\rangle$$.

If the state of the system is $$|ψ_j\rangle$$ with probability $$p_j$$, and you measure in a basis $$\{|x_i\rangle\}$$, the probability of getting outcome $$x_i$$ is

$$\sum_j p_j\,\left|\langle ψ_j|x_i\rangle\right|^2 = \sum_j p_j\, \langle x_i|ψ_j\rangle\langle ψ_j|x_i\rangle = \langle x_i| \left( \sum_j p_j\,|ψ_j\rangle\langle ψ_j| \right) |x_i\rangle$$

That holds for any choice of basis, so two systems are experimentally indistinguishable if the quantity in parentheses is the same for both of them.

"I am aware of finding the expected value using the $$Tr(\rho A)$$. But it is an average value over the statistical mixure"

Density matrix not only gives expected value, it also gives the probability of getting any measured outcome. Given any observable A, $$\langle a| \rho | a \rangle$$ is the probability of getting outcome a. You can write the density matrix explicitly thorugh the {up, down basis} or {+, - basis}. However, the density matrix itself is the same so any probablity from any measurement outcome is already determined. In this sense, as long as the density matrices of two situations, are the same, no amount of experiment can distinguish them.

I read this in a paper : "... we cannot distinguish them by making measurements because they have the same density matrix". The authors are referring to two different decompositions of the same density matrix. Example:
\begin{aligned} &\rho=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}| 1\rangle\langle 1|\\ &\rho=\frac{1}{2}|+\rangle\langle+|+\frac{1}{2}|-\rangle\langle-| \end{aligned}

These can not be distinguished because they are literally the same matrix: $$\rho=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}| 1\rangle\langle 1| = \left(\begin{matrix}1/2 & 0 \\ 0 & 1/2\end{matrix}\right)$$ $$\rho=\frac{1}{2}|+\rangle\langle +|+\frac{1}{2}|-\rangle\langle -| = \left(\begin{matrix}1/2 & 0 \\ 0 & 1/2\end{matrix}\right)$$

You can also see this more abstractly from expressions in terms of the the bras and kets: $$\rho=\frac{1}{2}|+\rangle\langle +|+\frac{1}{2}| -\rangle\langle -| =\frac{1}{2}\left( \frac{1}{2}\left(|0\rangle + |1\rangle\right)\left(\langle 0| + \langle 1 |\right) + \frac{1}{2}\left(|0\rangle - |1\rangle\right)\left(\langle 0| - \langle 1 |\right) \right)$$ $$=\frac{1}{4}\left( |0\rangle\langle 0| +|0\rangle\langle 1| +|1\rangle\langle 0| +|1\rangle\langle 1| +|0\rangle\langle 0| -|0\rangle\langle 1| -|1\rangle\langle 0| +|1\rangle\langle 1| \right)$$ $$=\frac{1}{4}\left( |0\rangle\langle 0| +|1\rangle\langle 1| +|0\rangle\langle 0| +|1\rangle\langle 1| \right)$$ $$=\frac{1}{2}\left( |0\rangle\langle 0| +|1\rangle\langle 1| \right)$$

Also, if ultimately the statement is true, then why to even write a density matrix in a concrete decomposed form, if that form and any other are physically indistinguishable?

Because we need to actually calculate things.