# What is meant with "different ensembles can give rise to the same density matrix?"

I am reading the Nielsen & Chuang section on density matrices and I don't understand the example given to demonstrate a concept. Here is what I am reading:

First, they said these two different ensembles of quantum states gave rise to the same density matrix. Are the two ensembles |a> and |b>? What do they mean by the same density matrix? It looks to me like they formed the density matrices for both of those states (which would be different if shown) and then added them, which equaled some random, artificial density matrix shown above.

Second, the probability of measuring the states is exactly what it says is not necessarily the case (3/4 and 1/4), so I am not sure what this example is demonstrating.

If anyone can clarify what is being said here that would be incredibly helpful. Thank you!

• – glS
Commented May 22 at 7:42

An ensemble in this context is a set of states with attached probabilities. In your example the ensembles would be written as $$\{(\frac12,|a\rangle\langle a|),(\frac12,|b\rangle\langle b|)\}$$ and $$\{(\frac34,|0\rangle\langle 0|),(\frac14,|1\rangle\langle 1|)\}$$.

The density matrix corresponding to a generic ensemble $$\{(p_i,\rho_i)\}_i$$ is $$\sum_i p_i \rho_i$$. So different ensembles can correspond to the same density matrix when this sum gives the same result, as is the case in your example: $$\rho = \frac12 |a\rangle\langle a|+\frac12|b\rangle\langle b|= \frac34|0\rangle\langle0| + \frac14 |1\rangle\langle1|.$$ If you measure this state in the basis $$\{|a\rangle,|b\rangle\}$$ you'll get $$(1/2,1/2)$$ as outcome probabilities, while if you measure it in the basis $$\{|0\rangle,|1\rangle\}$$ you'll get $$(3/4,1/4)$$ as outcome probabilities.

• If I measure in the {|𝑎⟩,|𝑏⟩} basis with outcome probabilities (1/2, 1/2), can I use the law of total probability to see what I would get in the {|0⟩,|1⟩} basis by using the equivalence relations for |𝑎⟩ and |𝑏⟩ and adding them? It doesn't seem to work because the |1⟩'s cancel out, leaving me with |0⟩ with probability 3/4, so the norm doesn't sum to 1. Commented May 22 at 16:03
• if you're asking whether the average probability of getting $|0\rangle$ after having first measured in the $\{|a\rangle,|b\rangle\}$ equals the probability of directly measuring in the latter basis, then actually yes, that is true. Both $|a\rangle$ and $|b\rangle$ have a 3/4 probability of giving $|0\rangle$ if measured in the computational basis, and each have probability 1/2 of being found when $\rho$ is measured in their basis. But note this is a different physical procedure altogether, that involves two subsequent measurements and an average over the first one
– glS
Commented May 22 at 16:35
• Oh, so is the point that only given the density matrix in the computational basis, we cannot recover the a, b ensemble without knowing the transformation between the computational basis and a,b basis? Commented May 22 at 20:40
• @researcher101 the density matrix isn't given in any basis. It's an operator that tells you the outcome probabilities with respect to any possible measurement. If you know the density matrix (regardless of how you wrote it) and you know what the elements of the a,b basis are, then you can compute the outcome probabilities
– glS
Commented May 23 at 5:48
• So if we were solely given the density matrix and two ordered bases: {|𝑎⟩,|𝑏⟩}, {|0⟩,|1⟩}. Can we always just read off the diagonals as the probabilities of measuring each element in the basis? so Pr(|𝑎⟩) = Pr(|0⟩) = 3/4 and Pr(|b⟩) = Pr(|1⟩) = 1/4. However, what if |𝑎⟩ is defined as |𝑎⟩ = 0.8|0> + 0.6|1>, now isn't there an inconsistency in the probabilities of measuring |0> if we take the direct route and the indirect route? Don't we need a specific definition of |𝑎⟩ = x|0> + y|1> such that the probabilities line up? I am pretty confused as you can see. Commented May 23 at 18:44

The two ensembles are $$|0\rangle, |1\rangle$$ and $$|a\rangle, |b\rangle$$. The point of this example is to show you that the same density matrix $$\rho = \begin{pmatrix} 3/4 & 0 \\ 0 & 1/4 \end{pmatrix}$$ cannot be interpreted as having the state $$|0\rangle$$ with probability $$3/4$$ and state $$|1\rangle$$ with probability $$1/4$$. This is because it can be also constructed by a different ensemble, and can be interpreted as having the state $$|a\rangle$$ with probability $$1/2$$ or the state $$|b\rangle$$ with probability $$1/2$$.

To sum up, each ensemble of quantum state can be represent by a density matrix in a closed form; but a density matrix does not have a unique decomposition into a specific ensemble.

• Oh, so is it basically saying that a downfall of density matrices is that they are matrices, and matrices don't specify the basis they are written in? Commented May 22 at 15:54