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

Question

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!

Answer 1

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.

Answer 2

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.