# Fallacy of special significance of eigenvalues and eigenvectors of density operator

This question is an addition to the following question. Nielsen and Chuang open the discussion of the unitary freedom on the ensemble for density matrices by pointing out the common fallacy to suppose that eigenvalues and eigenvectors of a density matrix have some special significance.

They illustrate this by showing that two distinct ensembles might give rise to the same density operator. I do understand this part of the argument. However, where I struggle is the comment at the beginning of the example, where they state

For example, one might suppose that a quantum system with a density matrix $$\rho = \frac{3}{4} |0\rangle \langle 0| + \frac{1}{4} |1\rangle \langle 1|.$$ must be in the state $$|0\rangle$$ with probability 3/4 and in the state $$|1\rangle$$ with probability 1/4. However, this is not necessarily the case.

They proceed at this point with the example and show another ensemble that gives rise to the same density matrix as above. My problem with the sentence is that the factors 3/4 and 1/4 are not related to the probabilities of the states $$|0\rangle$$ and $$|1\rangle$$.

The problem I have is that the very definition of a density operator states that for an ensemble $$\{p_i, |\psi _i\rangle\}$$ with quantum states $$|\psi _i\rangle$$ and corresponding probabilities $$p_i$$ the density matrix is given by $$\rho \equiv \sum_i p_i | \psi _i\rangle \langle \psi _i |$$

So in other words, the system with $$\rho = \frac{3}{4} |0\rangle \langle 0| + \frac{1}{4} |1\rangle \langle 1|.$$ must be in the state $$|0\rangle$$ with probability 3/4 and in the state $$|1\rangle$$ with probability 1/4 as given in the definition.

Nonetheless, as the example of Nielsen&Chuang shows, a density matrix is not uniquely associated with one ensemble, so it is also true that a system with $$\rho = \frac{1}{2} |a\rangle \langle a| + \frac{1}{2} |b\rangle \langle b| = \frac{3}{4} |0\rangle \langle 0| + \frac{1}{4} |1\rangle \langle 1|$$ must be in the state $$|a\rangle$$ with probability 1/2 and in the state $$|b\rangle$$ with probability 1/2.

In my opinion, this should not be a problem as those are distinct states and hence have different probabilities.

So why do they call it a common fallacy to think like that? Where lies my misunderstanding?

It is a "problem" because your statement might not reflect reality.

Consider the even simpler example: $$\rho = \frac12 |0\rangle \langle 0| + \frac12 |1\rangle\langle 1| = \frac12 |+\rangle \langle +| + \frac12 |-\rangle \langle -|$$

If I understand you correctly you have no problem with the following two statements both being true:

1. The state of the system is $$|0\rangle$$ with probability $$1/2$$ and $$|1\rangle$$ with probability $$1/2$$.
2. The state of the system is $$|+\rangle$$ with probability $$1/2$$ and $$|-\rangle$$ with probability $$1/2$$.

But suppose now the experimentalist tells you that they prepared the system in the $$\{|0\rangle, |1\rangle\}$$ basis by flipping a coin and preparing $$|0\rangle$$ if the coin was heads and $$|1\rangle$$ if the coin was tails. The experimentalist remembers the outcome of the coin, writes it down on a piece of paper and places it in a sealed envelope which they give to you along with the quantum state $$\rho$$.

Now if you measure $$\rho$$ in the $$\{|0\rangle, |1\rangle\}$$ basis you will ALWAYS see the outcome that agrees with the experimentalist's coin. This is because the statement 1 is correct. However, if you measured in the $$\{|+\rangle, |-\rangle\}$$ basis and you assign heads to a $$+$$ outcome and tails to a $$-$$ outcome then you will only agree with the experimentalists coin toss 50% of the time. This is because statement 2 is not correct, the state of the system was never $$|+\rangle$$ or $$|-\rangle$$.

• Thank you. So, with only the knowledge of the density operator, I cannot make any of the statements as in 1) or 2) since I do not know in which state the system actually has been prepared. Nonetheless, one of the infinitely many statements would be correct as the system must have been prepared in some state. Surely I do not see it yet, but why would we even work with the representation in terms of + and - states, then? It seems wrong to use it. Since the representation is not unique, we shouldn't use any representation as long as we do not know the "correct" prepared states for that system. Commented Aug 14 at 15:16
• Density matrices are a mathematical representation of your state of knowledge of the system. If you do not know how the system was prepared then from your perspective all preparations are a priori possible. The density matrices allow you to calculate probabilities and other interesting properties of the system which are valid irrespective of the preparation (I don't think I have ever had to care about which preparation was "correct" before). The preparation you choose to use is usually of mathematical convenience, often the representation from the spectral decomposition is useful. Commented Aug 14 at 15:25

I think it is mostly a matter of how your read their statement. Yes, it is true that if you construct the density operator from the following given ensemble:

$$\left\{p_i, |\psi_i\rangle \right \} = \left \{\left(\frac{3}{4}, |0\rangle \right) , \left(\frac{1}{4}, |1\rangle \right)\right \}$$

using the definition of the density matrix:

$$\rho = \frac{3}{4} |0\rangle \langle 0| + \frac{1}{4} |1\rangle \langle 1|$$

you are guaranteed to have state $$|0\rangle$$ with probability $$3/4$$, and state $$|1\rangle$$ with probability $$1/4$$ because you knew the ensemble to begin with.

But if you are given its density operator in the form of a matrix:

$$\rho = \begin{bmatrix} \frac{3}{4} & 0 \\ 0 & \frac{1}{4} \end{bmatrix},$$

without knowing how it was constructed (or you deduced it from measurements), then it is impossible for you to know what the underlying prepared states and their associated classical probabilities are. This is because, as you point out, there is an infinite combination of pure states and probabilities that can be used to construct the same matrix. An example would be the ensemble:

$$\left\{p_i, |\psi_i\rangle \right \} = \left \{\left(\frac{1}{2}, |a\rangle = \sqrt{\frac{3}{4}}|0\rangle + \sqrt{\frac{1}{4}}|1\rangle \right) , \left(\frac{1}{2}, |b\rangle = \sqrt{\frac{3}{4}}|0\rangle - \sqrt{\frac{1}{4}}|1\rangle \right)\right \}$$

which results in the operator:

$$ϙ = \frac{1}{2} |a\rangle \langle a| + \frac{1}{2} |b\rangle \langle b|.$$

And if you work out the math, this has the same density matrix as ensemble given above:

$$ϙ = \begin{bmatrix} \frac{3}{4} & 0 \\ 0 & \frac{1}{4} \end{bmatrix}.$$

• Thanks for pointing out the difference between having the density operator given as a sum or a matrix. However, does it mean then that when only given the matrix form, we could write the operator also in terms of any ensemble in sum format even though for a specific system every representation except one would be "false"? Commented Aug 14 at 15:21
• Well yes. You can write it in terms of any ensemble in sum form, but physically that representation you are assigning might not be the one really describing the system that was originally prepared. Now, whether that matters or not is a different story. The density matrix gives you all the information you need to compute statistics of measurements of the prepared system, but if you really need to know what ensemble was prepared, you won't be able to deduce it from the density matrix. Commented Aug 14 at 15:50