What does it mean for a result of a measurement to be known/unknown?

Question

Let me start the question with two examples.

First, I am reading Nielsen & Chuang section "8.3.3 Bit flip and phase flip channels". There is a description of a quantum operation

$\rho \to \mathcal{E}(\rho) = P_0 \rho P_0 + P_1 \rho P_1$, where $P_0 = |0 \rangle \langle 0|$, $P_1 = |1 \rangle \langle 1|$, which corresponds to a measurement of the qubit in the $|0 \rangle$, $|1 \rangle$ basis, with the result of the measurement unknown. [Italics is mine - A.P.]

Second, in the edX course "Quantum Information Science I, Part 3" there is a question that looks like this:

After quantum measurement <...>, if the measurement result is known, <...>. [Italics is mine - A.P.]

So, I do not understand what does that mean for a result of a measurement to be known/unknown? Moreover, how could that knowledge or an absense of knowledge further affect the quantum system once the measurement is performed? Would anything change in the examples if we replace "known" with "unknown" and vice versa? Is there a mathematical formalism for the "is known/unknown" expression?

I believe, the source of my confusion comes from the Schrödinger's cat paradox solution. My understanding is that the cat is strictly alive or dead once the "measurement" by a detector happens, regardless of whether we know the fact (i.e., result of the "measurement") or not. That is a knowledge of an experimenter, and it has no relation to the "measurement".

Answer 1

Forget about quantum mechanics for a second and consider two people predicting a coin flip. Alice flips a coin, covers it with her hand, and asks Bob to predict the result. Alice knows the coin is heads, but Bob is unsure if it is heads or tails. They will describe the state of the coin using different probability distributions.

The same situation can apply to quantum systems. Alice may know the state of the system, while Bob is unsure. When this happens, they each will describe the situation using a different density matrix: the matrix formalism that is used to describe a quantum system in a mixed state, meaning a statistical ensemble of several quantum states.

(On the other hand, if they describe the same system using different superpositions, that's bad. At least one of them is objectively wrong. Similarly, if Alice says a die roll was definitely 100% five, and Bob says the die roll was definitely 100% six, at least one of them is dead wrong.)

how could that knowledge or an absense of knowledge further affect the quantum system once the measurement is performed?

It doesn't affect the system, it determines your ability to accurately describe the system.

For example, suppose I have a qubit in the state $\frac{3}{5} |0\rangle + \frac{4}{5}|1\rangle$. I measure the state. I tell you all this, but don't tell you what the measurement outcome was.

If you're asked to predict the state of the system, the best you can do is bet 36% odds on $|0\rangle$ and 64% odds on $|1\rangle$. And a succinct way to describe your knowledge about the system is the density matrix $0.36 |0\rangle \langle 0| + 0.64 |1\rangle \langle 1| = \begin{bmatrix} 0.36 & 0 \\ 0 & 0.64\end{bmatrix}$. By contrast, I know the measurement result. It happens to have been $|1\rangle$. So the density matrix describing my knowledge is $|1\rangle \langle 1| = \begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}$.

Would anything change in the examples if we replace "known" with "unknown" and vice versa?

If you perform the same operations to multiple systems in the state $|\psi\rangle$, their outputs will be identically distributed regardless of which ones you knew started in the state $|\psi\rangle$.

However, your ability to do useful tasks with a quantum system often depends on knowing what state that system is in. In that sense it does matter if you know the state or not. For example, it's pretty hard to do magic state distillation if you keep forgetting whether or not you already did the distillation.

Is there a mathematical formalism for the "is known/unknown" expression?

Yes. Density matrices.

Answer 2

Suppose you have a qubit in a state $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle$. For simplicity let us assume $\alpha$ and $\beta$ are real.

Alternatively, the state can be described by density matrix $$\rho=|\psi\rangle\langle\psi|=\begin{pmatrix} \alpha^2 & \alpha\beta \\ \alpha\beta & \beta^2 \end{pmatrix} $$

If we measure the qubit in the standard basis but don't look an the measurement outcome, the qubit's state after measurement is

$$\rho_{out}=\begin{pmatrix} \alpha^2 & 0 \\ 0 & \beta^2 \end{pmatrix} $$

(measurement killed off-diagonal terms, and now it is a mixed state)

If we looked at the outcome of the measurement and found that the outcome is $|0\rangle$ state, then the qubit's state after measurement is

$$\rho_{out,0}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} $$

The relation between $\rho_{out}$ and $\rho_{out,i}$ is the same as relation between unconditional and conditional probabilities studied in probability theory, $$\rho_{out}=\alpha^2 \rho_{out,0}+\beta^2 \rho_{out,1} $$

as explained in Craig's answer.

Concerning the cat - yes, the cat is strictly dead or alive after the box is opened (cat is measured), but we don't know it before we looked into the box. This is purely classical situation, well known in probability theory.