Density matrix after measurement on density matrix

Question

Let's say Alice wants to send Bob a $|0\rangle$ with probability .5 and $|1\rangle$ also with probability .5. So after a qubit Alice prepares leaves her lab, the system could be represented by the following density matrix: $$\rho = .5 |0\rangle \langle 0| + .5 |1\rangle \langle 1|= \begin{bmatrix} .5 & 0 \\ 0 & .5 \end{bmatrix} $$Am I right?

Then, Bob would perform measurement in the standard basis. This is where I get confused. For example, he could get $|0\rangle$ with probability .5 and $|1\rangle$ with .5. So what is the density matrix representing the state after Bob performs his measurement in the standard basis accounting for both of his possible outcomes?

Answer 1

So, Bob is given the following state (also called the maximally-mixed state):

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

As you noticed, one nice feature of density matrices is they enable us to capture the uncertainty of an outcome of a measurement and account for the different possible outcomes in a single equation. Projective measurement is defined by a set of measurement operators $P_i$, one for each possible measurement outcome. For example, when measuring in the computational basis (collapsing to $|0\rangle$ or $|1\rangle$) we have the following measurement operators:

$P_0 = |0\rangle\langle0| = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, $P_1 = |1\rangle\langle1| = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$

where $P_0$ is associated with outcome $|0\rangle$ and $P_1$ is associated with outcome $|1\rangle$. These matrices are also called projectors.

Now, given a single-qbit density operator $\rho$, we can calculate the probability of it collapsing to some value with the following formula:

$p_i = Tr(P_i \rho)$

where $Tr(M)$ is the trace, which is the sum of the elements along the main diagonal of matrix $M$. So, we calculate the probability of your example collapsing to $|0\rangle$ as follows:

$p_0 = Tr(P_0 \rho) = Tr \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \right) = Tr \left( \begin{bmatrix} \frac 1 2 & 0 \\ 0 & 0 \end{bmatrix} \right) = \frac 1 2$

And the formula for the post-measurement density operator is:

$\rho_i = \frac{P_i \rho P_i}{p_i}$

which in your example is:

$\rho_0 = \frac{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}}{\frac 1 2} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

which is indeed the density matrix for the pure state $|0\rangle$.

We don't just want the density operator for a certain measurement outcome, though - what we want is a density operator which captures the branching nature of measurement, representing it as an ensemble of possible collapsed states! For this, we use the following formula:

$\rho' = \sum_i p_i \rho_i = \sum_i P_i \rho P_i$

in our example:

$\rho' = \left( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \right) + \left( \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right)$

$\rho' = \begin{bmatrix} \frac 1 2 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & \frac 1 2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{bmatrix} = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1|$

Your final density matrix is unchanged! This should actually be unsurprising, because we started out with the maximally-mixed state and performed a further randomizing operation on it.

Much of this answer copied from my previous detailed answer here, which in turn is based off of another answer here.

Answer 2

So Alice sends Bob a qubit with the density matrix

$$\rho = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \begin{bmatrix} .5 & 0 \\ 0 & .5 \end{bmatrix}$$

as you said. (I've fixed the notation to make it a density matrix, what you wrote was in the structure of a state, but with non-normalized coeffecients. It is important to note the distinction of pure states vs mixed states. What you are discussing is a mixed state and cannot be described as a single state $|\psi\rangle$, mixed states can only be described in the density matrix picture.)

After measurement the state Bob has becomes a pure state, as you described. He gets either

$$|0\rangle \rightarrow \rho = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ or $$|1\rangle \rightarrow \rho = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$$ with equal probability. As you can see once the qubit is measured you lose all indication that it was once in a mixed state or anything else about its history.

Answer 3

It is important to emphasise that a density matrix may not be absolute; it represents the state of somebody's knowledge of the system.

To see this, consider 3 parties: Alice, Bob and Charlie. Alice prepares a qubit in either $|0\rangle$ or $|1\rangle$. Now, Alice knows which state she prepared (let's assume it's $|0\rangle$), so Alice's description of the qubit is $|0\rangle\langle 0|$. However, Bob and Charlie only know the preparation procedure, not which choice Alice made. So, they both have to describe the state as 50:50 being 0 or 1. i.e. $$ \frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|). $$

Now, Bob receives the state and measures it in the Z basis. So long as Alice knows what measurement is being done, she knows that the state won't change, so her description of the state remains the same, $|0\rangle\langle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0\rangle$, so if he's using a density matrix, he describes it as $|0\rangle\langle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by $$ \frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|). $$