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

Then, Bob would perform measurement in the standard basis. This is where I get confused. For example, he could get $|0rangle$ with probability .5 and $|1rangle$ 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?

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

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

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 $|0rangle$ or $|1rangle$) we have the following measurement operators:

$P_0 = |0ranglelangle0| = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}$, $P_1 = |1ranglelangle1| = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$

where $P_0$ is associated with outcome $|0rangle$ and $P_1$ is associated with outcome $|1rangle$. 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 $|0rangle$ 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 $|0rangle$.

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:

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

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

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.

So Alice sends Bob a qubit with the density matrix

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

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 $|psirangle$, 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

$$|0rangle rightarrow rho = begin{bmatrix} 1 & 0 \ 0 & 0 end{bmatrix}$$
or $$|1rangle rightarrow rho = begin{bmatrix} 0 & 0 \ 0 & 1 end{bmatrix}$$
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.