## Density matrix after measurement on density matrix

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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?

2You're confusing kets and the associated pure density matrices. Should be like $|0\rangle \langle 0 |$ – AHusain – 2018-11-30T18:15:35.947

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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:

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

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

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.

to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer. – Hasan Iqbal – 2018-11-30T19:48:12.420

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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.

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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 $$\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|).$$