Density matrix after measurement on density matrix











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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?










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    You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
    – AHusain
    Nov 30 at 18:15

















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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?










share|improve this question




















  • 1




    You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
    – AHusain
    Nov 30 at 18:15















up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





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?










share|improve this question















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?







quantum-information measurement cryptography density-matrix






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edited Dec 1 at 6:55









Blue

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asked Nov 30 at 17:55









Hasan Iqbal

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3086








  • 1




    You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
    – AHusain
    Nov 30 at 18:15
















  • 1




    You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
    – AHusain
    Nov 30 at 18:15










1




1




You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
– AHusain
Nov 30 at 18:15






You're confusing kets and the associated pure density matrices. Should be like $|0rangle langle 0 |$
– AHusain
Nov 30 at 18:15












3 Answers
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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.






share|improve this answer























  • to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
    – Hasan Iqbal
    Nov 30 at 19:48


















up vote
4
down vote













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.






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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 $|0rangle$ or $|1rangle$. Now, Alice knows which state she prepared (let's assume it's $|0rangle$), so Alice's description of the qubit is $|0ranglelangle 0|$. 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.
    $$
    frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
    $$



    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, $|0ranglelangle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0rangle$, so if he's using a density matrix, he describes it as $|0ranglelangle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by
    $$
    frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
    $$






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      3 Answers
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      3 Answers
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      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.






      share|improve this answer























      • to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
        – Hasan Iqbal
        Nov 30 at 19:48















      up vote
      5
      down vote



      accepted










      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.






      share|improve this answer























      • to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
        – Hasan Iqbal
        Nov 30 at 19:48













      up vote
      5
      down vote



      accepted







      up vote
      5
      down vote



      accepted






      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.






      share|improve this answer














      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.







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 30 at 19:45

























      answered Nov 30 at 19:40









      ahelwer

      1,215112




      1,215112












      • to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
        – Hasan Iqbal
        Nov 30 at 19:48


















      • to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
        – Hasan Iqbal
        Nov 30 at 19:48
















      to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
      – Hasan Iqbal
      Nov 30 at 19:48




      to tell you the truth, i suspected something like this, but was lacking the solidity present in your answer.
      – Hasan Iqbal
      Nov 30 at 19:48












      up vote
      4
      down vote













      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.






      share|improve this answer

























        up vote
        4
        down vote













        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.






        share|improve this answer























          up vote
          4
          down vote










          up vote
          4
          down vote









          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.






          share|improve this answer












          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.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 30 at 18:25









          Dripto Debroy

          73519




          73519






















              up vote
              3
              down vote













              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 $|0rangle$ or $|1rangle$. Now, Alice knows which state she prepared (let's assume it's $|0rangle$), so Alice's description of the qubit is $|0ranglelangle 0|$. 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.
              $$
              frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
              $$



              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, $|0ranglelangle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0rangle$, so if he's using a density matrix, he describes it as $|0ranglelangle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by
              $$
              frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
              $$






              share|improve this answer

























                up vote
                3
                down vote













                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 $|0rangle$ or $|1rangle$. Now, Alice knows which state she prepared (let's assume it's $|0rangle$), so Alice's description of the qubit is $|0ranglelangle 0|$. 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.
                $$
                frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                $$



                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, $|0ranglelangle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0rangle$, so if he's using a density matrix, he describes it as $|0ranglelangle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by
                $$
                frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                $$






                share|improve this answer























                  up vote
                  3
                  down vote










                  up vote
                  3
                  down vote









                  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 $|0rangle$ or $|1rangle$. Now, Alice knows which state she prepared (let's assume it's $|0rangle$), so Alice's description of the qubit is $|0ranglelangle 0|$. 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.
                  $$
                  frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                  $$



                  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, $|0ranglelangle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0rangle$, so if he's using a density matrix, he describes it as $|0ranglelangle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by
                  $$
                  frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                  $$






                  share|improve this answer












                  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 $|0rangle$ or $|1rangle$. Now, Alice knows which state she prepared (let's assume it's $|0rangle$), so Alice's description of the qubit is $|0ranglelangle 0|$. 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.
                  $$
                  frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                  $$



                  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, $|0ranglelangle 0|$. Bob gets the measurement result, so now he knows that the state is in $|0rangle$, so if he's using a density matrix, he describes it as $|0ranglelangle 0|$. However, Charlie, who does not know what measurement result Bob got, still does not know any better than to describe the state by
                  $$
                  frac{1}{2}(|0ranglelangle 0|+|1ranglelangle 1|).
                  $$







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Dec 1 at 11:58









                  DaftWullie

                  11.5k1536




                  11.5k1536






























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