- #1
Yan Campo
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- TL;DR Summary
- I need to show that the probability when measuring a local $$O^{a}$$ observable is $$Pr(o_{j}^{(a)})=Tr(\rho_{ab}(\ket{o_{j}^{(a)}}\bra{o_{j}^{(a)}}\otimes \mathbb{I}_{2}))$$
I have information that $$\rho_{ab}=\sum_{j}p_{j}\ket{\Psi_{j}^{ab}}\bra{\Psi_{j}^{ab}}$$ and $$Pr(o_{j}^{(a)}|\Psi_{ab})=Tr_{ab}(\ket{\Psi_{ab}}\bra{\Psi_{ab}}(\ket{o_{j}^{(a)}}\bra{o_{j}^{(a)}}\otimes \mathbb{I}_{2})) \text{.}$$
I started by representing the density operator for pure states, such that $$\rho = \ket{\Psi^{ab}}\bra{\Psi^{ab}}\text{.}$$
Substituting directly into the equation that was given for the probability I arrive at a result $$Pr(o_{j}^{(a)}|\Psi_{ab})=Tr_{ab}(\ket{\Psi_{j}^{ab}}\bra{\Psi_{j}^{ab}}(\ket{o_{j}^{(a)}}\bra{o_{j}^{(a)}}\otimes \mathbb{I}_{2}))\text{.}$$
I believe this is not right, as I have not found a way to make this equal to what was asked.
Any clue what should I do? Any help is welcome.
I started by representing the density operator for pure states, such that $$\rho = \ket{\Psi^{ab}}\bra{\Psi^{ab}}\text{.}$$
Substituting directly into the equation that was given for the probability I arrive at a result $$Pr(o_{j}^{(a)}|\Psi_{ab})=Tr_{ab}(\ket{\Psi_{j}^{ab}}\bra{\Psi_{j}^{ab}}(\ket{o_{j}^{(a)}}\bra{o_{j}^{(a)}}\otimes \mathbb{I}_{2}))\text{.}$$
I believe this is not right, as I have not found a way to make this equal to what was asked.
Any clue what should I do? Any help is welcome.
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