Worked Problems in Physics

Nielsen and Chuang Exercise 2.66

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.66 Show that the average value of the observable {X_1Z_2} for a two-qubit system measured in the state {(\left|00\right>+\left|11\right>)/\sqrt{2}} is zero.

Twice the average value is given by

\displaystyle 2A=\left<00|XZ|00\right>+\left<00|XZ|11\right>+\left<11|XZ|00\right> +\left<11|XZ|11\right>

\displaystyle =\left<00|10-01+10-01\right>=0.

Nielsen and Chuang Exercise 2.65

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.65 Express the states {(\left|0\right>+\left|1\right>)/\sqrt{2}} and {(\left|0\right>-\left|1\right>)/\sqrt{2}} in a basis in which they are not the same up to a relative phase shift.

This is perhaps a trivial transformation, but, if we express the two states in the eigenbasis of the {X} operator, the states {(\left|0\right>+\left|1\right>)/\sqrt{2}} and {(\left|0\right>-\left|1\right>)/\sqrt{2}} correspond to the {\left|+\right>} and {\left|-\right>} states, which no longer differ by merely a relative phase shift.

Nielsen and Chuang Exercise 2.64

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.64 Suppose Bob is given a quantum state chosen from a set {\left|\psi_1\right>,\dots,\left|\psi_m\right>} of linearly independent states. Construct a POVM {\{E_1,E_2,\dots,E_{m+1}\}} such that if outcome {E_i} occurs, {1\leq i \leq m}, then Bob knows with certainty that he was given the state {\psi_i}. The POVM must be such that {\left<\psi_i|E_i|\psi_i\right>>0} for each {i}.

Since the states are linearly independent, we can define {E_i=\left|i\right>\left<i\right|}, for {i=1,\dots,m}, and {E_{m+1}=1-\sum_{i=1}^mE_i}, which satisfies {\sum_mE_m=1}, and insures if {E_i} is measured, that the system is in state {\left|i\right>}.

Nielsen and Chuang Exercise 2.63

Posted in Nielsen/Chuang by rpmuller on March 26, 2010

Exercise 2.63 Suppose a measurement is described by measurement operators {M_m}. Show that there exist unitary operators {U_m} such that {M_m=U_m\sqrt{E_m}}, where {E_m} is the POVM associated to the measurement.

We can confirm that the proposed operator works by considering that {M_m^\dagger M_m=\sqrt{E_m}U_m^\dagger U_m\sqrt{E_m}=E_m}. Thus, if {\sum_m E_m=1}, which is the case if the {\{E_m\}} represent a POVM, then {\sum_m M_m^\dagger M_m=1}, which demonstrates that the unitary transformation converts the measurement operators to the POVMs.