Exercise 4.32

Consider a two qubit density matrix \(\rho\)​, if we perform projective measurement to the second qubit with operator

(1)\[ P_0 = I\otimes |0\rangle\langle 0|, P_1 = I\otimes |1\rangle\langle 1| \]

The post-measurement state after measuring \(P_0\) is \(P_0\rho P_0/p_0\) where \(p_0\) is the probability of measuring \(0\), and the post-measurement state after measuring \(P_1\) is \(P_1\rho P_1/p_1\) where \(p_1\) is the probability of measuring \(1\). If an observer did not know the measurement result, then we will have probability \(p_0\) to get \(P_0\rho P_0/p_0\), and have probability \(p_1\) to get \(P_1\rho P_1/p_1\). So

(2)\[ \rho' =p_0\cdot \frac{P_0\rho P_0}{p_0} + p_1\cdot \frac{P_1\rho P_1}{p_1} = P_0\rho P_0+P_1\rho P_1 \]

We can calculate the partial trace for the first qubit as

(3)\[ {\rm Tr}_{2}(\rho) = (I\otimes \langle 0|) \rho (I\otimes | 0\rangle ) + (I\otimes \langle 1|) \rho (I\otimes | 1\rangle ) \]

Similarly, the partial trace of \(\rho'\)​ for the first qubit is

(4)\[ \begin{align}\begin{aligned}\begin{split} \begin{align} {\rm Tr}_{2}(\rho') =& (I\otimes \langle 0|) \rho' (I\otimes | 0\rangle ) + (I\otimes \langle 1|) \rho' (I\otimes | 1\rangle ) \\ =& (I\otimes \langle 0|) P_0\rho P_0 (I\otimes | 0\rangle ) +(I\otimes \langle 0|) P_1\rho P_1 (I\otimes | 0\rangle ) \\ &+ (I\otimes \langle 1|) P_0\rho P_0 (I\otimes | 1\rangle ) +(I\otimes \langle 1|) P_1\rho P_1 (I\otimes | 1\rangle )\\ =& (I\otimes \langle 0|) (I\otimes |0\rangle\langle 0|)\rho (I\otimes |0\rangle\langle 0|) (I\otimes | 0\rangle ) \\ &+(I\otimes \langle 0|) (I\otimes |1\rangle\langle 1|)\rho (I\otimes |1\rangle\langle 1|)(I\otimes | 0\rangle ) \\ &+ (I\otimes \langle 1|) (I\otimes |0\rangle\langle 0|)\rho (I\otimes |0\rangle\langle 0|) (I\otimes | 1\rangle ) \\ &+(I\otimes \langle 1|) (I\otimes |1\rangle\langle 1|)\rho (I\otimes |1\rangle\langle 1|)(I\otimes | 1\rangle )\\\end{split}\\\begin{split}=& (I\otimes \langle 0|0\rangle\langle 0|) \rho (I\otimes |0\rangle\langle 0|0\rangle ) \\ &+(I\otimes \langle 0|1\rangle\langle 1|) \rho (I\otimes |1\rangle\langle 1|0\rangle ) \\ &+ (I\otimes \langle 1|0\rangle\langle 0|) \rho (I\otimes |0\rangle\langle 0|1\rangle ) \\ &+(I\otimes \langle 1|1\rangle\langle 1|) \rho (I\otimes |1\rangle\langle 1|1\rangle ) \\ =& (I\otimes \langle 0|) \rho (I\otimes |0\rangle)+(I\otimes \langle 1|) \rho (I\otimes |1\rangle) \end{align} \end{split}\end{aligned}\end{align} \]

which is the same as eq. (3). Thus, we have \({\rm Tr}_{2}(\rho)={\rm Tr}_{2}(\rho')\).