# Reduced density matrix of the ensemble of two-qubit state that is possessed by Alice and Bob

Consider the entanglement-based protocol that in each round Alice prepares a Bell state $$\Phi_+= \frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)=\frac{1}{\sqrt{2}}(|++\rangle+|--\rangle)$$ and sends one qubit to Bob. Assume Alice and Bob store their qubit in perfect quantum memory. What is the reduced density matrix of the ensemble of two-qubit state that is possessed by Alice and Bob

• You can't store (= clone) a qubit. Apr 6 '21 at 5:46
• Ignoring clarifying questions means you don't need a sensible answer. Apr 6 '21 at 6:17
• @kludg Why can't you store a qubit (in theory)? As far as I'm aware, storage usually just refers to a system that does not evolve with time. Or equivalently evolution is governed by the identity map. Apr 6 '21 at 9:14
• I don't understand what OP means by "storing" a qubit. The only meaning that comes to my mind is storing=cloning, but that is impossible. I want OP to explain what he means by storing a qubit. Apr 6 '21 at 9:20
• @kludg The way I interpret the question is that Bob for example receives on round $i$ a state $\rho_{B_i} = \mathrm{tr}_A[\Phi_+]$ and this density operator will not evolve with time. Meaning that after all the rounds of sharing have been completed, the state of system $B_i$ is still $\rho_{B_i}$. But yes, the OP should clarify if it is not clear. Note that experimentalists do have rudimentary quantum memories which are something along the lines of send your state (photon) through a really long loop of optical fibre (hoping that this approximates the identity channel well). Apr 6 '21 at 9:53

The reduced density matrix is $$\begin{bmatrix}\frac{1}{2} & 0 \\ 0 & \frac{1}{2}\end{bmatrix}$$
This easily seen by taking the partial trace over the subsystem of A: $$Tr_{A}\begin{bmatrix}\frac{1}{2}&0&0&\frac{1}{2} \\ 0&0&0&0\\0&0&0&0\\\frac{1}{2}&0&0&\frac{1}{2}\end{bmatrix}=\frac{|0\rangle\langle0|\langle0|0\rangle}{2}+\frac{|1\rangle\langle1|\langle1|1\rangle}{2}+\frac{|1\rangle\langle0|Tr_{A}(|1\rangle\langle0|)}{2}+\frac{|0\rangle\langle1|Tr_{A}(|0\rangle\langle1|)}{2} =\frac{|0\rangle\langle0|}{2}+\frac{|1\rangle\langle1|}{2}$$