# Superdense coding - What are the individual qubits A and B?

With reference to this PDF discussing Superdense coding, it is mentioned that there are two qubits A and B whose superposition gives the system $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$. It is said that Alice has A while Bob has B. What are the individual qubits A and B?

Since the system is entangled, can there exist two qubits A and B such that their superposition is $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ ? If the answer is no, how is $\frac{1}{\sqrt{2}}|00\rangle + \frac{1}{\sqrt{2}}|11\rangle$ actually shared between Alice and Bob (or maybe how is it realized in practise)?

A better question could be whether there are two qubits such that a simple composition is the state you can see there. A qubit state would look like $|\psi\rangle = a|0\rangle + b|1\rangle$. The simplest form of composition of systems is taking the tensor product, i.e. if you have a qubit for system A called $|\psi_A\rangle$ and a qubit for system B called $|\psi_B\rangle$, then the tensor product would be written $|\psi_A\rangle \otimes |\psi_B\rangle$.
You could ask whether you can write your state like that and the answer is no. Your state is written in a superposition of two such states (namely $|0_A\rangle \otimes |0_B\rangle$ and $|1_A\rangle \otimes |1_B\rangle$), but you cannot have a simple tensor product. The reason is that such states cannot have entanglement and your state has maximal entanglement. I won't go through the math, but it's not too difficult.
But how does this look like in practice? Well, the system could consist of two photons, one is on Alice's side and one is on Bob's side. The combined state of the two photons is given by your state $1/\sqrt{2} (|00\rangle +|11\rangle)$. How does the state look like on Alice's side or Bob's side? This is exactly what the partial trace is for: The state on Alice's side looks like the partial trace over Bob's system (you'll find that it is the maximally mixed state), and vice versa.