# gate actions in classical (not in the physics sense) superdense coding narrative

I am currently reading through superdense coding and the setup is as follows:

Both Alice and Bob shares a fixed entangled state $$|\psi\rangle = \frac{1}{\sqrt{2}}[|00\rangle + |11\rangle]$$ with the convention $$|Alice, Bob\rangle$$.

As an example to what is causing confusion for me:

Then, if Alice wishes to send the bit string 00 to Bob, she does nothing at all to her qubit. - this means she applies an identity gate to her qubits.

From the state $$|\psi\rangle$$, the Alice's qubit is $$[\frac{1}{\sqrt{2}}|Alice = 0, Bob = 0 \rangle + |Alice = 1, Bob = 1\rangle]$$

But Alice has a qubit 1 as seen above. What am I misunderstanding here?

The point of the protocol is to establish a "dictionary" between quantum states and bit strings. It does not actually matter what the underlying quantum states are.

There are four orthonormal states that are maximally entangled "Bell" states of two qubits: $$|B_{00}\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$$, $$|B_{10}\rangle=(|00\rangle-|11\rangle)/\sqrt{2}$$, $$|B_{01}\rangle=(|01\rangle+|10\rangle)/\sqrt{2}$$, and $$|B_{11}\rangle=(|01\rangle-|10\rangle)/\sqrt{2}$$. Bob can always perform a single ["Bell-state"] measurement to uniquely determine which of the four states Alice sent him. Assuming they have established this dictionary ahead of time ("if you see state $$|B_{ij}\rangle$$ then you know I sent the message $$ij$$"), the superdense coding protocol works.

If the two parties don't share this relationship ahead of time, you are right that there is no intrinsic way for Bob to know that some state $$|B_{00}\rangle$$ corresponds to the bit string 00. In fact, we only decided to label this state $$B_{00}$$ because we assumed there is an established agreement that it corresponds to this bit string.

I.e., you are not misunderstanding any physics. There is just a convention built in for how each state corresponds to each bit string; you can use a different convention in any practical situation that you please, but the mainstream literature will probably stick to one.

I think the confusion arises because you might be looking at a part of the super-dense coding protocol and not completely. The complete superdense coding protocol is as follows:

1. Alice and Bob share an entangled pair of qubits. Call it $$|AB\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$.
2. Say Alice wants to send the bits $$b_1b_2$$ to Bob. If $$b_1=1$$, then Alice applies a $$H$$ gate to her qubit, and if $$b_2=1$$, then Alice applies a $$X$$ gate to her qubit.
3. Alice then sends her qubit to Bob. So, Bob has both the qubits now.
4. Bob performs a Bell measurement on the pair of qubits. More precisely, Bob applies a $$CNOT$$ gate with the first qubit as the control qubit and the second qubit as the target qubit. Bob then follows this with a $$H$$ gate on the first qubit.
5. Bob measures his qubits, and the measurement outcome is the bits that Alice wanted to send to Bob.

Now, say if Alice wants to send $$00$$ to Bob. Then, as per the algorithm, Alice just applies an identity gate to her qubit. If you see the state of the two qubits together, you can easily work it out as $$|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$. Alice then sends her qubit to Bob. So, at this point, Bob has both the qubits, and the state of the two qubits is $$|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$. Next, Bob applies the $$CNOT$$ and $$H$$ gates that transforms the state $$|\psi\rangle$$ to $$|00\rangle$$.

The core idea of superdense coding is that, depending on the two bits Alice wants to send to Bob, Alice should manipulate the qubit with her in such a way that when Bob finally gets hold of both the qubits, the state of the qubit pair for each of the four possibilities $$00, 01, 10$$ and $$11$$, should be orthogonal. So, even after performing Alice's operations, Alice's qubit is still in a superposition, it is the combined state of both the qubits that contains information as to what the bits that Alice wants to send are.

I hope this clears the confusion.