# Understanding shared Bell states from quantum entanglement

I'm trying to understand an entanglement swapping derivation provided in this PDF (pages 2 - 3)

I have several things about this process that I don't understand, and I was hoping someone could clarify.

Are the CNOT and Hadamard gates applied with respect to the $$A_c$$ bit? If so, am I correct in this math (starting from the same joint state given in the PDF)?

$$|\psi\rangle = \frac{1}{2}(|0_C0_{Ac}\rangle + |1_C1_{Ac}\rangle) \otimes (|0_A0_{B}\rangle + |1_A1_{B}\rangle)$$

$$|\psi\rangle = \frac{1}{2}(|0_C0_{Ac}0_A0_B\rangle + |0_C0_{Ac}1_A1_B\rangle + |1_C1_{Ac}0_A0_B\rangle + |1_C1_{Ac}1_A1_B\rangle)$$

Applying CNOT using $$Ac$$ as control bit and $$A$$ as target:

$$\mathrm{CNOT}|\psi\rangle = \frac{1}{2}(|0_C0_{Ac}0_A0_B\rangle + |0_C0_{Ac}1_A1_B\rangle + |1_C1_{Ac}1_A0_B\rangle + |1_C1_{Ac}0_A1_B\rangle)$$

Applying Hadamard onto $$Ac$$ bit:

$$\mathrm{H}.\mathrm{CNOT}|\psi\rangle = \frac{1}{2}(|0_C\rangle\otimes\frac{1}{\sqrt{2}}(|0_{Ac}\rangle + |1_{Ac}\rangle)\otimes|0_A0_B\rangle + |0_C\rangle\otimes\frac{1}{\sqrt{2}}(|0_{Ac}\rangle + |1_{Ac}\rangle)\otimes|1_A1_B\rangle + |1_C\rangle\otimes\frac{1}{\sqrt{2}}(|0_{Ac}\rangle - |1_{Ac}\rangle)\otimes|1_A0_B\rangle + |1_C\rangle\otimes\frac{1}{\sqrt{2}}(|0_{Ac}\rangle - |1_{Ac}\rangle)\otimes|0_A1_B\rangle)$$

Grouping and factoring, I end up with the following, which is different from the PDF derivation (and I'm not sure why):

$$\mathrm{H}.\mathrm{CNOT}|\psi\rangle = \frac{1}{2\sqrt{2}}|0_{Ac}0_A\rangle \otimes (|0_{C}0_B\rangle + |1_{C}1_B\rangle)$$

$$+ \frac{1}{2\sqrt{2}}|0_{Ac}1_A\rangle \otimes (|0_{C}1_B\rangle + |1_{C}0_B\rangle)$$

$$+ \frac{1}{2\sqrt{2}}|1_{Ac}0_A\rangle \otimes (|0_{C}0_B\rangle - |1_{C}1_B\rangle)$$

$$+ \frac{1}{2\sqrt{2}}|1_{Ac}1_A\rangle \otimes (|0_{C}1_B\rangle - |1_{C}0_B\rangle)$$

Given this final answer, as I understand it, if Alice has bits $$00$$, then Carol and Bob have the entangled bell state $$00$$; if Alice has $$01$$, then Carol and Bob have entangled bell state $$10$$; if Alice has $$10$$, then Carol and Bob have entangled bell state $$01$$, and lastly, if Alice has bits $$11$$, then Carol and Bob have entangled bell state $$11$$.

Is the general idea here that we're teleporting Carol's bit across Alice's entangled bits to then end up entangling Carol and Bob?