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?
