- The qubit to be sent is \(\ket{\psi} = \alpha\ket{0} + \beta\ket{1}\)
- The input to our system is the tensor product between \(\ket{\psi}\) and the EPR pair \(\ket{\beta_{00}}\):
- This is \(\ket{\psi}\ket{\beta_{00}} = \frac{1}{\sqrt{2}}[\alpha\ket{0}(\ket{00} + \ket{11}) + \beta\ket{1}(\ket{00} + \ket{11})]\)

- The first two qubits will be Alice's. That is, every qubit in the first two positions.
- Alice sends her two qubits through a CNOT gate (see quantum gates). So \(\ket{\psi}\) is acting as the control. We obtain
- \(\frac{1}{\sqrt{2}}[\alpha\ket{0}(\ket{00} + \ket{11}) + \beta\ket{1}(\ket{10} + \ket{01})]\)
- If it isn't clear how we got here, you can re-write the input state by distributing:
- \(\frac{1}{\sqrt{2}}[\alpha\ket{000} + \alpha\ket{011} + \beta\ket{100} + \beta\ket{111}]\)
- Then, see that you apply CNOT, taking the first two positions of each term as input

- She then sends her first qubit through a Hadamard gate (see quantum gates) to obtain:
- \(\frac{1}{2}[\alpha(\ket{0} + \ket{1})(\ket{00} + \ket{11}) + \beta(\ket{0} - \ket{1})(\ket{10} + \ket{01})]\)

- Let's regroup our terms:

\begin{align*}
\frac{1}{2}[& \ket{00}(\alpha\ket{0} + \beta\ket{1}) \\
+ & \ket{01}(\alpha\ket{1} + \beta\ket{0})] \\
+ & \ket{10}(\alpha\ket{0} - \beta\ket{1})] \\
+ & \ket{11}(\alpha\ket{1} - \beta\ket{0})] \\
\end{align*}

- This was done by factoring out the first two qubits
- Now, we see that Alice can measure her qubits. There are 4 possible outcomes for her measurement
- She will send this outcome to Bob
- Bob will then know what he has to do to achieve the desired state. If the measurement is \((00)\), then Bob has to do nothing. If the measurement is \((10)\), then Bob will have to pass his qubit through an \(X\) gate then a \(Z\) gate (see quantum gates) to achieve \(\alpha\ket{0} + \beta\ket{1}\)