# How is the joint state of these qubits derived?

Can someone show to me the steps to derive the joint state at the bottom of this image, please?

I tried to follow his explanation but I didn't get the same resultsâ€¦ This is taken from the lecture notes of Ronald de Wolf in case it may help

• Welcome to QCSE! Images are not text-searchable. It would be helpful if you take the trouble to type it out in MathJax. Here's a tutorial. – Sanchayan Dutta Jul 8 '18 at 15:38
• @Blue Thanks, I don't know MathJax and that tutorial will come in handy. I'll try to do that next time for sure. – E.s Jul 8 '18 at 20:58

Since this is a homework-type question, I'll just outline the method:

You begin in the state $(\alpha_0|0\rangle + \alpha_1|1\rangle) \otimes \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$.

It can be written as $(\alpha_0|0\rangle_{A1}) \otimes \frac{1}{\sqrt{2}}(|0\rangle_{A2} |0\rangle_B + |1\rangle_{A2} |1\rangle_{B}) + (\alpha_1|1\rangle_{A1}) \otimes \frac{1}{\sqrt{2}}(|0\rangle_{A2} |0\rangle_B + |1\rangle_{A2} |1\rangle_{B})$

Apply the CNOT to qubits $\text{A1}$ and $\text{A2}$. If $\text{A1}$ is in $|0\rangle$, $\text{A2}$ remains unchanged or else it flips.

You get to the state:

$(\alpha_0|0\rangle_{A1}) \otimes \frac{1}{\sqrt{2}}(|0\rangle_{A2} |0\rangle_B + |1\rangle_{A2} |1\rangle_{B}) + (\alpha_1|1\rangle_{A1}) \otimes \frac{1}{\sqrt{2}}(|1\rangle_{A2} |0\rangle_B + |0\rangle_{A2} |1\rangle_{B})$

Then apply the Hadamard gate on $\text{A1}$. Remember that the Hadamard gate maps $|0\rangle_{A1}$ to $\frac{|0\rangle + |1\rangle}{\sqrt 2}$ and $|1\rangle_{A1}$ to $\frac{|0\rangle - |1\rangle}{\sqrt 2}$.

You finally get to the state shown in the diagram.

Note: $\text{A1}$ refers to Alice's first qubit. $\text{A2}$ refers to Alice's second qubit. $\text{B}$ refers to Bob's qubit.

• Thank you! Somehow every time I tried I forgot to flip the last bit. From an algebraic point of view can I always write a joint state of qubits by multiplying their states and develop the expression by applying the proper transforms to the corresponding qubits? – E.s Jul 8 '18 at 21:12
• @E.s It's not really a multiplication. It's a tensor product of the qubit states. Say $|00\rangle$ is shorthand notation for the tensor product of the qubit states $|0\rangle$ and $|0\rangle$. However, yes, you can always separate it like that, and do the transformations. – Sanchayan Dutta Jul 8 '18 at 21:16