# Using hadamard transform after a CNOT transform

I saw that mathematica already had other quantum computer questions here, so I think this is the best stack-exchange to ask it.

Lets say you have two qubits. The first is hadamard transformed and the second is a 0.

When you then do a $CNOT$ transform, you get:

$\begin{bmatrix}1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{bmatrix} \begin{bmatrix}\sqrt{\frac 1 2}\\ 0\\ \sqrt{\frac 1 2}\\ 0 \end{bmatrix} = \begin{bmatrix}\sqrt{\frac 1 2}\\ 0\\ 0\\ \sqrt{\frac 1 2} \end{bmatrix}$

Note that the order in the vector from top to bottom is $|00\rangle, |01\rangle, |10\rangle, |11\rangle$

You can clearly see now that both qubits are entangled, since the outcome is either $|00\rangle$ or $|11\rangle$ when reading either qubit and either has a 50% chance to be the outcome.

My question is this: if you now apply a hadamard transform to the second qubit, what would happen to the state of the two qubits? How would you perform such a transformation on entangled qubits?

• physics.stackexchange.com/search?tab=newest&q=qubits is maybe better May 30, 2016 at 5:05
• see the answer here : physics.stackexchange.com/a/250416 you have to factor out the un-changed qubit to get $= \Big(a|0⟩ + c|1⟩\Big)|0⟩ + \Big(b|0⟩ + d|1⟩\Big)|1⟩$ and apply the Hadamard transform separately to $a|0⟩ + c|1⟩$ and to $b|0⟩ + d|1⟩$ May 30, 2016 at 5:56

(Cribbing from this answer on physics stackexchange.)

1. Group the states by the uninvolved qubits.

$|\psi\rangle = a|00⟩ + b|01⟩ + c|10⟩ + d|11⟩$

$= \Big(a|0⟩ + c|1⟩\Big)|0⟩ + \Big(b|0⟩ + d|1⟩\Big)|1⟩$

2. Apply the operation within each group.

$H_0 |\psi\rangle = \Big(H(a|0⟩ + c|1⟩)\Big)|0⟩ + \Big(H(b|0⟩ + d|1⟩)\Big)|1⟩$

$= \Big(\frac{a+c}{\sqrt 2}|0⟩ + \frac{a-c}{\sqrt 2}|1⟩\Big)|0⟩ + \Big(\frac{b+d}{\sqrt 2}|0⟩ + \frac{b-d}{\sqrt 2}|1⟩\Big)|1⟩$

3. Ungroup

$=\frac{a+c}{\sqrt 2}|00⟩ + \frac{b+d}{\sqrt 2}|01⟩ + \frac{a-c}{\sqrt 2}|10⟩ + \frac{b-d}{\sqrt 2}|11⟩$

In your case, where $a=d=\frac{1}{\sqrt 2}$ and $b=c=0$, the result is

$$\frac{1}{2}|00⟩ + \frac{1}{2}|01⟩ + \frac{1}{2}|10⟩ + \frac{-1}{2}|11⟩ = \frac{1}{2} \begin{bmatrix} 1\\1\\1\\-1 \end{bmatrix}$$

Which is easily confirmed with a circuit simulator: 