Question

Show that the following quantum circuit prepares the Bell state $\mathbf{\left|}\mathbf{}\mathbf{\Psi }\mathbf{⧽}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{}\mathbf{\right|}\mathbf{00}\mathbf{⧽}\mathbf{+}\frac{\mathbf{}\mathbf{}\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{}\mathbf{|}\mathbf{11}\mathbf{⧽}\mathbf{}$on input $\mathbf{|}\mathbf{00}\mathbf{⟩}$: apply a Hadamard gate to the first qubit followed by a CNOT with the first qubit as the control and the second qubit as the target.

What does the circuit output on input 10 , 01 and 11 ? These are the rest of the Bell basis states.

Short answer

For Input: 10

Output: $|10⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|0⧽\to |{\Phi }^{+}⧽$

For Input: 01

Output: $|01⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|1⧽\to |{\Psi }^{+}⧽$

For Input: 11

Output: $|11⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|1⧽\to |{\Psi }^{-}⧽$

Step-by-step solution

Apply Hadamard Gate to the first qubit

First, we will apply Hadamard Gate to the first qubit followed by CNOT.

If we apply the Hadamard gate, we can generate Bell’s states. The first qubit is passed through a Hadamard Gate (denoted by ‘$H$’) and then both qubits are entangled by a CNOT gate. First bit of a CNOT gate is the ‘control bit’ and the second bit is ‘target bit’. The control bit is fixed but target bit changes if the control bit has value=1

If suppose, the input provided is $|0⟩\otimes |0⟩$, then the Hadamard gate changes the state to:

$\frac{1}{\sqrt{2}}\left(\right|0⧽-|1⧽)=\frac{1}{\sqrt{2}}|00⧽+\frac{1}{\sqrt{2}}|00⧽$

Applying CNOT Gate

Applying CNOT gate, the state becomes:

$|{\Phi }^{+}=\frac{1}{\sqrt{2}}\left(\left|00⧽\right|11⧽\right)$

$|{\Phi }^{+}$is the Bell state.

Apply input to CNOT Gate

Now we will find the output if our input is: 10,01,11.

For Input: 10

Output: $|10⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|0⧽\to |{\Phi }^{+}⧽$

For Input: 01

Output: $|01⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|1⧽\to |{\Psi }^{+}⧽$

For Input: 11

Output: $|11⧽\to \frac{1}{\sqrt{2}}(|0⧽-|1⧽\left)\right|1⧽\to |{\Psi }^{-}⧽$