# Understanding Bell Inequality of Quantum Computing

Quantum Mechanics/Physics is rising and intersting area, proper?. When studying it you come throughout subject referred to as Bell Inequality. That is little information for it.

It’s a check which depends upon Quantum Mechanics to explain actual world experiment.

So let’s know what’s it and the way can we calculate it.

### Bell Inequality

John Stewart Bell is a Irish physicist. He devised an incredible check which distinguish between Bohr and Einstein fashions. Most superb function of this check shouldn’t be solely it’s philosophies but in addition testable. So, let’s check out it.

Earlier than we begin studying Bell Inequality, we now have to be taught it is primary concepts on which it relies upon. I hope you’ve gotten primary data about matrices and it is operations.

### Fundamental

Column vector characterize as “ket”

$ket{psi}$

Row vector characterize as “bra”

$bra{psi}$

You’ll be able to take a look at it like this

Often known as H gate. It’s single qubit operation that map qubit into equal superposition.

Merely it convert foundation state into equally superposition states like this


ket{0} = {ket{0} + ket{1} above{2pt}
sqrt{2}}


ket{1} = {ket{0} – ket{1} above{2pt} sqrt{2}}

Matrix for that is

Merely each component is 1 with scalar a number of of

${1 above{1pt} sqrt{2}}$

instance let’s calculate hadamard matrix for

$ket{0}$

At high we write matrix for

$ket{0}$

, take take a look at it as a result of it incorporates 1 at high and 0 and backside describing it’s at

$ket{0}$

state. Aspect of it’s Hadamard matrix (H). Under is solely multiplication of 2×2 matrix and 2×1 matrix. After calculating we get 2×1 matrix containing 1 at every place indicating it has each

$ket{0} ket{1}$

states
and on the finish we will write it in easy kind


ket{0} = {ket{0} + ket{1} above{2pt} sqrt{2}}

### CNOT Gate

It’s two qubit operation. So it take two qubit as enter, label it as first is “management” qubit and second is “goal” qubit. Foremost level to recollect is,

• If management qubit is 1 then it carry out Pauli-X gate on course qubit means convert 0 to 1 or vice-versa.
• If management qubit is 0, then goal qubit stay unchanged.

You’ll be able to contemplate this as classical NOT gate.
Instance


ket{00} = ket{00}


ket{01} = ket{01}


ket{10} = ket{11}


ket{11} = ket{10}

#### Image

That is it we have to perceive Bell Inequality, though you learn EPR and Bell’s story as further assets to your data.

### Bell Inequality

Earlier than getting output from Bell circuit (just like classical logic circuit) it first endure Hadamard transformation after which CNOT gate act on it and provides us output.

#### Course of

Now we now have hadamard gates and CNOT gate with us the circuit of Bell state is

So first hadamard gate act on it like


ket{0} = {ket{0} + ket{1} above{2pt} sqrt{2}}

And after changing

$ket{0}$

to it is equally superposition states we apply CNOT gate. As CNOT gate require two qubits as enter.


ket{00} = {(ket{0} + ket{1})ket{0} above{2pt} sqrt{2}}


ket{00} = {ket{00} + ket{10} above{2pt} sqrt{2}}

We all know CNOT gate will change goal qubit if management qubit is 1 else it stays unchanged.


ket{00} = {ket{00} + ket{11} above{2pt} sqrt{2}}

Equally we get one other 3 states.

for

$ket{01}$


ket{01} = {ket{01} + ket{10} above{2pt} sqrt{2}}

for

$ket{10}$


ket{10} = {ket{00} – ket{11} above{2pt} sqrt{2}}

for

$ket{11}$


ket{11} = {ket{01} – ket{10} above{2pt} sqrt{2}}

And above are Bell’s states/pairs