Preparing an entangled two qubit states

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It is known that any two qubit state (a1+a2 i) |00>+(b1+b2 i) |01>+(c1+c2 i) |10>+(d1+d2 i) |00> can be prepared using local gates and a most one controlled-Z gate. Here we provide a way to take any two qubit state into the state |00>. You do not need to normalize the two qubit.

Enter the amplidudes

(a1+a2 i) |00>+(b1+b2 i) |01>+(c1+c2 i) |10>+(d1+d2 i) |00>

Your two qubit state is: (1+(0) i) |00>+(1+(0) i) |01>+(1+(0) i) |10>+(1+(0) i) |11>

To normilize your qubit you need to divide it by 2

The normilize two qubit is:

(0.5) |00>+(0.5)|01>+(0.5)|10>+(0.5)|11>

If this number ad-bc = 0 es the number zero then this two qubit is not-entangled

If your two qubit is not-entangled, you do not need to use a controlled-Z gate to prepared the state, you only need two local gates. You can use the next link to prepared your non-entangled state

• Preparing a non entangled two qubit state

If your two qubit is entangled, then following instructions will give you a way to create a circuit that prepares your state.

In order to take your qubit state to the qubit state |00> we need to apply the local gate W to the fisrt qubit where

The entry 11 of the matrix W is 0.7071067811865475

The entry 12 of the matrix W is -0.7071067811865475

The entry 21 of the matrix W is 0.7071067811865475

The entry 22 of the matrix W is 0.7071067811865475

and then we apply the controlled-Z gate. Applying these two gates reduces your two qubit state to the state

(0) |00>+(0.7071067811865475)|01>+(0)|10>+(-0.7071067811865475)|11>

Now we apply the gate K1 to the second qubit where

The entry 11 of the matrix K1 is 0.7071067811865476

The entry 12 of the matrix K1 is -0.7071067811865476

The entry 21 of the matrix K1 is 0.7071067811865476

The entry 22 of the matrix K1 is 0.7071067811865476

After applying the local gate K1 to the second qubit, the state reduces to

(0) |00>+(1)|01>+(0)|10>+(0)|11>

Now we apply the gate K2 to the first qubit where

The entry 11 of the matrix K2 is 0

The entry 12 of the matrix K2 is 1

The entry 21 of the matrix K2 is -1

The entry 22 of the matrix K2 is 0

After applying the local gate K2 to the first qubit, the state reduces to

(1) |00>+(0)|01>+(0)|10>+(0)|11>

Which is the state |00>