CODEWITHSUNDEEP
quantum-computingq#qubit
Indrajit Banerjee
Indrajit Banerjee

Reputation: 428

Effect of S-gate on one qubit of a combined(maybe entangled) state of 3 qubits

Suppose I have a register(qs) of 3 qubits (first 2 being used solely for control, the last one is the input) . The first two control qubits are in the |+> state and the state of the 3rd input is unknown. Let it be a|0> + b|1>.

Now I apply CCNOT(qs[0],qs[1],qs[2]) so their combined state becomes 0.5(a,b,a,b,a,b,b,a) in Transposed matrix form [Please correct if I'm wrong here] . Now I apply S-gate to the 3rd qubit which transforms |1> -> i|1> .

I am unable to guess the state of the combined state of 'qs' now.

What I thought:

  1. One logic is to multiply every state by 'i' if it has the form|XY1> so the combined state becomes 0.5(a,ib,a,ib,a,ib,b,ia) [Transposed]
  2. Another logic is to find tensor product of (I x I x S) since I'm not changing the first 2 qubits. Performing this yields a different result which is 0.5(a,b,a,b,ia,ib,ib,ia) [Transposed] [Again, correct me if I'm wrong].

Which is the correct output after passing through S-gate (if any) ?

Upvotes: 2

Views: 140

Answers (1)

Mariia Mykhailova
Mariia Mykhailova

Reputation: 2032

The first two qubits can't start in |+> state, since |+> is a single-qubit state. I assume that the starting state of the first two qubits in the register is 0.5 (|00> + |01> + |10> + |11>).

Both approaches are correct, because they are different ways to represent the same transformation. The first answer 0.5(a,ib,a,ib,a,ib,b,ia) [Transposed] is correct. Your second answer 0.5(a,b,a,b,ia,ib,ib,ia) [Transposed] seems to be obtained by multiplying by S x I x I, i.e., applying S gate on the first qubit instead of the third one.

The tensor product I x I x S can be calculated as tensor product of I x I (which is just a 4x4 identity matrix) and S. The result is an 8x8 matrix which consists of 16 copies of S matrix, multiplied by corresponding elements of I x I:

1 0 | 0 0 | 0 0 | 0 0
0 i | 0 0 | 0 0 | 0 0
- -   - -   - -   - -
0 0 | 1 0 | 0 0 | 0 0
0 0 | 0 i | 0 0 | 0 0
- -   - -   - -   - -
0 0 | 0 0 | 1 0 | 0 0
0 0 | 0 0 | 0 i | 0 0
- -   - -   - -   - -
0 0 | 0 0 | 0 0 | 1 0
0 0 | 0 0 | 0 0 | 0 i

If you multiply the state of the qubits by this matrix, you'll get the same answer as in the first approach.

Upvotes: 1

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