What is the superposition that results if you apply the Hadamard transform H⊗n to the state 12√(|0n⟩+|1n⟩)? (Here, |0n⟩:=|00⋯0n⟩ and |1n⟩:=|11⋯1n⟩.)
To determine the superposition that results from applying the Hadamard transform H⊗n to the given state, we need to understand the effect of the Hadamard gate on individual qubits and then apply it to each qubit in the state.
The Hadamard gate H transforms a single qubit as follows:
H|0⟩ = 1/√2 (|0⟩ + |1⟩)
H|1⟩ = 1/√2 (|0⟩ - |1⟩)
In the given state 1/2√(|0n⟩ + |1n⟩), the subscript n represents the number of qubits. So, the state can be expanded as follows:
1/2√(|0⟩ ⊗ |0⟩ ⊗ ... ⊗ |0⟩ + |1⟩ ⊗ |1⟩ ⊗ ... ⊗ |1⟩)
Now, let's apply the Hadamard gate H to each qubit represented by |0⟩ and |1⟩:
H|0⟩ = 1/√2 (|0⟩ + |1⟩)
H|1⟩ = 1/√2 (|0⟩ - |1⟩)
Applying these transformations to each qubit in the given state, we get:
1/2√( H|0⟩ ⊗ H|0⟩ ⊗ ... ⊗ H|0⟩ + H|1⟩ ⊗ H|1⟩ ⊗ ... ⊗ H|1⟩)
Simplifying further, we have:
(1/2)^n * 1/√2^n (|0⟩ + |1⟩)^⊗n + (1/2)^n * 1/√2^n (|0⟩ - |1⟩)^⊗n
This can also be written as:
(1/2)^n * 1/√2^n (|000...0⟩ + |000...1⟩ + |000...1⟩ + ... + |111...1⟩)
So, the resulting superposition is (1/2)^n * 1/√2^n (|000...0⟩ + |000...1⟩ + |000...1⟩ + ... + |111...1⟩).