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 find the superposition that results from applying the Hadamard transform H⊗n to the state 12√(|0n⟩+|1n⟩), we need to understand how the Hadamard transform acts on individual basis states.
The Hadamard transform H is defined as follows:
H = 1/√2 * [[1, 1], [1, -1]]
When we apply the Hadamard transform to a single-qubit state, it results in a superposition of the basis states, with equal amplitude:
H|0⟩ = 1/√2 * (|0⟩ + |1⟩)
H|1⟩ = 1/√2 * (|0⟩ - |1⟩)
Now, let's apply this principle to each qubit in the n-qubit state 12√(|0n⟩+|1n⟩).
H⊗n (|0n⟩) = H|0⟩⊗H|0⟩⊗...⊗H|0⟩
= (1/√2 * (|0⟩ + |1⟩)) ⊗ (1/√2 * (|0⟩ + |1⟩)) ⊗ ... ⊗ (1/√2 * (|0⟩ + |1⟩))
= 1/(2^(n/2)) * ∑(|b0⟩ ⊗ |b1⟩ ⊗ ... ⊗ |bn-1⟩), where each |bn⟩ is either 0 or 1
H⊗n (|1n⟩) = H|1⟩⊗H|1⟩⊗...⊗H|1⟩
= (1/√2 * (|0⟩ - |1⟩)) ⊗ (1/√2 * (|0⟩ - |1⟩)) ⊗ ... ⊗ (1/√2 * (|0⟩ - |1⟩))
= 1/(2^(n/2)) * ∑((-1)^c|b0⟩ ⊗ |b1⟩ ⊗ ... ⊗ |bn-1⟩), where each |bn⟩ is either 0 or 1, and c is the number of 1's in the tensor product.
Finally, we can combine the results of applying the Hadamard transform to both the |0n⟩ and |1n⟩ states:
H⊗n (1/√2 * (|0n⟩ + |1n⟩)) = 1/2 * (H⊗n (|0n⟩) + H⊗n (|1n⟩))
= 1/2 * (1/(2^(n/2)) * ∑(|b0⟩ ⊗ |b1⟩ ⊗ ... ⊗ |bn-1⟩) + 1/(2^(n/2)) * ∑((-1)^c|b0⟩ ⊗ |b1⟩ ⊗ ... ⊗ |bn-1⟩))
= 1/(2^(n/2 + 1)) * ∑((1 + (-1)^c)|b0⟩ ⊗ |b1⟩ ⊗ ... ⊗ |bn-1⟩), where each |bn⟩ is either 0 or 1, and c is the number of 1's in the tensor product.
Therefore, the superposition that results from applying the Hadamard transform H⊗n to the state 12√(|0n⟩+|1n⟩) is 1/(2^(n/2 + 1)) times the sum of all tensor products of n qubits, where each qubit is either |0⟩ or |1⟩, and the coefficient in front of each tensor product is given by (1 + (-1)^c), where c is the number of 1's in the tensor product.