What is the superposition that results if you apply the Hadamard transform H⊗n to the state 12√(|0n⟩+|1n⟩)?
|0>^n
wrong
what is the correct answer?
1/sqrt(2^(n-1)) SUM |X> where X in {0,1}^n and contains an even number of 1's
To understand the superposition that results from applying the Hadamard transform H⊗n to the state 1/2√(|0n⟩+|1n⟩), let's break down the process step by step:
1. Start with the given state 1/2√(|0n⟩+|1n⟩), where |0n⟩ and |1n⟩ represent n-qubit states with all qubits initialized to 0 and 1 respectively.
2. The Hadamard transform H is a single-qubit operation that can be represented by the following matrix:
H = 1/√2 * [[1, 1],
[1, -1]]
3. When you apply the Hadamard transform H⊗n to a multi-qubit state, it means applying H to each individual qubit in the state.
4. In this case, we need to apply H⊗n to the state 1/2√(|0n⟩+|1n⟩). Since the state is composed of n qubits, we apply the Hadamard transform H to each qubit, resulting in a tensor product of n Hadamard transforms.
Therefore, the result of the Hadamard transform applied to this state will be:
H⊗n(1/2√(|0n⟩+|1n⟩)) = (1/2√(|0⟩+|1⟩))^⊗n
5. Expanding the tensor product, we get:
H⊗n(1/2√(|0n⟩+|1n⟩)) = 1/2√(|0⟩⊗|0⟩⊗...⊗|0⟩ + |1⟩⊗|0⟩⊗...⊗|0⟩ + |0⟩⊗|1⟩⊗...⊗|0⟩ + ... + |1⟩⊗|1⟩⊗...⊗|1⟩)
6. This can be simplified to:
H⊗n(1/2√(|0n⟩+|1n⟩)) = 1/2√(|0...0⟩ + |0...1⟩ + |1...0⟩ + ... + |1...1⟩)
7. Here, the state |0...0⟩ represents the all-zero state, and |1...1⟩ represents the all-one state. All other terms represent superpositions of these basis states.
So, the result of applying the Hadamard transform H⊗n to the state 1/2√(|0n⟩+|1n⟩) is a superposition of all possible n-bit strings, with equal amplitudes for each basis state.