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source: German language Wikipedia

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There is a game in this video that the Qcomputer always wins:

                                                       

Asked Gemini to explain to me why this is so...

                                                                            

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EXPLANATION

You

quantum computers; what does the Hadamard operator do

Copilot

The Hadamard gate (H) is a fundamental single-qubit gate in quantum computing. Its significance lies in its ability to perform two key operations:

Superposition Generation: The Hadamard gate transforms the basis states of a qubit as follows:

|0⟩ ⟼ (|0⟩ + |1⟩) / √2

|1⟩ ⟼ (|0⟩ - |1⟩) / √2

This means that when a qubit is measured after undergoing a Hadamard gate, it has an equal probability of collapsing to either |0⟩ or |1⟩. Essentially, the Hadamard gate creates a superposition of states, which is a crucial feature for quantum algorithms and protocols12.

Basis Transformations: The Hadamard gate also performs a basis transformation. It maps the computational basis states (|0⟩ and |1⟩) to a new basis set. This transformation is useful for various quantum operations, such as creating entanglement, implementing quantum search algorithms, and preparing states for quantum Fourier transforms31.

In summary, the Hadamard gate plays a pivotal role in quantum information processing by enabling superposition and facilitating basis transformations, making it an essential tool for quantum algorithms and circuits.

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GEMINI

Absolutely! Here's how we can work through applying a Hadamard gate to a 1,0 vector column:

1. Hadamard Gate Matrix:

The Hadamard gate is a fundamental quantum gate that operates on a single qubit. It can be represented by a 2x2 matrix:

H = 1/√2 * [[1, 1], [1, -1]]

2. State Vector:

Your input state is a 1,0 vector column, which represents a qubit in the basis state |0>. In quantum mechanics, state vectors are typically written as column vectors. So, here the state vector is:

|psi> = [1, 0]

3. Applying the Hadamard Gate:

To find the resulting state after applying the Hadamard gate, we perform matrix multiplication of the Hadamard gate matrix (H) with the state vector (|psi>):

Result = H * |psi> = 1/√2 * [[1, 1], [1, -1]] * [1, 0]

4. Matrix Multiplication:

Performing the multiplication, we get:

Result = [1/√2 * (1 + 0), 1/√2 * (1 - 0)]
        = [1/√2, 1/√2]

5. Interpretation:

The resulting vector, [1/√2, 1/√2], represents the state of the qubit after applying the Hadamard gate. This state is a superposition of the basis states |0> and |1>. In simpler terms, the qubit is now in a state where it's both |0> and |1> with equal probability (represented by the factor 1/√2).

Key takeaway: The Hadamard gate takes a qubit in the |0> state and puts it in a superposition of |0> and |1>.

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That's a great question! If you apply the Hadamard gate a second time (H * H * |0>), the surprising thing is that the qubit actually returns to its original state, |0>.

Here's why:

1. Hadamard Gate Property: The Hadamard gate is its own inverse. This means that applying it twice (H * H) is equivalent to applying the identity matrix (I), which leaves the state unchanged.

2. Matrix Multiplication:

• We know from the previous step that the first Hadamard gate transforms |0> to [1/√2, 1/√2].
• Applying the Hadamard gate again (H * [1/√2, 1/√2]) is essentially multiplying by the Hadamard matrix (H) once more.

3. Resulting State:
   • Performing the multiplication (H * [1/√2, 1/√2]), we get:

     [1/√2 * (1/√2 + 1/√2), 1/√2 * (1/√2 - 1/√2)]
     = [1, 0]
     

This final state [1, 0] represents the original |0> state the qubit was in.

Intuition:

Imagine the Hadamard gate as a rotation in the quantum world (Bloch sphere). Applying it once rotates the qubit to a superposition state. Applying it again performs another rotation, effectively bringing the qubit back to its initial state.