Hadamard Gate

Determine $(H^{\otimes 4})$ to $(|15⟩)$

To determine the result of applying ( $H^{\otimes 4}$ ) to $(|15⟩)$, let’s go through the steps methodically.

1. Represent ( \ket{15} ) in binary form: $15_{10}=1111_2$ So, $(|15⟩=|1111⟩)$.

2. Apply the Hadamard gate ( H ): The Hadamard gate ( H ) acts on a single qubit as follows: $H|0⟩=\frac{1}{\sqrt{2}}(|0⟩+|1⟩)$ $H|1⟩=\frac{1}{\sqrt{2}}(|0⟩-|1⟩)$

3. Apply ( $H^{\otimes 4}$ ) to ( $|1111⟩$): Applying ( H ) to each qubit in ( $|1111⟩$):

$H|1⟩=\frac{1}{\sqrt{2}}(|0⟩-|1⟩)$

So, for four qubits: $H^{\otimes 4}|1111⟩=H|1⟩\otimes H|1⟩\otimes H|1⟩\otimes H|1⟩$

$=\left(\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\right)\otimes \left(\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\right)\otimes \left(\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\right)\otimes \left(\frac{1}{\sqrt{2}}(|0⟩-|1⟩)\right)$ 4. Simplify the expression: $H^{\otimes 4}|1111⟩=\frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right)\otimes \frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right)\otimes \frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right)\otimes \frac{1}{\sqrt{2}}\left(|0⟩-|1⟩\right)$

$=\frac{1}{2^2}\left(|0⟩-|1⟩\right)\otimes \left(|0⟩-|1⟩\right)\otimes \left(|0⟩-|1⟩\otimes \left(|0⟩-|1⟩\right)\right)$ $=\frac{1}{4}{\sum }_{x_1,x_2,x_3,x_4\in \{0,1\}}(-1{)}^{x_1+x_2+x_3+x_4}|x_1{x}_2{x}_3{x}_4⟩$

Where each $x_i$ can be either 0 or 1.

Thus, the result of applying ( $H^{\otimes 4}$ ) to ( $|15⟩$ ) is: $H^{\otimes 4}|1111⟩=\frac{1}{4}{\sum }_{x\in \{0,1{\}}^4}(-1{)}^{x_1+x_2+x_3+x_4}|x⟩$

This result is a superposition of all possible 4-qubit states with phase factors determined by the parity of the number of $1s$ in the binary representation.