# How to apply the Hadamard gate to a given qubit state?

I have this qubit state:

$$H \left[ \frac{1}{\sqrt{2}} |0\rangle + \left( \sqrt{\frac{2}{7}}+\frac{1}{\sqrt{7}}i \right) |1\rangle \right]$$

How to solve this given Hadamard gate on qubit?

Hadamard matrix should be multiplied with vector 2x1 (1 and 1) but what to do with numbers in front of it?

• Welcome to the Quantum Computing Stack Exchange! Please update the title of your question to accurately reflect the problem you are asking about. ;) Nov 12, 2021 at 19:16
• possible duplicate: quantumcomputing.stackexchange.com/a/3895/55
– glS
Nov 13, 2021 at 1:38

Quantum operators are linear, meaning any operator $$\hatΩ$$ obeys the following relation: $$\hatΩ \left[ α|ψ⟩ + β|φ⟩ \right] = α \left(\hatΩ|ψ⟩\right) + β \left(\hatΩ|φ⟩\right)$$ In other words, you can just pull the constants out in front, calculate out the results of $$H|0⟩$$ and $$H|1⟩$$ as usual, then re-distribute your constants.
Generally $$α$$ and $$β$$ can be any complex number at all, but in your case $$α=\frac{1}{\sqrt{2}}$$ and $$β=\left(\sqrt\frac{2}{7} + \frac{i}{\sqrt{7}}\right)$$.
Matrix operations are designed to make these linear operations easier to think about. The vector you are trying to apply $$H$$ to can be written as: $$|Ψ⟩ = \left[ \begin{array}{c} α \\ β \end{array} \right]$$ Meanwhile, $$H$$ can be written as: $$H = \frac{1}{\sqrt{2}} \left[ \begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right]$$ The final result is: $$H|Ψ⟩ = \frac{1}{\sqrt{2}} \left[\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array} \right] \left[ \begin{array}{c} α \\ β \end{array} \right] = \frac{1}{\sqrt{2}} \left[ \begin{array}{c} α+β \\ α-β \end{array} \right]$$ So, the final answer is: $$H|Ψ⟩ = \frac{1}{\sqrt{2}} \left[ (α+β) |0⟩ + (α-β) |1⟩ \right]$$