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?

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

You can think about this in two ways:

Braket Notation

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 Notation

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: $$ H|Ψ⟩ = \frac{1}{\sqrt{2}} \left[ (α+β) |0⟩ + (α-β) |1⟩ \right] $$


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