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I’m reading up on Simon’s problem from the Qiskit textbook, but don’t understand the second Hadamard transform of the first qubit register.

Mathematically, they are stating that a state that looks like this

$$ \frac{1}{\sqrt{2}} ( | x \rangle + | x \oplus b \rangle ) $$

Becomes this after a Hadamard transform

$$ \frac{1}{\sqrt{2^{n+1}}} \sum_{z \in \{0, 1\}^n} \left [ (-1) ^ { x \cdot z } + (-1) ^ {y \cdot z} \right ] |z\rangle $$

But I don’t see how at all that step was made. Can someone explain how the $x$ and $y$ jumped into the power, and somehow got the bitwise product operation performed on them?

These are the steps I am referencing

steps 4 and 5

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2 Answers 2

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Can someone explain how the x and y jumped into the power, and somehow got the bitwise product operation performed on them?

It is helpful to write out a few examples to see what is happening.

1-bit Example: $$ H|0\rangle = \frac{1}{\sqrt{2}}|0\rangle + |1\rangle = \frac{1}{\sqrt{2}}\sum_z |z\rangle (-1)^{0\cdot z} $$ $$ H|1\rangle = \frac{1}{\sqrt{2}}|0\rangle - |1\rangle = \frac{1}{\sqrt{2}}\sum_z |z\rangle (-1)^{1\cdot z} $$


2-bit Example: $$ H|00\rangle = \frac{1}{\sqrt{2^2}}(|0\rangle + |1\rangle)(|0\rangle + |1\rangle) = \frac{1}{\sqrt{2^2}}(|00\rangle + |01\rangle + |10\rangle + |11\rangle) $$ $$ = \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[00]\cdot z} $$

$$ H|01\rangle = \frac{1}{\sqrt{2^2}}(|0\rangle + |1\rangle)(|0\rangle - |1\rangle) = \frac{1}{\sqrt{2^2}}(|00\rangle - |01\rangle + |10\rangle - |11\rangle) $$ $$ = \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[01]\cdot z} $$

$$ H|10\rangle = \frac{1}{\sqrt{2^2}}(|0\rangle - |1\rangle)(|0\rangle + |1\rangle) = \frac{1}{\sqrt{2^2}}(|00\rangle + |01\rangle - |10\rangle - |11\rangle) $$ $$ = \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[10]\cdot z} $$

$$ H|11\rangle = \frac{1}{\sqrt{2^2}}(|0\rangle - |1\rangle)(|0\rangle - |1\rangle) = \frac{1}{\sqrt{2^2}}(|00\rangle - |01\rangle - |10\rangle + |11\rangle) $$ $$ = \frac{1}{\sqrt{2^2}}\sum_z |z\rangle (-1)^{[11]\cdot z} $$


You see that, whatever the starting ket in our six above examples, and whether the number of qubits $n$ is 1 or 2, the expression on the far RHS is always the same, namely: $$ H|x\rangle = \frac{1}{\sqrt{2^n}}\sum_z |z\rangle(-1)^{x\cdot z}\;. $$

You can convince yourself that this is quite reasonable since wherever we have a single $|0\rangle$ ket in the direct product of the starting ket we get a $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and wherever we have a single $|1\rangle$ ket in the direct product of the starting ket we get a $\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$. So, we can only pick up (and must pick up) minus signs in the expansion of the direct product when we expand through a $|1\rangle$.


If you want to prove it, try induction.

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You can simply use the definition of the Hadamard gate: $$H|x\rangle=\frac{1}{\sqrt{2^n}}\sum_z(-1)^{x\cdot z}|z\rangle$$ Thus, we have: $$\begin{align} H\left(\frac{1}{\sqrt{2}}\left(|x\rangle+|y\rangle\right)\right)&=\frac{1}{\sqrt{2^{n+1}}}\left[\sum_{z}(-1)^{x\cdot z}|z\rangle+\sum_{z}(-1)^{y\cdot z}|z\rangle\right]\\ &=\frac{1}{\sqrt{2^{n+1}}}\sum_{z}\left[(-1)^{x\cdot z}+(-1)^{y\cdot z}\right]|z\rangle \end{align}$$

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