# Quantum Fourier transform question about the state after application of controlled-$R_{k}$ gate?

Quantum fourier transform "circuit" is this:

Each of the $R_{k}$ is this:

After the hadamard gate on first bit: The state is:

I don't know why we can see after the controlled $R_{2}$ gate the state becomes:

This representation of the state is right after checking all cases.

1. But how to derive this rigorously from the matrix representation of controlled $R_{2}$ gate in the basis of all the $n$ computational basis state (It should be a mess but let's assume $n =3$, then what should the matrix representation of controlled $R_{2}$ gate be in 8 by 8 matrix?)

2. Or can you tell me another way how to derive this state?

3. $0.j_{1}j_{2}$ is a binary fraction. But the matrix representation of $R_{k}$ is in decimal. So in the matrix representation of controlled-$R_{k}$ in the basis of $n$ qubits, should it be written in binary representation or decimal representation?

(This question is not related from my another question of controlled $R_{k}$. Thank you for understanding.)

The controlled-$R_2$ gate obviously does not affect, or depend on, qubits $3$, $4$, ..., $n$, so let's look just at qubits $1$ and $2$.
The result of a Hadamard gate $H_1$ can be written, for $j_1 = 0,1$, $$H_1|j_1 j_2\rangle = \frac{1}{\sqrt{2}}\left( |0_1\rangle + e^{i\pi j_1}|1_1\rangle \right)|j_2\rangle = \frac{1}{\sqrt{2}}\left( |0_1\rangle + e^{2\pi i \frac{j_1}{2}}|1_1\rangle \right)|j_2\rangle$$ The action of the controlled-$R_2$ on the computational basis states reads $$CR_2|0_1 0_2\rangle = |0_1 0_2\rangle \\ CR_2|1_1 0_2\rangle = |1_1 0_2\rangle \\ CR_2|0_1 1_2\rangle = |0_1 1_2\rangle \\ CR_2|1_1 1_2\rangle = e^{\frac{2\pi i}{2^2}}|1_1 1_2\rangle$$ or, in $j_1$, $j_2$ notation, $$CR_2|j_1 j_2\rangle = e^{2\pi i\frac{j_1 j_2}{2^2}}|j_1 j_2\rangle$$ Applying the controlled-$R_2$ to the result of the Hadamard gate gives $$CR_2 \frac{1}{\sqrt{2}}\left( |0_1\rangle + e^{2\pi i \frac{j_1}{2}}|1_1\rangle \right)|j_2\rangle = \frac{1}{\sqrt{2}}\left(e^{2\pi i\frac{0 \cdot j_2}{2^2}}|0_1 j_2\rangle + e^{2\pi i\frac{1 \cdot j_2}{2^2}}e^{2\pi i \frac{j_1}{2}}|1_1 j_2\rangle \right) =\\ = \frac{1}{\sqrt{2}}\left(|0_1 j_2\rangle + e^{2\pi i\left( \frac{j_1}{2} + \frac{j_2}{2^2}\right)}|1_1 j_2\rangle \right) = \frac{1}{\sqrt{2}}\left(|0_1\rangle + e^{2\pi i 0.j_1 j_2}|1_1\rangle \right) |j_2\rangle$$ where $0.j_1 j_2 = \frac{j_1}{2} + \frac{j_2}{2^2}$ uses binary notation in analogy to the decimal notation $0.n_1 n_2 ... = \frac{n_1}{10} + \frac{n_2}{10^2} + ...$. If we bring in the other qubit states, we retrieve the final result.
See above. The $0.j_1j_2...j_k$ notation for $\frac{j_1}{2} + \frac{j_2}{2^2} + ... + \frac{j_k}{2^k}$ gives a convenient and beautiful number-theory meaning to an otherwise cumbersome sum of fractions: it is a fractional number in binary representation. The "decimal" point is in this case a "binary" point, indicating that the powers of $2$ associated with each $j_i$ in the sequence are negative, $2^{-i}$.