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

Question

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.)

Answer 1

I am going to answer questions (2) and (3), and leave question (1) as exercise, since you'll have everything you need to construct the 8x8 matrix on your own.

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

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.

3. 0.j1j2 is a binary fraction. But the matrix representation of Rk is in decimal. So in the matrix representation of controlled-Rk in the basis of n qubits, should it be written in binary representation or decimal representation?

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, $2^{-i}$.