Intuitively, what does the quantum Fourier transform do?

Let's see what QFT does on two qubit (and then on three qubit) computational basis states and try to gain some insights. The QFT action on $|jrangle$ basis state:

$$QFT |jrangle = frac{1}{2^{frac{n}{2}}} sum_{k=0}^{2^n -1} e^{2 pi i frac{jk}{2^n}} |krangle$$

where $n$ is the qubit number. Now suppose $n=2$, then:

begin{align*} QFT |00rangle &= QFT |0rangle = frac{1}{2} sum_{k=0}^{3} e^{2 pi i frac{0 cdot k}{4}} |krangle = frac{1}{2}big( |0rangle + |1rangle + |2rangle + |3rangle big) QFT |01rangle &= QFT |1rangle = frac{1}{2} sum_{k=0}^{3} e^{2 pi i frac{1 cdot k}{4}} |krangle = frac{1}{2}big( |0rangle + i |1rangle - |2rangle - i|3rangle big) QFT |10rangle &= QFT |2rangle = frac{1}{2} sum_{k=0}^{3} e^{2 pi i frac{2 cdot k}{4}} |krangle = frac{1}{2}big( |0rangle - |1rangle + |2rangle - |3rangle big) QFT |11rangle &= QFT |3rangle = frac{1}{2} sum_{k=0}^{3} e^{2 pi i frac{3 cdot k}{4}} |krangle = frac{1}{2}big( |0rangle - i|1rangle - |2rangle + i|3rangle big) end{align*}

From here one can see that each $|j rangle$ after QFT becomes a superposition state of all basis states with equal probabilities (in this case the probability is equal to $frac{1}{4}$). And because QFT is a unitary operator, if $langle j | j'rangle= 0$ (when $j ne j'$), then $langle j |QFT^{dagger} QFT | j'rangle= 0$, so the states generated by $QFT | jrangle$ are different superposition states with equal probabilities that are orthogonal to each other.

Now three qubit case. I will write down only for three cases:

begin{align*} QFT &|000rangle = QFT |0rangle = frac{1}{2^{frac{3}{2}}} sum_{k=0}^{7} e^{2 pi i frac{0 cdot k}{2^n}} |krangle = &=frac{1}{2^{frac{3}{2}}}big( |0rangle + |1rangle + |2rangle + |3rangle + |4rangle + |5rangle + |6rangle + |7ranglebig) QFT &|001rangle = QFT |1rangle = frac{1}{2^{frac{3}{2}}} sum_{k=0}^{7} e^{2 pi i frac{1 cdot k}{8}} |krangle = &=frac{1}{2^{frac{3}{2}}}big( |0rangle + e^{i frac{pi}{4}}|1rangle + e^{i frac{pi}{2}}|2rangle +e^{i frac{3 pi}{4}} |3rangle + e^{i pi}|4rangle +e^{i frac{5pi}{4}} |5rangle + e^{i frac{3pi}{2}}|6rangle + e^{i frac{7 pi}{4}}|7ranglebig) QFT &|111rangle = QFT |7rangle = frac{1}{2^{frac{3}{2}}} sum_{k=0}^{7} e^{2 pi i frac{7 cdot k}{8}} |krangle = &=frac{1}{2^{frac{3}{2}}}big( |0rangle + e^{i frac{7 pi}{4}}|1rangle + e^{i frac{3pi}{2}}|2rangle +e^{i frac{5 pi}{4}} |3rangle + e^{i pi}|4rangle +e^{i frac{3pi}{4}} |5rangle + e^{i frac{pi}{2}}|6rangle + e^{i frac{ pi}{4}}|7ranglebig) end{align*}

This time also $QFT |jrangle$ generates superposition states with equal probabilities (note that $| frac{e^{ivarphi}}{2^{frac{3}{2}}}|^2 = frac{1}{8}$ for any given $varphi$ ) that are orthogonal to each other. The same logic works for arbitrary number of qubits $n$. $H$ can be regarded as one qubit QFT and note that $H |j rangle$ ($j = 0,1$), in the same manner, also produces superposition states with equal probabilities that are orthogonal to each other.

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If instead of computational basis $|j rangle$ we apply QFT on an arbitrary superposition state $sum_{j = 0}^{2^n -1} a_j |jrangle$ things get slightly complicated:

$$QFT sum_j a_j |jrangle = frac{1}{2^{frac{n}{2}}} sum_{l,k=0}^{2^n -1} e^{2 pi i frac{lk}{2^n}} | k rangle langle l | sum_{j = 0}^{2^n -1} a_j |jrangle = frac{1}{2^{frac{n}{2}}} sum_{j,k=0}^{2^n -1} a_j e^{2 pi i frac{jk}{2^n}} | k rangle $$

And the probability of measuring $|k rangle$ is equal to:

$$p_k = frac{1}{2^n} left|sum_{j = 0}^{2^n - 1} a_j e^{2 pi i frac{jk}{2^n}} right|^2$$

As an example let's apply QFT on this Bell state $| Phi^+ rangle = frac{1}{sqrt{2}} big(|00rangle + |11rangle big) = frac{1}{sqrt{2}} big(|0rangle + |3rangle big)$:

$$QFT frac{1}{sqrt{2}} big(|0rangle + |3rangle big) = frac{1}{2 sqrt{2}} big(2|0rangle + (1 - i)|1rangle + (1 + i)|3rangle big)$$

The probability of measuring $|0rangle$ state is equal to $frac{1}{2}$, but the probability of measuring $|1rangle$ or $|3rangle$ states are equal $frac{1}{4}$. Also, note that the probability of measuring $|2rangle$ state is zero in this case.