Question from exam:
Bob built a quantum computer wiht 10 qubits. All qubits are set to zeroes. Bob performed a quantum Fourier transform on the system and then measured the system.
what is the probability that the result will be 0 (all qubits are zeroes)?
By looking to the circuit for the QFT presented in the M. Nielsen and I. Chuang textbook (Figure 5.1.) we can notice that all controlled rotations can be neglected because for each control rotation gate the control qubits are in $| 0\rangle$ state (for the case described in the question). Here is the Figure from the book:
So effectively, in this case, we have only Hadamard gates acting on $n$ qubits. And, of course, Swap gates (will talk about it later). So
$$H^{\otimes n} |000...0\rangle = \frac{1}{\sqrt{2^n}} \sum_x{|x\rangle}$$
where $H^{\otimes n}$ is the notation of the $n$ Hadamard gates acting on $n$ qubits that are all in the $|0\rangle$ state, $\sum_x{|x\rangle}$ is the sum of all bit strings (from $|000...0\rangle$ to $|111...1\rangle$). Swap gates at the end of the QFT circuit (not presented in the Figure 5.1.) will not change this state: the final state after the swaps will stay the same. So all bit strings have equal amplitudes $\frac{1}{\sqrt{2^n}}$, and therefore equal probabilities $\left|\frac{1}{\sqrt{2^n}}\right|^2 = \frac{1}{2^n}$. Thus, the answer is: for 10 qubits we will measure all qubits in the $|0\rangle$ state with $\frac{1}{2^{10}} \approx 0.001$ probability.
When input state to QFT is $|00\dots0\rangle$ then no controlled gates in QFT work as each control qubit is set to $|0\rangle$. However, there are also Hadamard gates on each qubit. As a result, QFT behaves as $H^{\otimes n}$ and produces superposition consisting of all basis state with same probability $\frac{1}{2^n}$, where $n$ is number of input qubits.
For your case $n=10$, hence probability that state $|00\dots0\rangle$ is preserved after QFT application is $\frac{1}{1024}.$
The unitary matrix of the Fourier transform is $$ U_{QFT}=\frac{1}{\sqrt{2^n}}\sum_{x,y=0}^{2^n-1}e^{2\pi i\frac{xy}{2^n}}|y\rangle\langle x|. $$ So, if I apply this to the input state $|0\rangle$, I get the answer $$ U_{QFT}|0\rangle=\frac{1}{\sqrt{2^n}}\sum_{y=0}^{2^n-1}|y\rangle. $$ The amplitude for finding the system is state $y=0$ is $1/\sqrt{2^n}$. Hence the probability of this outcome is $1/2^n$ with $n=10$, so the probability is $1/1024$.
