= Solution
Use most-significant-bit-first ordering, so $l=\sum_{j=1}^n b_j2^{n-j}$ and $|l\rangle=|b_1\rangle\otimes\cdots\otimes|b_n\rangle$. Then the coefficient factors as $e^{i\phi_ml}=\prod_je^{i\phi_m b_j2^{n-j}}$. Expanding a <tensor product> over all bit strings proves the <product decomposition of a Fourier phase state>:
$$
\boxed{|\psi_m\rangle=\bigotimes_{j=1}^n
\frac{|0\rangle+e^{i\phi_m2^{n-j}}|1\rangle}{\sqrt2}.}
$$
Each factor is a normalized one-<qubit> state. The whole register is therefore a <product state>, hence a <separable state>; no claim that arbitrary outputs of the Fourier transform are separable is needed. Reversing the bit convention reverses the order of these factors.
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