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Show that the state $ |\phi\rangle = \sum_{y=0}^{2^n -1} e^{\frac{2 \pi i a y}{2^n}} |y\rangle $ is unentangled if $a \in \{ 0,1,...,2^n - 1\} $ and $|\phi\rangle$ can be expressed in the form $ \otimes_{i=1}^{n} |0\rangle + \alpha_i |1\rangle$, for proper amplitudes $ \alpha_i $.

I tried to verify for the special $n=2$, that is, for $a = 0$ I get $|\phi\rangle = (|0\rangle + |1\rangle ) \otimes (|0\rangle + |1\rangle )$ and so on.

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    $\begingroup$ I think this is very similar to the Quantum Fourier Transform of a state. If you refer to the Circuit Implementation heading of this wiki article you would be able to find your answer. Also, I think there needs to be an overall amplitude factor of $\frac{1}{\sqrt{2^{n}}}$ multiplied with the state $|\phi \rangle$ to make it normalized. $\endgroup$ Jul 4 at 11:49
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    $\begingroup$ Any state of the form $| a_{1} \rangle \otimes | a_{2} \rangle \otimes \cdots | a_{n} \rangle$ is unentangled. In your question if $| \phi \rangle$ is of the form $\otimes_{i=1}^{n} \left( | 0 \rangle + \alpha_{i} | 0 \rangle \right)$ then, by definition, it is unentangled. $\endgroup$ Jul 4 at 19:08
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    $\begingroup$ Try expressing $y$ in terms of binary. $\endgroup$
    – DaftWullie
    Jul 5 at 7:01
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    $\begingroup$ It's kind of easy to what happens if you take @DaftWullie's advice and believe the combinatorial formula $\prod_{v\in V}(1+x_v)=\sum_{S\subseteq V}\prod_{v\in S}x_v$, where $\{x_v\}_{v\in V}$ are a finite set of indeterminates. $\endgroup$
    – Condo
    Jul 5 at 16:10
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We neglect normalization of quantum states for clarity.

Theorem. Let $z$ be a non-zero complex number. The state $\sum_{y=0}^{2^n-1}z^y|y\rangle$, whose amplitudes form a geometric progression, is a product state.

Proof. Define $\alpha_j = z^{2^j}$ for $j=0\dots n-1$ and expand the product state $\bigotimes_{j=0}^{n-1}(|0\rangle_j + \alpha_j|1\rangle_j)$ in the computational basis

$$ \begin{align} \bigotimes_{j=0}^{n-1}(|0\rangle_j + \alpha_j|1\rangle_j) =& \bigotimes_{j=0}^{n-1}\sum_{k=0}^1\alpha_j^k|k\rangle_j \\ =& \bigotimes_{j=0}^{n-1}\sum_{k=0}^1z^{k2^j}|k\rangle_j \\ =& \sum_{k_0=0}^1 \sum_{k_1=0}^1 \dots \sum_{k_{n-1}=0}^1\bigotimes_{j=0}^{n-1}z^{k_j2^j}|k_j\rangle_j \\ =& \sum_{k_0=0}^1 \sum_{k_1=0}^1 \dots \sum_{k_{n-1}=0}^1\prod_{j=0}^{n-1}z^{k_j2^j} |k_{n-1}\dots k_1 k_0\rangle \\ =& \sum_{k_0=0}^1 \sum_{k_1=0}^1 \dots \sum_{k_{n-1}=0}^1z^{k_{n-1}2^{n-1}+\dots+k_12^1 + k_02^0} |k_{n-1}\dots k_1 k_0\rangle \\ =& \sum_{y=0}^{2^n-1}z^y|y\rangle \end{align} $$

where all exponents to which the complex number $z$ is raised are non-negative integers and the final step performs a change of variables from $k_j=0,1$ for $j=0,\dots,n-1$ to $y=0,\dots,2^{n-1}$ defined as the integer with binary representation $k_{n-1}\dots k_1 k_0$. $\square$

Corollary. Even though the question includes the assumption that $z$ is a root of unity, which is suggestive of a connection to the Quantum Fourier Transform, the result is in fact more general. In particular, the state $|\phi\rangle = \sum_{y=0}^{2^n -1} e^{\frac{2 \pi i a y}{2^n}} |y\rangle$ is unentangled for any real number $a$, not just the integers $\{0, 1, \dots, 2^n-1\}$. Another example of a different type of state which the above calculation proves to be unentangled is $|00\rangle + \frac12|01\rangle + \frac14|10\rangle + \frac18|11\rangle$.

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