# Find the conditions under which the state $|\phi\rangle = \sum_{y=0}^{2^n -1} e^{\frac{2 \pi i a y}{2^n}} |y\rangle$ is unentangled

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.

• 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. Jul 4 at 11:49
• 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. Jul 4 at 19:08
• Try expressing $y$ in terms of binary. Jul 5 at 7:01
• 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. Jul 5 at 16:10

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 $$|00\rangle + \frac12|01\rangle + \frac14|10\rangle + \frac18|11\rangle$$.