Is $\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi ixj \left/\phantom\vert\!{N} \right.}|j\rangle |j\rangle$ a valid entangled quantum state?

Question

$$\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i x j \left/ {\phantom\vert\!\!} N \right.}\,\left|j\right\rangle_h\left|j\right\rangle_t$$

For a valid state we should have sum of probabilities = 1. However, when I compute the sum of the squares of the amplitudes of this state, I get zero. Am I making a mistake?

Answer 1

For $\left|\psi\right> = \frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{2\pi i\frac{x_1}{N}j}\left|j\right>_h\left|j\right>_t$ to be normalized. It must satisfy $\left<\psi|\psi\right> = 1$. Let's check, $\left<\psi|\psi\right> = \frac{1}{N}\sum_{j=0}^{N-1}e^{-2\pi i\frac{x_1}{N}j}\left<j\right|_h\left<j\right|_t\sum_{j=0}^{N-1}e^{2\pi i\frac{x_1}{N}j}\left|j\right>_h\left|j\right>_t$. Assuming the $\left|j\right>_i$ states are orthonormal, then $\left<\psi|\psi\right> = \frac{1}{N}N = 1$

Answer 2

Another approach is $\exp(ix) = \cos x + i\sin x$, so $\left|\cos x + i\sin x\right| = \sqrt{\left(\cos x + i\sin x\right)\left(\cos x - i\sin x\right)} = \sqrt{\cos^2x + \sin^2x} = 1$. so each term in $|ψ⟩$ has coefficient amplitude $1/√N$, so sum of square of all the terms $N\times(1/√N)^2 =1$.