# 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?

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

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?

• You have to take the sum of squares of the moduli of the coefficients. In your example every coefficient has modulus $1/\sqrt{N}$. Squaring and summing gives 1. Sep 3 '19 at 11:02

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, $$\left<\psi|\psi\right> = \frac{1}{N}\sum_{j=0}^{N-1}e^{-2\pi i\frac{x_1}{N}j}\left_h\left|j\right>_t$$. Assuming the $$\left|j\right>_i$$ states are orthonormal, then $$\left<\psi|\psi\right> = \frac{1}{N}N = 1$$
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 $$N\times(1/√N)^2 =1$$.