Equation (6) of the paper titled, Multi-player and Multi-choice quantum game has left me puzzled-after many hours-as to how it is being derived. My working begins from the generic form seen just after equation (5) - useful information begins from two paragraph above equation (3) and concludes at equation (6).

Multi-player and Multi-choice quantum game

Here is my working in an attempt to arrive at (6) from just after (5).

There are $N$ drivers and $N$ routes available to each drivers. The labelling of the drivers begins from $0$ to $N-1$ and the routes are label beginning from $0$ and ending with $N-1$

The initial state of a quantum game system is described by

$|\psi_{f}\rangle = \frac{1}{\sqrt{N}}\sum_{k=0}^{N-1}\omega_{N}^{k \cdot p} \ |k \cdot \cdot \cdot k\rangle$

$U = (u_{ij})_{N\times N} = \frac{1}{\sqrt{N}}(\omega_{N})^{ij}$


$\ |\psi_{final}\rangle = U^{\otimes N} | \psi_{initial} \rangle = (U\otimes \cdot \cdot \cdot \otimes U) \ |\psi_{initial} \rangle $

$= (U \otimes \cdot \cdot \cdot \otimes U) \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} \omega_{N}^{k \cdot p} (\ |k\rangle \otimes \cdot \cdot \cdot \otimes \ |k\rangle)$

$= \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1}\omega_{N}^{k \cdot p} (U\otimes \cdot \cdot \cdot \otimes U)(\ |k\rangle \otimes \cdot \cdot \cdot \otimes \ |k\rangle)$

$=\frac{1}{\sqrt{N}} \sum_{k=0}^{N-1}\omega_{N}^{k \cdot p}(\frac{1}{\sqrt{N}}(\omega_{N})^{ij}\otimes \cdot \cdot \cdot \otimes \frac{1}{\sqrt{N}}(\omega_{N}^{ij}))$

factoring out the reciprocal of a square root gives,

$=(\frac{1}{\sqrt{N}})^{N+1}\sum_{k=0}^{N-1}\omega_{N}^{k \cdot p}((\omega_{N})^{ij}\otimes \cdot \cdot \cdot \otimes (\omega_{N}^{ij}) (\ |k\rangle \otimes \cdot \cdot \cdot \ |k\rangle))$

I do not see how this would arrive at equation (6).

Any help is greatly appreciated.


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