# The action of tensor product over N terms on a ket.

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

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

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

Now,

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

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

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

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

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.