# Nash Equilibria of Quantum Games in the Special Two-Parameter Strategy Space
## 4 Games without Pure Nash Equilibria  
In this section, we will discuss classes of games in which there are no Nash equilibria in the strategy set (3). We begin our considerations by proving two technical lemmas concerning the scheme (2). They will be useful in proving the main statements of this section. **Lemma 1**. : _For a fixed strategy \((\theta^{*}_{b},\phi^{*}_{b})\) of player \(b\) and a fixed strategy \((\theta^{*}_{a},\phi^{*}_{a})\) of player \(a\), the following equations are satisfied:_  
\[\max_{\theta_{a},\phi_{a}}(P_{0,0}(\theta_{a},\phi_{a},\theta^{*}_{b},\phi^{*}_{ b})+P_{i,j}(\theta_{a},\phi_{a},\theta^{*}_{b},\phi^{*}_{b}))=1 \tag{27}\]  
_and_  
\[\max_{\theta_{b},\phi_{b}}(P_{0,0}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi _{b})+P_{i,j}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi_{b}))=1. \tag{28}\]  
_for \((i,j)\in\{(0,1),(1,0)\}\)._  
Proof. : As a first step, we prove (27) for \((i,j)=(1,0)\). Let us assume that \(\theta^{*}_{b}\
eq 0\) and \(\phi^{*}_{b}\
eq\phi_{a}-\pi\). Consider player \(a\)'s strategy \((\theta_{a},\phi_{a})\) that satisfies the following equations:  
\[\cot\frac{\theta_{a}}{2}=\cot\frac{\theta^{*}_{b}}{2}\tan\left(\frac{\phi_{a} }{2}-\frac{\phi^{*}_{b}}{2}\right),\quad\cos\phi_{a}=\cos(\cos\theta^{*}_{b} \cos\phi^{*}_{b}). \tag{29}\]  
Letting  
\[\phi^{*}_{a}=\arccos\left(\cos\theta^{*}_{b}\cos\phi^{*}_{b}\right) \tag{30}\]  
and  
\[\theta^{*}_{a}=2\arccos\left(\cot\frac{\theta^{*}_{b}}{2}\tan\left(\frac{ \arccos\left(\cos\theta^{*}_{b}\cos\phi^{*}_{b}\right)}{2}-\frac{\phi^{*}_{b}} {2}\right)\right) \tag{31}\]  
we conclude that  
\[P_{0,0}(\theta^{*}_{a},\phi^{*}_{a},\theta^{*}_{b},\phi^{*}_{b})+P_{1,0}( \theta^{*}_{a},\phi^{*}_{a},\theta^{*}_{b},\phi^{*}_{b})=1. \tag{32}\]  
For the case \(\theta^{*}_{b}=0\) and \(\phi_{b}\in[0,2\pi)\), Equation (32) is satisfied when player \(a\) chooses \((\theta^{*}_{a},\phi^{*}_{a})=(0,-\phi_{b})\). Finally, for \(\phi^{*}_{b}=\phi^{*}_{a}-\pi\), Equation (30) implies \(\phi^{*}_{a}=\pi/2\), \(\phi^{*}_{b}=-\pi/2\). Then, Equation (32) holds true for \(\theta^{*}_{a}=0\). To sum up, for any strategy of player \(b\), player \(a\) can choose \((\theta^{*}_{a},\phi^{*}_{a})\) so that Equation (32) holds true. In the same manner, we prove Equation (27) for \((i,j)=(0,1)\). Let us first assume that  
\[\sin\frac{\theta_{b}}{2}\sin\frac{\phi_{a}-\phi_{b}}{2}-\cos\frac{\theta_{b}} {2}\sin\frac{\phi_{a}+\phi_{b}}{2}\
eq 0. \tag{33}\]  
Consider a strategy \((\theta^{*}_{a},\phi^{*}_{a})\) such that  
\[\phi^{*}_{a}=-\arcsin(\cos\theta^{*}_{b}\sin\phi^{*}_{b}) \tag{34}\]  
and  
\[\theta^{*}_{a}=-2\arccos\left(\frac{\cos\frac{\theta^{*}_{a}}{2}\cos\frac{ \theta^{*}_{a}-\phi^{*}_{b}}{2}+\sin\frac{\theta^{*}_{b}}{2}\cos\frac{\phi^{* }_{a}+\phi^{*}_{b}}{2}}{\sin\frac{\theta^{*}_{a}}{2}\sin\frac{\phi^{*}_{a}- \phi^{*}_{b}}{2}}\right).\]  
Then  
\[P_{0,0}(\theta^{*}_{a},\phi^{*}_{a},\theta^{*}_{b},\phi^{*}_{b})+P_{0,1}( \theta^{*}_{a},\phi^{*}_{a},\theta^{*}_{b},\phi^{*}_{b})=1. \tag{35}\]  
If  
\[\max_{\theta_{b},\phi_{b}}(P_{0,0}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi _{b})+P_{i,j}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi_{b}))=1. \tag{28}\]0  
then the expression \(P_{1,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+P_{1,1}(\theta_{a},\phi_{a}, \theta_{b},\phi_{b})\) can be written as  
\[\max_{\theta_{b},\phi_{b}}(P_{0,0}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi _{b})+P_{i,j}(\theta^{*}_{a},\phi^{*}_{a},\theta_{b},\phi_{b}))=1. \tag{28}\]1Letting \(\theta_{a}=0\) yields \(P_{1,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+P_{1,1}(\theta_{a},\phi_{a}, \theta_{b},\phi_{b})=0\). As a result, we obtain \(P_{0,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+P_{0,1}(\theta_{a},\phi_{a}, \theta_{b},\phi_{b})=1\). Similar arguments apply to Equation (28). Let \((\theta_{a},\phi_{a},\theta_{b},\phi_{b})\) be a strategy profile in (2), and \(P_{i,j}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})\) is given by (6).