# Nash Equilibria of Quantum Games in the Special Two-Parameter Strategy Space
## 4 Games without Pure Nash Equilibria
The equilibrium is neither at \((0,\phi_{a}^{*},\pi,\phi_{b}^{*})\) nor at \((\pi,\phi_{a}^{*},0,\phi_{b}^{*})\) because according to (52) \(\Pi_{p}(0,\phi_{a}^{*},\pi,\phi_{b}^{*})=\Pi_{p}(\pi,\phi_{a}^{*},0,\phi_{b}^ {*})=\gamma\), whereas there exist \(\phi_{a}\) such that \(\Pi_{p}(0,\phi_{a},0,\phi_{b}^{*})\geq(\alpha+\beta)/2>\gamma\). We can therefore conclude that \((\theta_{a}^{*},\theta_{b}^{*})\
eq(0,\pi)\) and \((\theta_{a}^{*},\theta_{b}^{*})\
eq(\pi,0)\). The second line in Formula (52) can be expressed by a single cosine function because  
\[\cos(\phi_{a}+\phi_{b})(\cos\theta_{a}+\cos\theta_{b})+\sin(\phi_{a}+\phi_{b}) \sin\theta_{a}\sin\theta_{b}=C\cos(\phi_{a}+\phi_{b}+\varphi), \tag{53}\]  
where \(C^{2}=(\cos\theta_{a}+\cos\theta_{b})^{2}+\sin^{2}\theta_{a}\sin^{2}\theta_{b}\). The phase \(\varphi=\arctan\Bigl{(}-\frac{\sin\theta_{a}\sin\theta_{b}}{\cos\theta_{a}+ \cos\theta_{b}}\Bigr{)}\) or \(\varphi=\pm\frac{\pi}{2}\) if \(\cos\theta_{a}+\cos\theta_{b}=0\). Therefore, the payoff (52) can be transformed to a form in which it depends on \(\phi_{a}+\phi_{b}\) only through one cosine function  
\[\Pi_{p}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})=\frac{\alpha+\beta}{4}(1+\cos \theta_{a}\cos\theta_{b})+\frac{\gamma}{2}(1-\cos\theta_{a}\cos\theta_{b})\pm \frac{\beta-\alpha}{4}C\cos(\phi_{a}+\phi_{b}+\varphi). \tag{54}\]  
From the definition of NE, it follows that \(\Pi_{a}\bigl{(}\theta_{a}^{*},\theta_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}\bigr{)} \geq\Pi_{a}\bigl{(}\theta_{a}^{*},\phi_{a},\theta_{b}^{*},\phi_{b}^{*}\bigr{)}\) for all \(\phi_{a}\in[0,2\pi)\). Note that equality holds only at \(\phi_{a}=\phi_{a}^{*}\). This is because the cosine in (54) reaches its maximum at only one point because \(\alpha>\beta\) and \(C\
eq 0\) for \((\theta_{a}^{*},\theta_{b}^{*})\
eq(0,\pi)\) and \((\theta_{a}^{*},\theta_{b}^{*})\
eq(\pi,0)\), as we assumed. Due to the sinusoidal term, \(\Pi_{a}\bigl{(}\theta_{a}^{*},\phi_{a},\theta_{b}^{*},\phi_{b}^{*}\bigr{)}\) has its strict maximum at \(\phi_{a}=\phi_{a}^{*}\). However, at the same time, the payoff \(\Pi_{b}\bigl{(}\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}\bigr{)}\) has its minimum at \(\phi_{b}=\phi_{b}^{*}\), because the last term of (54) has the opposite sign for player \(b\). This shows that \(\Pi_{b}\bigl{(}\theta_{a}^{*},\theta_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}\bigr{)} <\Pi_{b}\bigl{(}\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}\bigr{)}\) for some \(\phi_{b}\), and consequently, \((\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\) is not a Nash equilibrium of (51). Note that the presented proof of Proposition 4 can be generalized to show the following corollaries. Corollary 3: _There are no pure Nash equilibria of the game_  
\[ \begin{pmatrix}(\alpha,\beta)&(\gamma_{1},\gamma_{2})\\ (\gamma_{1},\gamma_{2})&(\beta,\alpha)\end{pmatrix} ,\quad\alpha>\beta,\quad( \alpha+\beta)/2>\gamma_{1},\gamma_{2}. \tag{55}\]  
Using the symmetry of the payoff function with respect to the swapping of the second player strategies, one can prove also the next corollary. **Corollary 4**. : _There are no pure Nash equilibria of the game_  
\[ \begin{pmatrix}(\gamma_{1},\gamma_{2})&(\alpha,\beta)\\ (\beta,\alpha)&(\gamma_{1},\gamma_{2})\end{pmatrix} ,\quad\alpha>\beta,\quad( \alpha+\beta)/2>\gamma_{1},\gamma_{2}. \tag{56}\]