# Nash Equilibria of Quantum Games in the Special Two-Parameter Strategy Space
## 4 Games without Pure Nash Equilibria
If \(P_{i,j}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})=1\) for some \(i,j\in\{0,1\}\), then \(\theta_{a}^{*}=\theta_{b}^{*}\in\{0,\pi\}\). Proof. : There is no loss of generality in assuming \(P_{0,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})=1\). Then, \(P_{1,1}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})=0\), and from (6) it follows that  
\[ \begin{split}&\left(\cos\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2} \cos\left(\frac{\phi_{a}+\phi_{b}}{2}\right)+\sin\frac{\theta_{a}}{2}\sin\frac {\theta_{b}}{2}\sin\left(\frac{\phi_{a}+\phi_{b}}{2}\right)\right)^{2}=1,\\ &\left(\cos\frac{\theta_{a}}{2}\cos\frac{\theta_{b}}{2}\sin( \frac{\phi_{a}+\phi_{b}}{2})-\sin\frac{\theta_{a}}{2}\sin\frac{\theta_{b}}{2} \cos\left(\frac{\phi_{a}+\phi_{b}}{2}\right)\right)^{2}=0.\end{split}  \tag{38}\]  
By squaring and adding both sides of Equation (38), we obtain  
\[\cos^{2}\frac{\theta_{a}}{2}\cos^{2}\frac{\theta_{b}}{2}+\sin^{2}\frac{\theta _{a}}{2}\sin^{2}\frac{\theta_{b}}{2}=1 \tag{39}\]  
which is equivalent to  
\[\cos^{2}\frac{\theta_{a}}{2}\sin^{2}\frac{\theta_{b}}{2}+\sin^{2}\frac{\theta _{a}}{2}\cos^{2}\frac{\theta_{b}}{2}=0. \tag{40}\]  
The last equation is satisfied only for \(\theta_{a}^{*}=\theta_{b}^{*}\in\{0,\pi\}\), as claimed. Let us consider a symmetric game of the type  
\[ \begin{pmatrix}(\gamma,\gamma)&(\beta,\alpha)\\ (\alpha,\beta)&(\delta,\delta)\end{pmatrix} ,\quad\alpha>\gamma>\delta,\quad \alpha>\beta,\quad\text{and}\quad(\gamma,\gamma)\
otin Conv\big{(}\{(\alpha, \beta),(\delta,\delta),(\beta,\alpha)\}\big{)}, \tag{41}\]  
where \(Conv(\cdot)\) denotes the convex hull of the set of points. There are no pure Nash equilibria of the game (41) in parameterization (3). Proof. : Let us consider the payoff functions  
\[\Pi_{a}(\theta_{a},\phi_{a},\theta_{b},\phi_{b}) =\gamma P_{0,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+\beta P_{ 0,1}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+\alpha P_{1,0}(\theta_{a},\phi_{ a},\theta_{b},\phi_{b})+\delta P_{1,1}(\theta_{a},\phi_{a},\theta_{b},\phi_{b}), \tag{42}\] \[\Pi_{b}(\theta_{a},\phi_{a},\theta_{b},\phi_{b}) =\gamma P_{0,0}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+\alpha P _{0,1}(\theta_{a},\phi_{a},\theta_{b},\phi_{b})+\beta P_{1,0}(\theta_{a},\phi_{ a},\theta_{b},\phi_{b})+\delta P_{1,1}(\theta_{a},\phi_{a},\theta_{b},\phi_{b}). \tag{43}\]  
Suppose that there exists a Nash equilibrium \((\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\). By Lemma 1, we conclude that player \(a\)'s response \((\theta_{a}^{*},\phi_{a}^{*})\) to an equilibrium strategy \((\theta_{b}^{*},\phi_{b}^{*})\) of player \(b\) implies  
\[\Pi_{a}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}) =\gamma P_{0,0}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{ b}^{*})+\alpha P_{1,0}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\geq\gamma. \tag{44}\]  
Similarly, if \((\theta_{a}^{*},\phi_{a}^{*})\) is an equilibrium strategy of player \(a\), we can see that  
\[\Pi_{b}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}) =\gamma P_{0,0}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b }^{*})+\alpha P_{0,1}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*}) \geq\gamma. \tag{45}\]  
Inequalities (44) and (45) show that a Nash equilibrium for each player would give each player at least \(\gamma\). Since \((\gamma,\gamma)\
otin Conv((\alpha,\beta),(\delta,\delta),(\beta,\alpha))\), a strict inequality of (44) or (45) would mean that the other player gets less than \(\gamma\) in the equilibrium. Therefore, there is no Nash equilibrium that would give a payoff vector other than \((\gamma,\gamma)\). Suppose that there exists a Nash equilibrium \((\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})\) with the outcome \((\gamma,\gamma)\). Thus, a strategy profile \(((\theta_{a}^{*},\phi_{a}^{*}),(\theta_{b}^{*},\phi_{b}^{*}))\) that generates the payoff \(\gamma\) obeys the following condition:  
\[P_{0,0}(\theta_{a}^{*},\phi_{a}^{*},\theta_{b}^{*},\phi_{b}^{*})=1. \tag{46}\]By Lemma 2, it follows that \(\theta_{a}^{*}=\theta_{b}^{*}\in\{0,\pi\}\).