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$$\exp(i\pi/4* \begin{matrix} 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{matrix} )$$ pretty sure the answer is in the form: $$ (a+bi)*\begin{matrix} 1 & 0 & 0 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ \end{matrix} $$ where a and b are infinite sums I can't solve. I think they are: $$ a= \sum_{n=1}^\infty (-1)^n(\pi/4)^{2n}/(2n)! $$ $$ b= \sum_{n=1}^\infty (-1)^{n-1}(\pi/4)^{2n-1}/(2n-1)! $$ trying to make sense of a paper on quantum game theory. (Jens Eisert, Martin Wilkens, and Maciej Lewenstein. “Quantum Games and Quantum Strategies.” Phys. Rev. Lett. 83, 3077 – Published 11 October 1999.)

Thanks in advance!

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You’re close. Calling the matrix in your expression $A$ and setting $q=i\frac\pi4$, we have $A^2=I$, so that $$\begin{align} \exp(qA) &= I + qA + {q^2\over2!}A^2+{q^3\over3!}A^3+{q^4\over4!}A^4+\cdots \\ &=\left(1+{q^2\over2!}+{q^4\over4!}+\cdots\right)I + \left(q+{q^3\over3!}+{q^5\over5!}+\cdots\right)A \\ &= (\cosh q) I + (\sinh q) A.\end{align}$$ Now, $$\cosh{i\frac\pi4} = \cos{\frac\pi4} = \frac1{\sqrt2}$$ and $$\sinh{i\frac\pi4} = i\sin{\frac\pi4} = i\frac1{\sqrt2},$$ therefore $$\exp\left(i\frac\pi4 A\right) = \frac1{\sqrt2}(I+iA) = \frac1{\sqrt2}\begin{bmatrix}1&0&0&i\\0&1&-i&0\\0&-i&1&0\\i&0&0&1\end{bmatrix}.$$

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