346: 2 x 2 Special Unitary Matrix Can Be Expressed with Sine and Cosine of Angle and Imaginary Exponentials of 2 Angles

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A description/proof of that 2 x 2 special unitary matrix can be expressed with sine and cosine of angle and imaginary exponentials of 2 angles

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Topics

About: matrix

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of matrix.

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Target Context

• The reader will have a description and a proof of the proposition that any 2 x 2 special unitary matrix can be expressed with the sine and the cosine of an angle and the plus and minus imaginary exponentials of another angle and the plus and minus imaginary exponentials of another angle.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Description

Any 2 x 2 special unitary matrix, $M$, can be expressed as $\left(\begin{array}{cc}sin\theta e^{\lambda i}& -cos\theta e^{-\mu i}\\ cos\theta e^{\mu i}& sin\theta e^{-\lambda i}\end{array}\right)$ where $\theta$ is an angle such that $0\le \theta <2\pi$, $\lambda$ is an angle such that $0\le \lambda <\pi$, and $\mu$ is an angle such that $0\le \mu <\pi$.

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2: Proof

Let $M$ be $\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)$. $M^{\ast }M=\left(\begin{array}{cc}{|a|}^2+{|b|}^2& a^{\ast }c+b^{\ast }d\\ a{c}^{\ast }+b{d}^{\ast }& {|c|}^2+{|d|}^2\end{array}\right)=I$ and $detM=ad-bc=1$.

$a=sin\theta e^{\lambda i},b=cos\theta e^{\mu i}$ where $0\le \theta <2\pi$ and $0\le \lambda <\pi$ and $0\le \mu <\pi$ where $\lambda$ or $\mu$ does not need to be to $2\pi$ because $a=sin\theta e^{(\lambda +\pi )i},b=cos\theta e^{\mu i}$ can be realized by choosing a new $\theta$ which changes the sign of the sine while it leaves the cosine the same, which is obviously possible in $0\le \theta <2\pi$, $a=sin\theta e^{\lambda i},b=cos\theta e^{(\mu +\pi )i}$ can be realized by choosing a new $\theta$ which changes the sign of the cosine while it leaves the sign of the sine, which is obviously possible in $0\le \theta <2\pi$, and $a=sin\theta e^{(\lambda +\pi )i},b=cos\theta e^{(\mu +\pi )i}$ can be realized by choosing a new $\theta$ which changes the sign of the sine and the sign of the cosine.

Likewise, $c=-cos{\theta }^{\prime }{e}^{-{\mu }^{\prime }i},d=sin{\theta }^{\prime }{e}^{-{\lambda }^{\prime }i},$ where $0\le {\theta }^{\prime }<2\pi$ and $0\le {\lambda }^{\prime }<\pi$ and $0\le {\mu }^{\prime }<\pi$.

$ad-bc=sin\theta sin{\theta }^{\prime }{e}^{(\lambda -{\lambda }^{\prime })i}+cos\theta cos{\theta }^{\prime }{e}^{(\mu -{\mu }^{\prime })i}=1$.

$a{c}^{\ast }+b{d}^{\ast }=-sin\theta cos{\theta }^{\prime }{e}^{(\lambda +{\mu }^{\prime })i}+cos\theta sin{\theta }^{\prime }{e}^{(\mu +{\lambda }^{\prime })i}=0$. $sin\theta cos{\theta }^{\prime }=cos\theta sin{\theta }^{\prime }$ or $sin\theta cos{\theta }^{\prime }=-cos\theta sin{\theta }^{\prime }$, because regarding $-sin\theta cos{\theta }^{\prime }{e}^{(\lambda +{\mu }^{\prime })i}$ and $cos\theta sin{\theta }^{\prime }{e}^{(\mu +{\lambda }^{\prime })i}$ as vectors, any 2 vectors of different lengths cannot cancel.

For the former case, $cos\theta =cos{\theta }^{\prime }=0$ or $tan\theta =tan{\theta }^{\prime }$. When $cos\theta =cos{\theta }^{\prime }=0$, $sin\theta =sin{\theta }^{\prime }$ or $sin\theta =-sin{\theta }^{\prime }$, $ad-bc=si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}=e^{(\lambda -{\lambda }^{\prime })i}=1$ or $ad-bc=-si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}=-e^{(\lambda -{\lambda }^{\prime })i}=1$, $\lambda -{\lambda }^{\prime }=0$ or impossible ($\lambda -{\lambda }^{\prime }$ cannot be $\pi$ or $-\pi$), so, $\theta ={\theta }^{\prime }$ and $\lambda ={\lambda }^{\prime }$ and $\mu ={\mu }^{\prime }$ (in fact, $\mu$ and ${\mu }^{\prime }$ do not matter, so, they can be taken to equal). When $tan\theta =tan{\theta }^{\prime }$, ($cos\theta =cos{\theta }^{\prime }$ and $sin\theta =sin{\theta }^{\prime }$) or ($cos\theta =-cos{\theta }^{\prime }$ and $sin\theta =-sin{\theta }^{\prime }$), $ad-bc=si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}+co{s}^2\theta e^{(\mu -{\mu }^{\prime })i}=1$ or $ad-bc=-si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}-co{s}^2\theta e^{(\mu -{\mu }^{\prime })i}=1$, ($\lambda -{\lambda }^{\prime }=0$ and $\mu -{\mu }^{\prime }=0$) or impossible ($\lambda -{\lambda }^{\prime }$ or $\mu -{\mu }^{\prime }$ cannot be $\pi$ or $-\pi$), so, $\theta ={\theta }^{\prime }$, $\lambda ={\lambda }^{\prime }$, and $\mu ={\mu }^{\prime }$.

For the latter case, $cos\theta =cos{\theta }^{\prime }=0$ or $tan\theta =-tan{\theta }^{\prime }$. The $cos\theta =cos{\theta }^{\prime }=0$ case is included in the last paragraph case. When $tan\theta =-tan{\theta }^{\prime }$, ($cos\theta =cos{\theta }^{\prime }$ and $sin\theta =-sin{\theta }^{\prime }$) or ($cos\theta =-cos{\theta }^{\prime }$ and $sin\theta =sin{\theta }^{\prime }$), $ad-bc=-si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}+co{s}^2\theta e^{(\mu -{\mu }^{\prime })i}=1$ or $ad-bc=si{n}^2\theta e^{(\lambda -{\lambda }^{\prime })i}-co{s}^2\theta e^{(\mu -{\mu }^{\prime })i}=1$, impossible anyway ($\lambda -{\lambda }^{\prime }$ or $\mu -{\mu }^{\prime }$ cannot be $\pi$ or $-\pi$).

So, ${\theta }^{\prime }=\theta$, ${\lambda }^{\prime }=\lambda$, and ${\mu }^{\prime }=\mu$.

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References

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