Quantum Computing "Not" Operation - Mathematics Steps For Deriving It

[Sparky_]

TL;DR

Math steps in deriving Quantum Computing Not Operation MIT 8.04

Hello,

I was watching a video lecture from MIT 8.04 (Allan Adams)– lecture #24 (around the 38 minute mark give or take)

The topic is quantum computing, Dr. Adams is deriving / explaining how to get various computing operations. For the “NOT” operation he explains that the operator

$$ U_{Not} = \begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix} \\
$$
Performs the not operation

Next (and to my question)

He states, “I can write this as
$$ U_{Not} = \begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix} \\ = -ie^{i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}
}
$$
... I mean you can't stop me"

Next, he says, "expanding this out with the exponential we get 1 plus the thing and then all the other terms"

He writes:
$$ = -i(cos(\frac{\pi}{2}) \mathbb1 + isin(\frac{\pi}{2})
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix})
$$

His point is to get this where one can see Schrödinger evolution with a magnetic field..

My question is I cannot fill in the steps to go from the expansion of the exponential to his result

Is there a little hand-waving?

Just straight series expansion:
$$
-ie^{i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}} = -i\sum_{n=0}^\infty \frac{i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}^n
} {n!} = -i(1 +
i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix} + higher terms)
$$

He says this equals
$$
= -i(cos(\frac{\pi}{2}) \mathbb1 + isin(\frac{\pi}{2})
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix})
$$

Notes: The bold "1" is the unitary matrix - he wrote the result with the unit matrix and the other matrix outside of the cos and sin terms

Can you help clarify the step or steps I'm missing?

Is he simply throwing in a "0" for and a "1" with the cos and sin terms? If so, I don't quite see what he is showing?

Thanks
-Sparky_

 

[George Jones]

Just straight series expansion:
$$
-ie^{i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}} = -i\sum_{n=0}^\infty \frac{i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}^n
} {n!} = -i(1 +
i\frac{\pi}{2}
\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix} + higher terms)
$$

In order to see the patterns, you have to calculate a few more terms. What is
$$\begin{pmatrix}

0& 1 \\

1 &0 \end{pmatrix}^n$$
for $n=2,3,4$?

 

[Sparky_]

Thank you so much!

Yep - I (now) see it after writing out a few more terms (and looking up the expansion for sin and cos), and squaring the matrix gives the unit matrix - so some terms have the original matrix while the others have the unit matrix

I see that "half" of the series gives sin and the other "half" gives the cos

I was expecting (assuming) a different path, meaning something like the higher terms not relevant and / or the Euler equation present itself .. (idiot)

Again - thank you for the help