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

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The discussion centers on the derivation of the quantum NOT operation in quantum computing, specifically how to express the operator using exponential functions. Dr. Adams presents the NOT operation as a matrix and expands it using the exponential series, leading to a combination of sine and cosine terms. The confusion arises regarding the transition from the series expansion to the final expression involving the unit matrix and the NOT matrix. After further exploration, it becomes clear that the series expansion separates into terms representing sine and cosine, with higher-order terms contributing to the overall expression. The clarification highlights the importance of understanding matrix powers in the context of quantum operations.
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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 Schrodinger 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_
 
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Sparky_ said:
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##?
 
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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
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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