Exercise 2.1
Assume thatand $\ket{d}$ are normalized basis vectors. And let:
$\ket{l} = \frac{1}{\sqrt{2}}\ket{u} - \frac{1}{\sqrt{2}}\ket{d}$
Prove that $\ket{r}$ is orthogonal to $\ket{l}$
pf/ $\ket{r}$ and $\ket{l}$ are orthogonal iff $\braket{r}{l} = 0$
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Now $\bra{r} = \overline{\ket{r}}$
And $\overline{\ket{r}} = \overline{\dfrac{1}{\sqrt{2}}\ket{u} + \dfrac{1}{\sqrt{2}}\ket{d}}$
$= \overline{\dfrac{1}{\sqrt{2}}\ket{u}} + \overline{\dfrac{1}{\sqrt{2}}\ket{d}} $ \hfill $(\overline{a+b} = \overline{a}+\overline{b})$
$= \overline{\dfrac{1}{\sqrt{2}}}\overline{\ket{u}} + \overline{\dfrac{1}{\sqrt{2}}}\overline{\ket{d}} $ \hfill $(\overline{ab} = \overline{a}\overline{b})$
So $ \bra{r} = \dfrac{1}{\sqrt{2}}\bra{u} + \dfrac{1}{\sqrt{2}}\bra{d} $ \hfill $(\overline{\ket{a}} = \bra{a})$
\end{tabular}
Now, $\braket{r}{l} = (\dfrac{1}{\sqrt{2}}\bra{u} + \dfrac{1}{\sqrt{2}}\bra{d}) \cdot (\dfrac{1}{\sqrt{2}}\ket{u} - \dfrac{1}{\sqrt{2}}\ket{d})$
\begin{tabular}{p{13cm}}
= $\dfrac{1}{2}\braket{u}{u} - \dfrac{1}{2} \braket{l}{l}$
= $ \dfrac{1}{2} *1 - \dfrac{1}{2}*1 = 0$ \hfill (Since $\bra{u}$ is normalized, $\braket{u}{u} = 1$)
So therefore $\braket{r}{l} = 0$
\end{tabular}
Exercise 2.2
There is nothing unique about up/down that is not true for left/right and in/out. So apply the previous exercise. All its trying to say is convince us that the current definition of left and right, are mutually orthogonal. And could work to represent spin states.
Exercise 2.3
Exercise 2.3
Let $\ket{i} = a\ket{u} + b\ket{d}$ and $\ket{o} = c\ket{u} + e\ket{d}$
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a) $a^*a = \braket{u}{i}\braket{i}{u} = 0.5$ \hfill (b/c when the spin is set to in, the prob of up is 0.5)
b)
c)
Say that $a^*b$ was real. Then $a^*b = (a^* b)^* = ba^*$. (real number are equal to their complex conjugates.) But then $a^*b + ba^* = 0 = 2a^*b$. But then either $a^*$ or $b$ must be 0. But then either $a^*a$ or $b^*b$ must be equal to 0 which contradicts.
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