Exercise 3.8 Discussion
$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.8}$
a) Show that $\frac{1}{\sqrt{2}}(\ket{0}\ket{0} + \ket{1}\ket{1})$ and $\frac{1}{\sqrt{2}}(\ket{+}\ket{+} + \ket{-}\ket{-})$ refer to the same quantum state.
b) Show that $\frac{1}{\sqrt{2}}(\ket{0}\ket{0} - \ket{1}\ket{1})$ refers to the same state as $\frac{1}{\sqrt{2}}(\ket{\i}\ket{\i} + \ket{-\i}\ket{-\i})$.
page revision: 0, last edited: 11 Dec 2011 23:17
$\frac{1}{\sqrt2}(|+\rangle|+\rangle+|-\rangle|-\rangle) =$
$\frac{1}{\sqrt2}\left(\frac{1}{2}\left(|0\rangle+|1\rangle\right)\otimes\left(|0\rangle+|1\rangle\right)+\frac{1}{2}\left(|0\rangle-|1\rangle\right)\otimes\left(|0\rangle-|1\rangle\right)\right)=$
$\frac{1}{2\sqrt2}\left(|0\rangle|0\rangle + |0\rangle|1\rangle + |1\rangle|0\rangle + |1\rangle|1\rangle + |0\rangle|0\rangle - |0\rangle|1\rangle - |1\rangle|0\rangle + |1\rangle|1\rangle\right)=$
$\frac{1}{\sqrt2}(|0\rangle|0\rangle+|1\rangle|1\rangle)$