Exercise 3.8 Discussion / Discussion

This is the discussion related to the wiki page Exercise 3.8 Discussion.

a)

twhittle 21 Dec 2013 03:59

$\frac{1}{\sqrt2}(|+\rangle|+\rangle+|-\rangle|-\rangle) =$
$\frac{1}{\sqrt2}\left(\frac{1}{2}\left(|0\rangle+|1\rangle\right)\otimes\left(|0\rangle+|1\rangle\right)+\frac{1}{2}\left(|0\rangle-|1\rangle\right)\otimes\left(|0\rangle-|1\rangle\right)\right)=$
$\frac{1}{2\sqrt2}\left(|0\rangle|0\rangle + |0\rangle|1\rangle + |1\rangle|0\rangle + |1\rangle|1\rangle + |0\rangle|0\rangle - |0\rangle|1\rangle - |1\rangle|0\rangle + |1\rangle|1\rangle\right)=$
$\frac{1}{\sqrt2}(|0\rangle|0\rangle+|1\rangle|1\rangle)$

Solution to part (b)

FlippingBits 31 May 2014 03:27

(1)

\begin{align} \frac{1}{\sqrt{2}}\big(\ket{\mathbf{i}}\ket{\mathbf{i}}+\ket{-\mathbf{i}}\ket{-\mathbf{i}}\big) & = \frac{1}{\sqrt{2}}\left[\frac{1}{2}\big(\ket{0}+\mathbf{i}\ket{1}\big) \otimes \big(\ket{0}+\mathbf{i}\ket{1}\big) + \frac{1}{2}\big(\ket{0}-\mathbf{i}\ket{1}\big) \otimes \big(\ket{0}-\mathbf{i}\ket{1}\big) \right] \\ \\ & = \frac{1}{2\sqrt{2}}\big(\ket{0}\ket{0} + \mathbf{i}\ket{0}\ket{1} + \mathbf{i}\ket{1}\ket{0} - \ket{1}\ket{1} + \ket{0}\ket{0} - \mathbf{i}\ket{0}\ket{1} - \mathbf{i}\ket{1}\ket{0} - \ket{1}\ket{1}\big) \\ \\ & = \frac{1}{\sqrt{2}}\big(\ket{0}\ket{0} - \ket{1}\ket{1}\big)\,. \quad \square \end{align}

Last edited on 31 May 2014 03:28 by FlippingBits Show more