Exercise 2.8 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 2.8}$

Alice is confused. She understands that $\ket 1$ and $-\ket 1$ represent the same state. But she does not understand why that does not imply that $\frac{1}{\sqrt 2}(\ket 0 + \ket 1)$ and $\frac{1}{\sqrt 2}(\ket 0 - \ket 1)$ would be the same state. Can you help her out?

page revision: 1, last edited: 12 Nov 2011 05:44

In the first case the coefficient "-1" represent a global phase. this coefficient does not alter in any way the probability of this qubit in any basis.

in the second case the coefficient is altering the relative phase of those qubits hence they are not the same state.

The global phase is always cancel out during the inner product calculation when changing basis, so it carries no useful information.

The relative phase on the other hand will not.

In the above example $\frac{1}{\sqrt{2}}(\ket 0 + \ket 1) = \ket +$ and $\frac{1}{\sqrt{2}}(\ket 0 - \ket 1) = \ket -$

So:

$\frac{1}{\sqrt{2}}(\ket 0 + \ket 1)\ \rightarrow\ P_{\ket +} = 1 , \ P_{\ket -} = 0$

$\frac{1}{\sqrt{2}}(\ket 0 - \ket 1)\ \rightarrow\ P_{\ket +} = 0 , \ P_{\ket -} = 1$

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