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?

bernardobl 19 Jul 2017 17:46

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

Last edited on 20 Jul 2017 21:22 by bernardobl Show more

A higher level answer

shb65 11 Jun 2018 12:23

|1⟩ and −|1⟩ are different representations of the same state. When you carry out an operation on a state, you have to do so with respect to the particular representation you're using.

Here's a really simple analogy. Consider (a) an isosceles triangle pointing upwards, and (b) the same size isosceles triangle rotated through 180 degrees, i.e. pointing downwards. We can say that these two images are different representations of the same shape.

Now let's carry out an operation "put a circle on" to each of these shapes. Position a circle directly above the triangle on the page, just touching the triangle. Are the shapes still the same? No.

To keep the shape the same after the operation, we need to make sure that the "put a circle on" operation is carried out in the right way, i.e. with respect to the particular representation we initially chose.