anwser to 2.5 a)

bernardobl 19 Jul 2017 16:14

**a)**

### $\frac{1}{\sqrt{2}}(\ket + + e^{i\theta}\ket -) \Rightarrow \frac{1}{\sqrt{2}}( \frac{1}{\sqrt{2}}(\ket 0 + \ket 1) + e^{i\theta}\frac{1}{\sqrt{2}}(\ket 0 - \ket 1))$

### $\frac{1}{\sqrt{2}}(\ket + + e^{i\theta}\ket -) \Rightarrow \frac{1 + eî\theta}{2}\ket 0 + \frac{1-e^{}i\theta}{2}\ket 1$

Coefficient of $\ket 0$ must be zero in order for the states to be equivalent, so $\frac{1 + eî\theta}{2} = 0$.

By Euler formula

### $e^{i\theta }= cos(\theta) + i sen(\theta)$

### $e^{i\theta} = cos(\theta) + i sen(\theta) = -1$

therefore $\theta = \pi + 2K\pi, K \in \mathbb{N}$

**b)**

$e^{i\theta} = 1$ and $e^{-i\theta} = 1$

$cos(\theta) + i sen(\theta) = 1$ and $cos(\theta) - i sen(\theta) = 1$

therefore $\theta = 0 + 2K\pi, K \in \mathbb{N}$

**c)**

$e^{i\theta}$ is a global phase therefore any value of $\theta$ make the two states equivalent