Nielsen&Chuang Exercise2.65: Express the states (|0 + |1)/ √ 2 and (|0−|1)/ √ 2 in a basis in which they are not the same up to a relative phase shift.

Consider an orthnormal basis :$\begin{cases}|w_1\rangle=\sqrt{\frac23}|0\rangle+\sqrt{\frac13}|1\rangle\\|w_2\rangle=\sqrt{\frac13}|0\rangle-\sqrt{\frac23}|1\rangle\end{cases}$

In this basis, the states can be expressed as:$\begin{cases}\frac{|0\rangle+|1\rangle}{\sqrt2}=\frac{\sqrt2+1}{\sqrt6}[|w_1\rangle-(1-\sqrt2)^2|w_2\rangle]\\\frac{|0\rangle-|1\rangle}{\sqrt2}=\frac{\sqrt2-1}{\sqrt6}[|w_1\rangle+(1+\sqrt2)^2|w_2\rangle]\end{cases}$

It's easy to verify that $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$. But in the Nielsen & Chuang, it says that:

we say that two amplitudes, a and b, differ by a relative phase if there is a real θ such that a = exp(iθ)b.

so clearly the amplitudes must satisfy the equation: $|\frac{a}b|=|exp(i\theta)|=1$. So $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$ violates this equation. The question is that can we say in this basis $|w_1\rangle,\ |w_2\rangle$, these two states are not the same up to a relative phase shift? What's more, if it is true, is my proof right? Additonally, could you please explain the phrase 'up to a relative phase shift' explicitly? I have difficulty grasping this.

Nielsen&Chuang Exercise2.65: Express the states (|0 + |1)/ √ 2 and (|0−|1)/ √ 2 in a basis in which they are not the same up to a relative phase shift.

Consider an orthnormal basis :$\begin{cases}|w_1\rangle=\sqrt{\frac23}|0\rangle+\sqrt{\frac13}|1\rangle\\|w_2\rangle=\sqrt{\frac13}|0\rangle-\sqrt{\frac23}|1\rangle\end{cases}$

In this basis, the states can be expressed as:$\begin{cases}\frac{|0\rangle+|1\rangle}{\sqrt2}=\frac{\sqrt2+1}{\sqrt6}[|w_1\rangle-(1-\sqrt2)^2|w_2\rangle]\\\frac{|0\rangle-|1\rangle}{\sqrt2}=\frac{\sqrt2-1}{\sqrt6}[|w_1\rangle+(1+\sqrt2)^2|w_2\rangle]\end{cases}$

It's easy to verify that $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$. But in the Nielsen & Chuang, it says that:

we say that two amplitudes, a and b, differ by a relative phase if there is a real θ such that a = exp(iθ)b.

so clearly the amplitudes must satisfy the equation: $|\frac{a}b|=|\exp(i\theta)|=1$. So $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$ violates this equation. The question is that can we say in this basis $|w_1\rangle,\ |w_2\rangle$, these two states are not the same up to a relative phase shift? What's more, if it is true, is my proof right? Additonally, could you please explain the phrase 'up to a relative phase shift' explicitly? I have difficulty grasping this.

Nielsen&Chuang Exercise2.65: Express the states (|0 + |1)/ √ 2 and (|0−|1)/ √ 2 in a basis in which they are not the same up to a relative phase shift.

Consider an orthnormal basis :$\begin{cases}|w_1\rangle=\sqrt{\frac23}|0\rangle+\sqrt{\frac13}|1\rangle\\|w_2\rangle=\sqrt{\frac13}|0\rangle-\sqrt{\frac23}|1\rangle\end{cases}$

In this basis, the states can be expressed as:$\begin{cases}\frac{|0\rangle+|1\rangle}{\sqrt2}=\frac{\sqrt2+1}{\sqrt6}[|w_1\rangle-(1-\sqrt2)^2|w_2\rangle]\\\frac{|0\rangle-|1\rangle}{\sqrt2}=\frac{\sqrt2-1}{\sqrt6}[|w_1\rangle+(1+\sqrt2)^2|w_2\rangle]\end{cases}$

It's easy to verify that $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$. But in the Nielsen & Chuang, it says that:

we say that two amplitudes, a and b, differ by a relative phase if there is a real θ such that a = exp(iθ)b.

so clearly the amplitudes must satisfy the equation: $|\frac{a}b|=|exp(i\theta)|=1$. So $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$ violates this equation. The question is that can we say in this basis $|w_1\rangle,\ |w_2\rangle$, these two states are not the same up to a relative phase shift? What's more, if it is true, is my proof right? Additonally, could you please explain the phrase 'up to a relative phase shift' explicitly? I have difficulty grasping this.

Nielsen&Chuang Exercise2.65: Express the states (|0 + |1)/ √ 2 and (|0−|1)/ √ 2 in a basis in which they are not the same up to a relative phase shift.

Consider an orthnormal basis :$\begin{cases}|w_1\rangle=\sqrt{\frac23}|0\rangle+\sqrt{\frac13}|1\rangle\\|w_2\rangle=\sqrt{\frac13}|0\rangle-\sqrt{\frac23}|1\rangle\end{cases}$

In this basis, the states can be expressed as:$\begin{cases}\frac{|0\rangle+|1\rangle}{\sqrt2}=\frac{\sqrt2+1}{\sqrt6}[|w_1\rangle-(1-\sqrt2)^2|w_2\rangle]\\\frac{|0\rangle-|1\rangle}{\sqrt2}=\frac{\sqrt2-1}{\sqrt6}[|w_1\rangle+(1+\sqrt2)^2|w_2\rangle]\end{cases}$

It's easy to verify that $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$. But in the Nielsen & Chuang, it says that:

we say that two amplitudes, a and b, differ by a relative phase if there is a real θ such that a = exp(iθ)b.

so clearly the amplitudes must satisfy the equation: $|\frac{a}b|=|exp(i\theta)|=1$. So $|\frac{\frac{\sqrt2+1}{\sqrt6}}{\frac{\sqrt2-1}{\sqrt6}}|\neq1$ violates this equation. The question is that can we say in this basis $|w_1\rangle,\ |w_2\rangle$, these two states are not the same up to a relative phase shift? What's more, if it is true, is my proof right? Additonally, could you please explain the phrase 'up to a relative phase shift' explicitly? I have difficulty grasping this.