What is the solution to Nielsen and Chuang Exercise 2.65?

Question

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.

Answer 1

For 2 amplitudes $\alpha$ and $\beta$ to be equivalent up to a relative phase means that the states associated with $\alpha$ and $\beta$ have the same probability of being measured. For example, if you have the state:

$$\vert \psi\rangle = \alpha \vert 0\rangle + e^{i\theta}\alpha \vert 1\rangle$$

Then the probability of measuring $\vert0\rangle$ is $\|\alpha\|^2 = \|e^{i\theta}\alpha\|^2$ which is equal to the probability of measuring $\vert1\rangle.$ So the phase $e^{i\theta}$ is in a sense "relative".

Now for 2 states $\vert \psi\rangle$ and $\vert \phi\rangle$ to only differ by a relative phase it means:

$$\vert \psi\rangle = e^{i\theta}\vert \phi\rangle$$

Now, this condition is stronger than saying their measurement outcomes will be the same. For instance:

$$\vert \psi \rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle -\vert 1\rangle)$$ $$\vert \phi \rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle +\vert 1\rangle)$$

Both will have the same probability of giving you 0 and 1, but they don't just differ by a relative phase. In other words, having the same output probability is a necessary condition but not sufficient. In your case, the 2 states don't even have the necessary condition as you've shown yourself so they definitely can't be the same up to a relative phase.