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.


1 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.

  • $\begingroup$ Most of my confusion has been removed thanks to your answer. And by referring to your answer I conclude that the two states expressed as $|\psi\rangle=\alpha|0\rangle+\beta|1\rangle,\ |\phi\rangle=\alpha e^{i\theta_1}|0\rangle+\beta e^{i\theta_2}|1\rangle$ where the relative phase $\theta_1,\ \theta_2$ are different differ by more than one relative phase shift. Is it correct? $\endgroup$
    – Reedlye
    Jul 28, 2022 at 12:27
  • $\begingroup$ @Reedlye Yes! Basically, if you can factor a phase out of all the elements such that one state equals the other, then the 2 states differ by relative phase and you can safely ignore this "global phase" since any subsequent operations on the two states will yield the same measurement result. However, this wouldn't be the case if the phase is not global, in the example above for example $H\vert\psi\rangle = \vert 1\rangle$ while $H\vert\phi\rangle = \vert 0\rangle$ . $\endgroup$
    – Dani007
    Jul 28, 2022 at 19:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.