I have looked at the following:

What is the difference between a relative phase and a global phase? In particular, what is a phase?

Global and relative phases of kets in QM

Global phases and indistinguishable quantum states, mathematical understanding

If two states differ by a scalar of magnitude of 1, then they are indistinguishable. Consider: \begin{align} \vert \psi_1 \rangle &= \dfrac{1}{\sqrt{2}} \vert 0 \rangle + \dfrac{i}{\sqrt{2}} \vert 1 \rangle\\ \vert \psi_2 \rangle &= \color{red}{i}\left(\dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle\right). \end{align}

Which of the following is true about $\vert \psi_1 \rangle$ and $\vert \psi_2 \rangle$?

  1. $\vert \psi_1 \rangle = \vert \psi_2 \rangle$
  2. $\vert \psi_1 \rangle \neq \dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$
  3. $\vert \psi_1 \rangle = \dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$ up to global phase.
  4. If we just ignore the global phase in $\vert \psi_2 \rangle$ and only deal with $\dfrac{-i}{\sqrt{2}} \vert 0 \rangle + \dfrac{1}{\sqrt{2}} \vert 1 \rangle$ , do we still have the state vector on a Bloch sphere yields the same projection as $\vert \psi_1 \rangle$?

Lastly, since the global phase is not physically observable, is it mathematically evident?

  • 2
    $\begingroup$ Does this answer your question? What is the difference between a relative phase and a global phase? In particular, what is a phase? $\endgroup$
    – glS
    Jun 13, 2020 at 5:31
  • $\begingroup$ @glS Thank you for the link. It is very helpful and I am also looking for a complete example showing why the global phase is not obervable, not just a rigorous (and complete) definition. This is why I chose a random example included in the question. $\endgroup$ Jun 15, 2020 at 20:16
  • 1
    $\begingroup$ it is not observable because the only measurable aspect of states are squared amplitudes of the form $|\langle\phi|\psi\rangle|^2$. There isn't really much more than that. We write something like $|\psi\rangle=|0\rangle+i|1\rangle$ because it is convenient to do so, but we should more precisel regard states as elements of a projective space, that is, vectors defined up to their amplitude and global phase: $|0\rangle$ and $e^{i\phi}|0\rangle$ are different as vectors, but both vectors model the same physical state $\endgroup$
    – glS
    Jun 15, 2020 at 20:39

1 Answer 1


1) So $|\psi_1\rangle \neq |\psi_2\rangle$, but it effectively is since they give the exact same distributions for any measurement in any basis.

2) Same discussion as above.

3) True

4) States in the Bloch sphere are of the form

$$|\psi\rangle = \cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle,$$

so the state you describe technically is not directly on the Bloch sphere. I think a better way of thinking about global phase is that it's an infinite equivalence class of states with the exact same physical properties, and one representative (the one with a real coefficient in front of $|0\rangle$) is on the Bloch sphere.

You cannot measure the global phase. Phases are only relevant when they are relative, and consequently affect superpositions/measurements.

  • $\begingroup$ Thank you for the answer. I have a couple of follow-up questions: the first is given points #2 and #3 in the question, how are they different but the same (up to global phase)? Does that mean it is obervable (mathematically at least?). Also, if I really want to see the state on the Bloch sphere, where would it be (I thought a Bloch sphere represents all single-qubits), or does that mean a Bloch sphere only represents the qubits in the form you gave and other single-qubit states might not be visually appropriate given their form? $\endgroup$ Jun 15, 2020 at 20:22
  • 2
    $\begingroup$ So a global phase is never observable, which is why we say it's ok to ignore. One cannot differentiate a gate $X$ and a gate $e^{i\alpha}X$, although it's important to note that this does not hold for controlled gates $C-X$ gate vs a $C-e^{i\alpha}X$ gate, since in that case the phase could become a relative phase between two states in superposition. $\endgroup$ Jun 16, 2020 at 17:13
  • 1
    $\begingroup$ The Bloch sphere represents all single qubit states up to a global phase, but the exact points on it correspond directly to states in the form I described above, it's just that we essentially never make the distinction between those states and ones that differ by a global phase. $\endgroup$ Jun 16, 2020 at 17:15
  • $\begingroup$ The original answer and these follow-up answers resolve everything! Thank you so much!! $\endgroup$ Jun 16, 2020 at 17:19

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.