1
$\begingroup$

Given two qubits, let's say it's in state $|01\rangle = |0\rangle \otimes |1\rangle$. These two qubits are not entangled.

Applying a $\theta$ angle phase change (sorry if this is the wrong terminology) to the first qubit gives $(e^{i\theta}|0\rangle) \otimes |1\rangle$. Applying the same phase change to the second qubit gives $|0\rangle \otimes (e^{i\theta}|1\rangle)$. Both of the results are equal, since they are equal to $e^{i\theta} |01\rangle$.

What is the physical interpretation of this? One way to implement two qubits would be having two unrelated photons. And this phase change would presumably some operation that causes some change to a single photon (presumably this done by delaying the photon wave somehow so the electromagnetic wave phase is different). Why would causing some change on one photon result in the same state as the same change on the other photon, and this state is not the same as the initial state? And how does this work with more qubits?

$\endgroup$
1

1 Answer 1

3
$\begingroup$

The underlying idea here is that the global phase of quantum state doesn't matter and is unobservable. For example with the state $e^{i\theta}|01\rangle$ you mentioned, if you measure it in any orthonormal basis one of the basis vectors being $|\psi\rangle$, the probability that you measure the state $|\psi\rangle$ is $|\langle \psi|(e^{i\theta}|01\rangle)|^2 = |\langle \psi|01\rangle|^2$, as if you didn't have a global phase.

On the other hand, relative phases matter a great deal. As mentioned before, there's no way to measure the phase of a single qubit, but you can measure the relative phase between states in superposition. For example, with one qubit, the difference between $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)$ since the phase difference of $-1 = e^{i \pi}$ is between the states $|0\rangle$ and $|1\rangle$. Similarly, the phase of a state like $|10\rangle+e^{i\theta}|01\rangle$ is measurable by interferometry.

Does that help answer your question?

$\endgroup$
2
  • 1
    $\begingroup$ Does this mean that the global phase doesn't correspond to anything physical? In other words, is global phase just an artifact of how the math is defined? So... the "set of possible physical states" is the equivalence class over global phase changes? $\endgroup$ Jan 4, 2023 at 11:03
  • $\begingroup$ Correct, all states that differ only by a global phase are in the same equivalence class corresponding to a single physical state. $\endgroup$
    – Chris E
    Jan 4, 2023 at 19:02

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.