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According to Wikipedia article on wave interference the probability of observing an object at location x in quantum mechanics for wave function $\Psi(x)=\Psi_1(x)+\Psi_2(x)$ is:
$P(x)=|\Psi(x)|^2=|\Psi_1|^2+|\Psi_2|^2+(\Psi_1^*\Psi_2+\Psi_1\Psi_2^*)$ and the last two terms are what is called quantum interference term.
Now in the case of a qubit we have:
$|\Psi\rangle=\frac{|u\rangle-i|d\rangle}{\sqrt{2}}$. In that case is the probability going to be
$P=\frac{\langle u|+i\langle d|}{\sqrt{2}} \frac{|u\rangle-i|d\rangle}{\sqrt{2}}= 1/2[\langle u|u\rangle+\langle d|d\rangle+(-i\langle u|d\rangle+i\langle d|u\rangle)]$ ?
Because if so then the quantum interference term is 0 and vanishes although we clearly have interference in such state!

I found the following two relevant questions but none of the answers were satisfying and clear to me!
Can there be an interference term in a two-state quantum system?
What is meant by the term “quantum interference”?

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    $\begingroup$ $P$ is the probability of finding the qubit in state $\left|\Psi\right\rangle$, given that it is in state $\left|\Psi\right\rangle$, which is 1. The notation here is a bit confusing. In the former case you are looking at wave functions depending on some parameter $x$ and you can assign probability to locations in space of finding the particle there. In the latter case, you have prepared the quibit in some state and are trying to calculate the probability of finding the qubit in said state. $\endgroup$
    – schade96
    Aug 16, 2021 at 15:15
  • $\begingroup$ @schade96 Thanks for the answer. And if we want to see the "quantum interference term" in qubit context, like what we have for $P(x)$ how should we represent it? $\endgroup$
    – al pal
    Aug 16, 2021 at 15:18
  • $\begingroup$ @schade96 $P(x)$ is the probability density of finding the system in state $|\Psi\rangle$ in configration state $x$. This is not one as you say. Indeed, $\langle \Psi |\Psi\rangle =1$ but $P(x) = \langle x|\Psi\rangle \neq \langle \Psi |\Psi\rangle$. $\endgroup$
    – Hldngpk
    Aug 16, 2021 at 16:39
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    $\begingroup$ @Hldngpk I was referring to "$P$" and not "$P\!\left(x\right)$". $\endgroup$
    – schade96
    Aug 16, 2021 at 17:05
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    $\begingroup$ @schaden96 ah I see. Sorry about the confusion. I indeed thought you were refereing to $P(x)$. $\endgroup$
    – Hldngpk
    Aug 16, 2021 at 17:07

1 Answer 1

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Take your qubit state $|\psi\rangle$ and let $|i\rangle$ some element in $\{|u\rangle, |d\rangle\}$ since this probability space only has two state. We thus have

$$\langle i| \psi\rangle = \frac{1}{\sqrt{2}}\big\{\langle i|u\rangle - \langle i| d\rangle\big\}.$$

Now,

$$P(i):= |\langle i| \psi\rangle|^{2} = \frac{1}{2}\big\{|\langle i|u\rangle|^{2} +|\langle i| d\rangle\big\}|^{2}+i\langle i|u\rangle\langle i| d\rangle^{*}-i\langle i|u\rangle^{*}\langle i| d\rangle.$$

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  • $\begingroup$ Thanks for the answer and comment. Here is what I'm confused about. We clearly have interference in state $|\Psi\rangle$ since it is a superposition of two other states (pls correct me if I'm wrong). But then in your calculations above if we substitute $|i\rangle$ with lets says $|u\rangle$, the interference term would vanish. So my question is what interference (interference between which two things) are we talking about in both $P(x)$ and $P(i)$? $\endgroup$
    – al pal
    Aug 16, 2021 at 16:48
  • $\begingroup$ You are correct, $|\Psi\rangle$ exhibits "interference". In the theory of quantum open systems one expects that the "interference" terms will decay and you would be left with a probability density $P(x)$ that would allow you to compute the probability that your system is in state $|x\rangle$. $\endgroup$
    – Hldngpk
    Aug 16, 2021 at 17:03
  • $\begingroup$ For the qubit case, these object $P(i)$ also spits out values whenever $i \in \mathbb{C}^{2} $ i.e. any vector in $\mathbb{C}^{2}$ not just $|u\rangle$ and $|d\rangle$. The fact that $P(i)$ is non zero for $|\phi\rangle = \alpha |u\rangle +\beta|d\rangle$ in general, for $\alpha $ and $\beta$ no equal to $0$ or $1$ qualifies this state as exhibiting "inteference". $\endgroup$
    – Hldngpk
    Aug 16, 2021 at 17:03
  • $\begingroup$ Let me rephrase my question. If by producing the interference term when calculating $P(i)$ we can show that there is interference between $|\Psi\rangle$ and $|I\rangle$, and knowing that for state $|\Psi\rangle=\frac{|u\rangle-i|d\rangle}{\sqrt{2}}$ there is interference between the two parts, how can we show the existence of such interference between the two parts of $|\Psi\rangle$ using the former $P(i)$ way(language, method)? $\endgroup$
    – al pal
    Aug 16, 2021 at 17:36
  • $\begingroup$ Or maybe if I want to put it another way is that in $|\Psi(x)\rangle$ we see the interference pattern on the screen, where is the physical manifestation of interference pattern in qubit case (or in Stern Gerlach, spin half case)? $\endgroup$
    – al pal
    Aug 16, 2021 at 21:12

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