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I've come across this relation numerous times, textbooks use it as if it is obvious. But I have never come across a proof or an intuitive explanation about why is it true.

It would be helpful if someone helps me with what exactly to refer to in order to understand this.

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  • $\begingroup$ This is more or less a postulate of quantum mechanics. Stated implicitly when talking about the probability to measure a certain eigenvalue given by the projection squared of a state onto the corresponding eigenstate. See postulate IV at en.wikipedia.org/wiki/… $\endgroup$
    – Hans Wurst
    Jul 7 at 7:58
  • $\begingroup$ If you consider a scattering problem in QM, you will see that the scatterinng cross section is expressed via $|f(\theta)|^2$. $\endgroup$ Jul 7 at 13:33
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This follows from a fundamental posulate in quantum mechanics, that $\langle f \mid i \rangle$ is literally the probability amplitude that given the system is initially in the state $\mid i\rangle$, we find it in the state $\mid f\rangle$.

Since we are dealing with probability amplitudes, the probability for this process is then the square of this. That is, $$P_{i\rightarrow f}=\langle f \mid i \rangle ^2$$

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    $\begingroup$ I think the OP's confusion is a bit different. For example, textbooks often talk about "the amplitude to go from $x_i$ to $x_f$". But in terms of position eigenstates, $\langle x_i | x_f \rangle = 0$ if $x_i \neq x_f$, which seems contradictory. In this case there is some tricky notation that many textbooks do a bad job of explaining... $\endgroup$
    – knzhou
    Jul 7 at 5:29
  • $\begingroup$ I don't know that that is actually applicable here, in that the position operator has a purely continuous spectrum. We could try a position operator such that $\hat X$ and $\hat X \delta(x-x_0) = x_0 \delta(x-x_0)$ though the delta function isn't actually a state in H - it is not a square-normalizable function (or a function at all). $\endgroup$
    – joseph h
    Jul 7 at 5:42
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Linear algrebra offers an intuitive explanation. Recall, that for two normalized vectors in $\mathbb R^2$, \begin{equation} \overline v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, \enspace \overline w = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix} \end{equation} the inner product is related to the angle between the vectors $\overline v \cdot \overline w = \cos \theta$. Hence the inner product measures the similarity between $\overline v$ and $\overline w$. Alternatively, the inner product $\overline v\cdot \overline w$ expresses the projection of $\overline v$ onto $\overline w$.

This interpretation of the inner product is not restricted to $\mathbb R^2$ but generalizes to other vector spaces. In quantum mechanics a physical state is described by a state vector $| i \rangle$ which lives in some abstract Hilbert space $\mathcal H$. Again, we should interpret the inner product $\langle i|f\rangle$ as measuring the similarity between $|i\rangle$ and $|f\rangle$. Thus, it seems very natural that $|\langle i|f\rangle|^2$ is the probability of finding $|i\rangle$ in the state $|f\rangle$.

While the explanation is by no means a rigorous proof, I find this intuition very useful. I hope it helps you understand the topic.

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  • $\begingroup$ what i meant by "ends up in state |f>" was that what are the chances that |i> will time evolve into |f>. $\endgroup$ Jul 7 at 11:08
  • $\begingroup$ In that case the answer is more or less the same except you remember that both states are time dependent $|i(t)\rangle$ and $|f(t)\rangle$. $\endgroup$ Jul 7 at 12:37

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