# Why is the probability that one state $|i\rangle$ ends up in the state $|f\rangle$ given by $|\langle i|f\rangle|^2$? [duplicate]

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.

• 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/… Jul 7 '21 at 7:58
• If you consider a scattering problem in QM, you will see that the scatterinng cross section is expressed via $|f(\theta)|^2$. Jul 7 '21 at 13:33

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, $$P_{i\rightarrow f}=\langle f \mid i \rangle ^2$$

• 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... Jul 7 '21 at 5:29
• 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). Jul 7 '21 at 5:42

Linear algrebra offers an intuitive explanation. Recall, that for two normalized vectors in $$\mathbb R^2$$, $$$$\overline v = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}, \enspace \overline w = \begin{pmatrix} w_1 \\ w_2 \end{pmatrix}$$$$ the inner product is related to the angle between the vectors $$\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.

• what i meant by "ends up in state |f>" was that what are the chances that |i> will time evolve into |f>. Jul 7 '21 at 11:08
• In that case the answer is more or less the same except you remember that both states are time dependent $|i(t)\rangle$ and $|f(t)\rangle$. Jul 7 '21 at 12:37