It's a small problem that is bothering me for a long time. Suppose I have a system in a state $ \psi_i$. It is a linear superposition of several states, let's say $$\psi_i = a\psi_1 + b\psi_2$$ with $|a|^2 ≤1$, $|b|^2 ≤1$ and $|a|^2 + |b|^2 =1$ . Now this system has a probability of interacting with another system which has a definite state and the probability of interaction is $$e,$$ and the former system ( after interaction ) reaches the final state $$\psi_f = c\psi_1 + d\psi_2$$ with $|c|^2 ≤1$, $|d|^2 ≤1$ and $|c|^2 + |d|^2 =1$. So my question is :

Is there any meaning attached to the statement $$\langle\psi_f|e|\psi_i\rangle ?$$

If so, then what is the meaning?

  • 1
    $\begingroup$ If you consider "e" as begin a probability, then this quantity is just a real number so in the last statement e can get in front... $\endgroup$ Jul 25, 2017 at 7:58
  • $\begingroup$ Okay, then what happens? What's the meaning of the rest? And if it has a meaning, how do I calculate it out explicitly? $\endgroup$ Jul 25, 2017 at 7:59
  • $\begingroup$ Are $\psi_1$ and $\psi_2$ orthogonal? $\endgroup$ Jul 25, 2017 at 8:10
  • $\begingroup$ Yeah, they are orthogonal base pairs. $\endgroup$ Jul 25, 2017 at 8:11
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    $\begingroup$ What does "probability of interaction" mean? That's not standard terminology in quantum mechanics. $\endgroup$
    – ACuriousMind
    Jul 25, 2017 at 9:59

1 Answer 1


Since $e$ is a real number (i.e. a probability), the quantity $\langle\psi_f|e|\psi_i\rangle$ is actually just $e\langle\psi_f|\psi_i\rangle$. The quantity $\langle\psi_f|\psi_i\rangle$ is typically called the "overlap" of the state $|\psi_f\rangle$ with the state $|\psi_i\rangle$. It's an analogue of the dot product (it's actually an inner product, which behaves similarly). The inner product operation is conjugate-linear in the first argument and linear in the second argument by definition,* and so we can separate components as follows:

\begin{align*} \langle\psi_f|\psi_i\rangle &= \langle c\psi_1+d\psi_2|a\psi_1+b\psi_2\rangle\\ &=\langle c\psi_1|a\psi_1+b\psi_2\rangle+\langle d\psi_2|a\psi_1+b\psi_2\rangle\\ &=\langle c\psi_1|a\psi_1\rangle+\langle c\psi_1|b\psi_2\rangle+\langle d\psi_2|a\psi_1\rangle+\langle d\psi_2|b\psi_2\rangle\\ &=c^*a\langle\psi_1|\psi_1\rangle+c^*b\langle\psi_1|\psi_2\rangle+d^*a\langle\psi_2|\psi_1\rangle+d^*b\langle\psi_2|\psi_2\rangle \end{align*}

Assuming $|\psi_1\rangle$ and $|\psi_2\rangle$ are an orthonormal basis, then $\langle \psi_1|\psi_1\rangle=\langle\psi_2|\psi_2\rangle=1$ and $\langle\psi_1|\psi_2\rangle=\langle\psi_2|\psi_1\rangle=0$, meaning that $\langle\psi_f|\psi_i\rangle=c^*a+d^*b$.

*This is not the usual definition adopted by mathematicians, but is common in physics. In mathematics, the first argument is linear and the second argument is conjugate-linear.

  • $\begingroup$ Well, thank you for this beautiful answer. But can you clear one doubt? Why did we take c'*a and not a'*c ? Is it because of the arrangement? And why do we even take the conjugate? $\endgroup$ Jul 25, 2017 at 8:29
  • $\begingroup$ $c$ and $a$ are complex numbers, which commute, so $c^*a=ac^*$. We take the conjugate because that is how the inner product is defined; see en.wikipedia.org/wiki/Inner_product_space. $\endgroup$ Jul 25, 2017 at 8:31

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