# Bra-ket notation average measured value

Let's consider an operator $A$ with eigenket $|a^\prime\rangle$. Then the average measured value according to 1.4.6 is $$\langle A\rangle=\sum_{a^\prime} \sum_{a^{\prime\prime}} \langle \alpha|a^{\prime\prime}\rangle\langle a^{\prime\prime}|A|a^\prime\rangle\langle a^\prime |\alpha\rangle =\sum_{a^\prime} a^\prime |\langle a^\prime|\alpha\rangle|^2$$ where $|a^{\prime\prime}\rangle$ is another eigenket. What I don’t understand is how this transformation works. If I let $A$ act on $|a^\prime\rangle$ I get the $a^\prime$ in front of the $|\langle a^\prime|\alpha\rangle|^2$ part.

$\textbf{Question:}$ Why does this mean that the sum over $a’’$ vanishes? Maybe my problem is also that I don't fully understand the difference between expectation value and this average measured value.

(Source: Sakurai, Modern Quantum Mechanics Revised Edition, formula 1.4.7)

• Hint: $\langle a^{\prime\prime}|A|a^\prime\rangle=a'\langle a^{\prime\prime}|a^\prime\rangle=a' \delta_{a'',a'}$ – user12029 Nov 4 '16 at 9:59

It vanishes because the eigenstates are assumed to be orthogonal and normalized and then $\langle a' | a''\rangle = \delta_{a'a''}$

\begin{eqnarray*} \langle \alpha | A | \alpha \rangle &=& \sum_{a'a''} \langle \alpha | a'' \rangle\langle a'' | A | a' \rangle \langle a' | \alpha \rangle \\ &=& \sum_{a'a''} a'\langle \alpha | a'' \rangle\langle a'' | a' \rangle \langle a' | \alpha \rangle \\ &=&\sum_{a'a''} a' \delta_{a'a''}\langle \alpha | a'' \rangle \langle a' | \alpha \rangle \\ &=&\sum_{a'} a' \langle \alpha | a' \rangle \langle a' | \alpha \rangle \\ &=& \sum_{a'} a'|\langle a | a'\rangle|^2 \end{eqnarray*}

• You've got an extra a'' in the second to last line. Also the alphas morph into as at the end. – Emilio Pisanty Nov 4 '16 at 10:07
• @caverac If you could fix the things Emilio Pisanty mentioned, I will mark it as answered. Thanks. – Quasar Nov 4 '16 at 11:23
• @Quasar Sorry, just a type – caverac Nov 4 '16 at 11:53

I don't fully understand the difference between expectation value and this average measured value.

This (the first expression) is just the expectation of $A$ on the state $|\alpha\rangle$ written in an expansion on the eigenbasis of $A$.

The fact that the eigenbasis of an observable $A$ is orthonormal and complete is expressed as

$$\langle a | a' \rangle = \delta_{a\,a'}$$

$$1 = \sum_a |a\rangle \langle a |$$

Now, starting with the expectation

$$\langle \alpha | A | \alpha \rangle$$

insert the completeness identity once

$$\langle \alpha | 1 \cdot A | \alpha \rangle = \langle \alpha |\sum_a |a\rangle \langle a | A | \alpha \rangle = \sum_a \langle \alpha |a\rangle \langle a | A | \alpha \rangle$$

and then insert the completeness identity again

$$\sum_a \langle \alpha |a\rangle \langle a | A \cdot 1 | \alpha \rangle = \sum_a \langle \alpha |a\rangle \langle a | A \sum_{a'} |a'\rangle \langle a' | \alpha \rangle = \sum_a \sum_{a'} \langle \alpha |a\rangle \langle a | A |a'\rangle \langle a' | \alpha \rangle$$

But, using the orthonormal property of the eigenbasis, we have

$$\langle a | A |a'\rangle = a' \langle a | a'\rangle = a' \delta_{a\,a'}$$

Since $\delta_{a\,a'}$ is zero unless $a' = a$,

• the sum over $a'$ has only one non-zero term

which is the term when $a' = a$.

$$\sum_a \sum_{a'} \langle \alpha |a\rangle a' \delta_{a\,a'} \langle a' | \alpha \rangle = \sum_a \langle \alpha |a\rangle a \langle a | \alpha \rangle = \sum_a a |\langle a | \alpha \rangle|^2$$