I'm currently reading "The principles of quantum mechanics" by Dirac, and I'm having some trouble understanding some of his assumptions, because in the quantum mechanics course I'm following at university we are taking a different approach (a more modern one I guess).
I got to the point where Dirac explains why he decides to define the average value of the observable $\xi$ in the state $|x\rangle$ as $\langle x|\xi|x\rangle$.
What I don't get is the following part, where he writes that the probability $P_a$ of $\xi$ having the value $a$ is given by the average value of the Kronecker delta $\delta_{\xi a}$:
$$P_a=\langle x|\delta_{\xi a}|x\rangle$$
I don't understand if he's deciding to define the probability of $\xi$ having the value $a$ in this way, or if he's deducing this formula.
In the following similar question: Why does the probability of obtaining a value of a measurement follow from Dirac's general assumption? By Symmetry writes:
The Projection Operator $\delta_{\varepsilon\, a}$ is $1$ for some particular $|a\rangle$ and is $0$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by \begin{equation} \left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = \Pr(|x\rangle = |a\rangle)\end{equation}
So, from this answer it seems to me that he's saying that $P_a=\langle x|\delta_{\xi a}|x\rangle$ by definition, but from Dirac's book I have the impression that he's deducing it from previous assumptions.
So my question is: is the formula for the probability of $\xi$ having value $a$ in the state $|x\rangle$ just a definition or can it be deduced by previous assumptions in Dirac's book? If he's deducing it, how is he doing it?
Best Answer
Based on his Section 16 at Equation 16, Dirac is using the $\delta$ symbol with two suffixes (like $\delta_{ab}$ or $\delta_{\xi a}$) to denote a Kronecker delta, not a Dirac delta.
Next, note that Dirac denotes "the normalized ket" by the symbol $|x\rangle$. He seem to sometimes use $|x\rangle$ in his text for kets with a continuous label, but he also refers to this ket as "the normalized ket," so in this case he may be using the notation differently.
Anyways, instead of calling the normalized ket "$|x\rangle$," I will call it "$|\Psi\rangle$."
And, instead of calling the observable "$\xi$," I will call it "$\hat A$."
Let $\hat A$ have eigenvectors $\{|a\rangle\}$.
Let the expansion of $|\Psi\rangle$ in terms of the $|a\rangle$ be: $$ |\Psi\rangle = \sum_{a}\Psi_a |a\rangle\;. $$
The expectation value of $\hat A$ in the state $|\Psi\rangle$ is: $$ \langle\Psi|\hat A|\Psi\rangle = \sum_{a,b}\Psi_a\Psi_b^*\langle b|\hat A|a\rangle = \sum_{a}|\Psi_a|^2 a $$
The expectation value of $f(\hat A)$ in the state $|\Psi\rangle$ is: $$ \langle\Psi|f(\hat A)|\Psi\rangle =\sum_{a,b}\Psi_a\Psi_b^*\langle b|f(\hat A)|a\rangle = \sum_{a}|\Psi_a|^2 f(a) $$
The expectation value of a function: $$ f_{a_0}(a) \equiv \delta_{aa_0} = \left\{ \begin{matrix} \;1 \qquad a=a_0\\ 0 \qquad \text{else}\end{matrix}\right. $$ is $$ \sum_a |\Psi_a|^2 \delta_{aa_0} = |\Psi_{a_0}|^2\;, $$ which, as usual, is the probability to measure $a_0$ when the state is $|\Psi\rangle$.
If the $a$ are continuous, $|\Psi(a)|^2$ is a probability density rather than a probability.
Update:
To spell out the above explanation again, maybe a bit more clearly:
Regarding why the expectation value of a function $f_{a_0}(a)$ that is zero for all $a$ other than $a_0$ and is $1$ at $a_0$ is the probability of $a_0$:
Consider the general definition of an expectation value (not specific to quantum mechanics, but completely general) for a given discrete set of probability values $P(a)$: $$ \langle f \rangle = \sum_a P(a) f(a)\;. $$ Clearly $$ \langle f_{a_0}\rangle = P(a_0)\;. $$