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I got stuck at the following bit from Dirac's book. In page 47, he introduces the probability concepts for the observables. He says that $\langle x|f(\zeta)|x\rangle$ denotes the average value of the function $f$ of the observable $\zeta$ at state (normalised) $|x\rangle$. He then defines: let $f(\zeta)$ be 1 iff $\zeta = a$ and zero otherwise. This is written as: $f(\zeta) = \delta_{\zeta a}$. He says now that $P_a = \langle x|\delta_{\zeta a}|x\rangle$ denotes the probability of $\zeta$ having the value $a$. Now, he states that if $a$ is not an eigenvalue of $\zeta$, $\delta_{\zeta a}$ multiplied to any eigenket of $\zeta$ is 0 and hence $\delta_{\zeta a} = 0, \implies P_a = 0$.

MY QUESTION: How is a $\delta_{\zeta a}$ ($a$ is not an eigenvalue) multiplied to an eigenket always zero? For an eigenvalue $b$, it is a solution of $b|x\rangle = \zeta |x\rangle$ for some eigenket $|x\rangle$. I don't quite follow how the previous statement follows from the definition of an eigenvalue. Can someone just provide some hints only?

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To be clear, the statement is the following:

Let $|\psi_b\rangle$ be an eigenket of the operator $\zeta$ with eigenvalue $b$ - i.e. $\zeta |\psi_b\rangle = b |\psi_b\rangle$. Define an operator $\delta_{\zeta,a}$ such that $\delta_{\zeta,a}|\psi_b\rangle = \delta_{b,a}|\psi_b\rangle$, where $$\delta_{b,a} = \cases{ 1 & $a=b$ \\ 0 & $a\neq b$}$$ Then, if $a$ is not an eigenvalue of $\zeta$, it follows that $\delta_{\zeta,a}|\Psi\rangle = 0$ for any arbitrary ket $|\Psi\rangle$.

If $\zeta$ is an observable, then we can expand any arbitrary ket in a basis formed by its orthonormal eigenkets: $$|\Psi\rangle = \sum_b c_b|\psi_b\rangle$$ where the expansion coefficients are $c_b = \langle \psi_b | \Psi\rangle$. Applying the delta operator, $$ \delta_{\zeta,a}|\Psi\rangle = \sum_b c_b\cdot \delta_{\zeta,a}|\psi_b\rangle = \sum_b c_b \cdot \delta_{b,a}|\psi_b\rangle$$

However, because $a\neq b$ for all $b$ (because $a$ is not an eigenvalue of $\zeta$), all of those $\delta_{b,a}$'s vanish, and therefore $$ \delta_{\zeta,a}|\Psi\rangle = 0$$

I implicitly assumed that the eigenvalue spectrum of $\zeta$ is discrete - this can be generalized to the continuous case by replacing the sums with integrals and the same idea applies.

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  • $\begingroup$ Ok thanks. I didn't realise the relation between the delta function of the observable and it's eigenvalues. $\endgroup$ – Lelouch Oct 4 '17 at 7:16

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