# Explanation of Dirac's proof of arbitrary ket being expressible with eigenkets of observable

In P.A.M. Dirac's The Principles of Quantum Mechanics, Chapter 10 (Observables), pp. 40, at the end of the chapter there is a proof that I don't understand at all.

The proof in question is on pp. 50 using the pdf reader's numbering and on pp. 40 using the books original numbering. I'm curious about the part starting with "We can now see..." until the end of the chapter.

Would somebody be so kind to explain me what happens there?

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Dirac being opaque and hard to follow? Well I never...

In Chapter 10 Dirac argues on physical grounds that the eigenkets of an observable must form a complete set. His argument goes that say we have an observable with eigenstates $|\varepsilon\rangle$ and some general state $|P\rangle$. Then I can write $|P\rangle$ as $$|P\rangle = \sum a_\varepsilon|\varepsilon\rangle + \sum b_\gamma |\gamma\rangle$$ Where $|\gamma\rangle$ are some state that cannot be written as a combination of the $|\varepsilon\rangle$s. Unlike Dirac I have normalized my eigenstates so that $\langle \varepsilon|\varepsilon\rangle = 1$ by pulling out a complex factor $a_\gamma$ and have not bothered to write down the integral, as the argument is essentially the same and it just complicates things. Now if we make a measurement of $\varepsilon$ for this state there is a probability $|b_\gamma|^2$ that we will find the system in each of the states $|\gamma\rangle$. But $|\gamma\rangle$ does not correspond to any allowed value of $\varepsilon$, so clearly does not make sense for the result of a measurement of $\varepsilon$! Therefore $b_\gamma = 0$ for all $\gamma$ and the $|\varepsilon\rangle$s form a complete set.

Edit: It is also worth noting that requiring an operator to have a complete set of real, orthogonal, eigenvalues is equivalent to requiring the operator to be Hermitian, which is how this requirement on observables is normally stated.

In the part you are asking about he is trying to prove that this way of writing $|P\rangle = \sum a_\varepsilon|\varepsilon\rangle$ is unique, i.e. for a given state $|P\rangle$ there is only one possible set of coefficients $a_\varepsilon$. Now we have from orthogonality that $$\langle\varepsilon^\prime|\varepsilon\rangle = \left\{\begin{array}{lc} 1& \varepsilon = \varepsilon^\prime \\ 0 & \varepsilon\ne\varepsilon^\prime\end{array}\right.$$ Let us say that there is another set of coefficients $a_\varepsilon^\prime$ such that $|P\rangle = \sum a_\varepsilon^\prime|\varepsilon\rangle$. Subtracting these two expressions for $|P\rangle$ we find $$0 = \sum (a_\varepsilon - a_\varepsilon^\prime)|\varepsilon\rangle$$ Multiplying by $\langle\varepsilon^\prime|$ we find that all the terms go to 0 except $$0 = (a_{\varepsilon^\prime}-a_{\varepsilon^\prime}^\prime)$$ So it terns out $a_{\varepsilon^\prime} = a_{\varepsilon^\prime}^\prime$ and since this must be true for each $\varepsilon$, the expression for $|P\rangle$ was unique.

With the integral included the discrete coefficients are replaced by functions in the integral $a(\varepsilon)$ and the normalisation is $\langle\varepsilon^\prime|\varepsilon\rangle = \delta(\varepsilon^\prime - \varepsilon)$ but otherwise the argument is unchanged.

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Good answer with physical intuition. One could, in more modern terms, mention that the completeness of the eigenstates is the reason why we demand observables to be self-adjoint. – ACuriousMind Jul 14 '14 at 13:47
@ACuriousMind Wasn't that due to the requirement that the eigenvalues be real? – Danu Jul 14 '14 at 14:55
@Danu: It's a side effect. But if the operator is not self-adjoint, you have no guarantee that it will be diagonalizable at all, even by states with complex eigenvalues. You really need self-adjointness to be sure that the eigenvectors form an orthonormal basis of the Hilbert space. – ACuriousMind Jul 14 '14 at 15:04
@ACuriousMind Forgive my mathematical pedantry, but self-adjointness alone is still not a sufficient condition to have an orthonormal basis of eigenvectors...you need also that the operator is either compact or with compact resolvent. – yuggib Jul 15 '14 at 21:10
@yuggib: Forgiven and duly noted. :) – ACuriousMind Jul 15 '14 at 21:13