I'm reading Dirac's "The principles of quantum mechanics" and I'm having some trouble understanding a specific notation choice: when he introduces the fact that any ket can be expressed as an integral plus a sum of eigenkets of an operator, he writes on p.37 the following

$$|P\rangle=\int|\xi'c\rangle d\xi'+ \sum_r |\xi^rd\rangle\tag{II.25}$$

After that he writes

"the labels $c$ and $d$ being inserted to distinguish the eigenvectors when the eigenvalues $\xi'$ lind $\xi^r$ are equal".

This is not very clear to me, and a particular occasion in which I noticed this was in the demonstration discussed in the following topic: Explanation of Dirac's proof of arbitrary ket being expressible with eigenkets of observable.

tneulinger's answer in that topic is very clear and it explains Dirac's demonstration step by step, but in the end (the last paragraph starting just before the last two equations) something becomes unclear to me. Let me call the integral term in the first equation A, the sum term B and do the same with the terms of the second equation by calling them C and D. I understand why A and B must be non vanishing and finite. However, then he implies that, since B is finite, then C must vanish and this is what confuses me: I would assume this is because B ($\langle\xi^ua|\xi^ub\rangle$) being finite implies $\langle\xi^ub|\xi^ua\rangle$ being finite as well, and this is the same as what's under the integral in C ($\langle\xi^ub|\xi'a\rangle$); this would explain why C vanishes, but I don't understand why those two are the same, and I think I'm having a problem understanding what the indexes and letters refer to.

I hope you can help me understand, also because I guess he'll go on using this notation throughout the whole book.


1 Answer 1


The notation’s purpose is to accommodate for degeneracy. To construct an orthonormal basis, the eigenvalues of the operator are not enough to distinguish the eigenvectors. You need to add some arbitrary labels to distinguish them within an eigenspace. This typically happens when the Hamiltonian in invariant by a non abelian group of symmetries.

The most common case is the group of rotations $SO(3)$. Take for example the Hamiltonian of the hydrogen atom. The resolution of identity gives: $$ |\psi\rangle = \int_{E>0} dE\sum_{l,m}\psi_{E,l,m}|E,l,m\rangle+\sum_n\sum_{l,m}\psi_{n,l,m}|E_n,l,m\rangle $$ The indices $c,d$ correspond to the pair $(l,m)$, and the principal quantum number $n$ corresponds to the $r$. Upon proper nomalization, you have an orthonormal basis: $$ \begin{align} \langle E,l,m|E',l',m'\rangle &= \delta(E-E')\delta_{ll'}\delta_{mm'} \\ \langle E,l,m|E_n,l',m'\rangle &= 0 \\ \langle E_n,l,m|E_{n'},l',m'\rangle &= \delta_{nn'}\delta_{ll'}\delta_{mm'} \end{align} $$

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.