I'm pretty new to quantum mechanics and after reading the Susskind's book I dived into the Dirac's one. I've managed to understand until this theorem has been enunciated (in my edition at page 32):

<<There are so many eigenkets of $\xi$ that any ket whatever can be expressed as a sum of such eigenkets.>>

I had several doubts about Dirac's procedure therefore I'm trying to ask for someone explining me how have this theorem been proved. For example when Dirac defines $\chi_r (c_r)$, he says that it is the number obtained by replacing $\xi$ in the algebric expression for $\chi_r (\xi)$. This last has been defined as:

$$\chi_r (c_r) = {\phi(\xi) \over {\xi - c_r}}$$

Therefore for the definition of $\phi (\xi)$:

$$\chi_r (\xi) = {{(\xi - c_1)(\xi - c_2) \cdots (\xi-c_n)} \over {\xi - c_r}} \Rightarrow \chi_r (c_r) = {{(c_r - c_1)(c_r - c_2) \cdots (c_r-c_n)} \over {c_r - c_r}}$$

but doing this, there has been obtained an expression divided by zero... Could anyone explain me this proof step by step, please?

P.S.: Thank you very much for your time and excuse me for my English, I'm still practising it.


1 Answer 1


Dirac says that $\phi(\xi)$ has the following factorization: $$ \phi(\xi) = (\xi -c_1)\dots(\xi -c_n) $$

Hence, the factor $(\xi -c_r)$, with $1 \leq r \leq n$, divides $\phi$. Then, he goes on to prove that the $c_i$ are all different. This in turn implies that $\chi_r(c_r) \neq 0$.

  • $\begingroup$ thank you very much for your response... but, can you explain also the rest of the prof please? $\endgroup$
    – Luke__
    Commented Feb 24, 2021 at 14:05
  • $\begingroup$ @LucaMattioni you have to be more specific for me to give you an answer. Is there something in particular that is not clear in that proof? You can edit your question so it can be helpful to other users. $\endgroup$
    – fresh
    Commented Feb 25, 2021 at 19:29

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