On pg 104 of "Introduction to Quantum Mechanics" by Griffiths, we are asked to find the eigenfunctions of the $x$ operator. Hence, we have to find functions such that $$x f(x)=\lambda f(x)$$ I have used the notation $\lambda$ instead of $y$ because it is less confusing for me. Clearly, any function that satisfies $f(x)=0$ for $x\neq \lambda$ will be an eigenfunction. However, Griffiths claim that the only eigenfunction is $\delta(x-\lambda)$. Why is this true?

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    $\begingroup$ what about the normalization condition? $\endgroup$ – GiorgioP Jun 13 at 21:39
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos Jun 14 at 10:44
  • $\begingroup$ You forgot that a condition for an eigenvector to be such is that its norm must be non-zero. What is the norm of $f(x)=0$ almost everywhere? $\endgroup$ – gented Jun 14 at 12:44
  • $\begingroup$ @gented And what's the norm of $\delta(x-\lambda)$? I wonder what could be objected if the OP had answered "There are none". $\endgroup$ – Elio Fabri Jun 14 at 13:01
  • $\begingroup$ @ElioFabri "There are none" would be the right answer :). However, I was pointing out that $f(x)=0$ almost everywhere is definitely not an eigenfunction (and then one can discuss if the deltas are). $\endgroup$ – gented Jun 14 at 13:14

Here is a formal wisecrack to reassure you: work in momentum space.

Up to normalization constants that do not matter that much for your un-normalizable wave function, consider $$ f_\lambda (x)=\langle x| f_\lambda\rangle= \int dp \langle x|p\rangle \langle p|f_\lambda\rangle = \int \frac{dp}{\sqrt{2\pi \hbar}} e^{ixp/\hbar} \langle p|f_\lambda\rangle ~. $$

Now your strarting point was $$ \hat x | f_\lambda\rangle = \lambda |f_\lambda \rangle , $$ and the momentum representation of $\hat x$ is but $$ \hat x= \int dp ~|p\rangle ( i\hbar \partial_p )\langle p| ~, $$ so that $$ \int dp ~|p\rangle ( i\hbar \partial_p )\langle p|f_\lambda\rangle =\lambda |f_\lambda\rangle.$$

Multiply on the left by $\langle p'|$, collapse the δ-function, and relabel p' to p, to get $$ i \hbar \partial_p \langle p|f_\lambda\rangle= \lambda \langle p|f_\lambda\rangle. $$

You may solve this by $$ \langle p|f_\lambda\rangle \propto e^{-i \lambda p/\hbar } , $$ readily leading to your $$ f_\lambda (x)= \int \frac{dp}{\sqrt{2\pi \hbar}} e^{i(x-\lambda) p/\hbar} \propto \sqrt{\hbar }~~\delta (x-\lambda) ~. $$

Dirac, sublimely slyly, all but does something equivalent in his book, on the basis of his magnificent standard ket, the translationally invariant momentum-space ket. I reckon Griffiths should be more humble in his implicit characterizations there.


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