0
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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