# Finding the eigenfunctions of the operator $x$

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?

• what about the normalization condition? – GiorgioP Jun 13 at 21:39
• – Cosmas Zachos Jun 14 at 10:44
• 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? – gented Jun 14 at 12:44
• @gented And what's the norm of $\delta(x-\lambda)$? I wonder what could be objected if the OP had answered "There are none". – Elio Fabri Jun 14 at 13:01
• @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). – gented Jun 14 at 13:14

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 $$f_\lambda (x)= \int \frac{dp}{\sqrt{2\pi \hbar}} e^{i(x-\lambda) p/\hbar} \propto \sqrt{\hbar }~~\delta (x-\lambda) ~.$$