# Eigenstates of position and momentum operators in QM

In Griffiths pages 103-105 "Introduction to Quantum Mechanics" 2nd editiion he states that the eigenfunctions of the position and momentum operators are $$g_y(x) = \delta(x-y)$$ where the eigenvalue equation is $$xg_y(x) = yg_y(x)$$

and for the momentum operator it is $$f_p(x) = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{i p x}{\hbar}}$$ where the eigenvalue equation is $$\frac{\hbar}{i}\frac{d}{dx}f_p(x) = p f_p(x).$$

In other literature, this is not stated, why is this the case? Are these in fact the eigenfunctions of the position and momentum operators?

• Yes, they are the (not normalizable) eigenstates of the operators. And some things are not stated somewhere because not every book lists all true statements. – Luboš Motl May 24 '16 at 11:34
• @LubošMotl Yeah but that seems like quite an important omission don't you think?...Especially for popular introductory literature like "Modern Quantum Mechanics" 2nd edition by N.Zettili. Griffiths also generally tries to avoid very advanced discussions. – user100411 May 24 '16 at 11:38
• Maybe I agree (I don't know which book omits this thing so I can't feel "involved") - the plane wave is really an important wave function, the free particle is an important physics problem in QM - but there are lots of other operators with their eigenstates and eigenvalues, too. From some fundamental viewpoint, any operator is as good as any other. – Luboš Motl May 24 '16 at 11:44
• "In other literature, this is not stated, why is this the case?" - How are we supposed to know what choices the textbook authors made of whatever book you're referring to here? – ACuriousMind May 24 '16 at 16:07
• There needs to be a canonical reference on how to treat $|x\rangle$ and $|p\rangle$ properly -- i.e., with rigged Hilbert spaces -- so that whenever books sweep the details under the rug, they can at least tell you where the rug is if you want to go look for them. – Robin Ekman May 24 '16 at 16:20

These relations are found in every book on QM, but the usual notation is $$X|x\rangle=x|x\rangle$$ and $$P|p\rangle=p|p\rangle$$
To go from these equations to the ones you've written, you just have to project them into the position basis $|x'\rangle$ (and use $\langle x'|x\rangle=\delta(x-x')$ and $\langle x'|p\rangle\sim\exp[ipx]$).
• It might be worth it to add a caveat about $\lvert x\rangle$ and $\lvert p \rangle$ not actually being states... – ACuriousMind May 24 '16 at 16:08