# Why position operator is non-degenerated?

In quantum mechanics one can assume position operator $\hat{X}$ must have continuous spectrum, as experiments say it is possible to find a quantum particle at any point of the space. The question is why it is always assumed to be "non-degenerate" for a spinless particle? I mean one could have $$\hat{X}|x,i\rangle=x|x,i\rangle$$ for several $i's$. However this is always taken to be $$\hat{X}|x\rangle=x|x\rangle$$ which leads to the position representation $\langle x|\psi\rangle=\psi(x)$ for the wavefunction of a spinless particle.

• What is $i$ in thiscontext – Slereah Dec 26 '17 at 10:28
• @Slereah, $i$ is just a label to distinguish several eigenvectors $|x,i\rangle$ with the same eigenvalue $x$ (of course in distributional sense as $\hat{X}$ has continuous spectrum). – NessunDorma Dec 26 '17 at 10:41

It isn't assumed to be non degenerate: For example, for the position operator in X-Direction, you have: $$\hat{x} |x, y, z \rangle = x |x, y, z \rangle$$ But aswell: $$\hat{x} |x, \tilde{y}, \tilde{z} \rangle = x |x, \tilde{y}, \tilde{z} \rangle$$
Looking at my example, this smaller Hilbert space could be the space of square integrable functions defined over $\Re$, compared to the space of square integrable functions defined over $\Re^3$, that could be choosen to represent non-degenerated eigenstates of $\hat{x}$, $\hat{y}$, and $\hat{z}$.
Edit: Eigenstates of the position operator are strictly no part of the hilbert space (since in the said representation, they wouldn't be square integrable functions, but instead Dirac $\delta$-distributions). I still think it is clear what I meant there.