How $[x,p]=i \hbar$ implies that $x$ and $p$ do not have simultaneous eigenstates?

I am reading Quantum Comuputing Explained by David McMohan. Here is a portion which I am not able to understand.

Let $$x$$ be the position and $$p$$ be the momentum of the particle, then we know that $$[x,p]=i \hbar$$. Above equation tells us, $$x$$ and $$p$$ do not have simultaneous eigenstates.

Can someone please explain how we can conclude from the above equation that $$x$$ and $$p$$ do not have simultaneous eigenstates?

• Use a different book. – Norbert Schuch Mar 5 at 21:02

Observe that since $$[X,P]=i\hbar$$ then $$[X,P]|\psi\rangle\neq 0$$ for all $$|\psi\rangle$$. Now suppose that there exists a simultaneous eigenstate of both $$X$$ and $$P$$, call it $$|\phi\rangle$$. Then we have

$$X|\phi\rangle=x|\phi\rangle,\quad P|\phi\rangle=p|\phi\rangle,\quad x,p\in \mathbb{R}$$

It follows that $$[X,P]|\phi\rangle=XP|\phi\rangle-PX|\phi\rangle=(xp-px)|\phi\rangle=0,$$

but this is contradiction with the fact that $$[X,P]|\psi\rangle \neq 0$$ for all $$|\psi\rangle$$, since we have found one state violating this!

Finally observe that nothing has specifically to do with $$i\hbar$$. If two operators have a commutator which applied to any state gives a non-zero result the same argument shows that they cannot have an eigenstate in common.

Suppose there was a simultaneous eigenstate $$|px\rangle$$ such that $$P|px\rangle=p|px\rangle$$ and $$X|px\rangle=x|px\rangle$$. (Note: I'm using capital $$X$$ and $$P$$ for the operators, and lower case $$x$$ and $$p$$ for the eigenvalues, which are just numbers.) Then $$PX|px\rangle=P\Big(x|px\rangle\Big)=xP|px\rangle=xp|px\rangle$$ and similarly $$XP|px\rangle=xp|px\rangle.$$ (Remember that $$x$$ and $$p$$ commute because they are just numbers.) Therefore $$(XP-PX)|px\rangle=0$$. However, $$[X,P]=i\hbar$$ implies $$(XP-PX)|px\rangle=i\hbar|px\rangle$$. This is a contradiction.

Assume that there would exist some eigenstate $$\psi$$ of both operators, such that $$\hat{x}\,\psi = x\,\psi$$ and $$\hat{p}\,\psi = p\,\psi$$ holds. We then find

$$\hat{x}\,\hat{p} \psi = \hat{x}\, p\, \psi = p \,\hat{x}\,\psi = p\, x\, \psi = \hat{p}\,\hat{x}\, \psi \quad .$$

Consequently $$[\hat{x},\hat{p}]\psi = 0$$, which contradicts that $$[\hat{x},\hat{p}] \Psi = i\hbar \,\Psi$$ for all $$\Psi$$. All in all, there is no state which is an eigenstate of both the momentum and position operator.