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?

  • $\begingroup$ Use a different book. $\endgroup$ Mar 5, 2021 at 21:02

3 Answers 3


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.


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