2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Use a different book. $\endgroup$ – Norbert Schuch Mar 5 at 21:02
9
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.