3
$\begingroup$

What is the value of this difference of anticommutators $$\{x^2,p^2\}-(\{x,p\}^2)/2$$ if the commutator $$[x,p]=i\hbar ~?$$

I have tried and obtained a value $$-3\hbar^2/2 - 2i\hbar px.$$ But the answer given is $-3\hbar^2/2$. I cannot decide whether I am wrong or the answer given is wrong.

$\endgroup$
3

1 Answer 1

2
$\begingroup$

You made a mistake somewhere. My approach was to write $xp=i\hbar +px$ and $px=xp-i\hbar$ on the side of my paper and just substitute away. I first wrote $$\{x^2,p^2\}-\frac{(\{x,p\})^2}{2}=x^2p^2+p^2x^2-\frac{xpxp+xp^2x+px^2p+pxpx}{2}$$ The terms in the fraction are $$xpxp=x^2p^2-i\hbar xp$$ $$xp^2x=x^2p^2-2i\hbar xp$$ $$px^2p=x^2p^2-2i\hbar xp$$ $$pxpx=x^2p^2-2i\hbar xp-i\hbar px$$ So the numerator is $$xpxp+xp^2x+px^2p+pxpx=4x^2p^2-7i\hbar-i\hbar px$$ Now add and subtract $i\hbar xp$ so we get a commutator $$4x^2p^2-8i\hbar+(i\hbar xp-i\hbar px)=4x^2p^2-8i\hbar xp-\hbar^2$$ We thus get $$-x^2p^2+p^2x^2+4i\hbar xp+\frac{\hbar^2}{2}$$ Some more substitutions give $$x^2p^2=2i\hbar xp+2i\hbar px+p^2x^2$$ So all the quadratic terms cancel. We end up with $$2i\hbar xp -2i\hbar px+\frac{\hbar^2}{2}=-2\hbar^2+\frac{\hbar^2}{2}=-\frac{3\hbar^2}{2}$$ as was to be shown.

$\endgroup$
2
  • $\begingroup$ $2i\hbar-2i\hbar px=-2\hbar^2$. How? $\endgroup$ Apr 3, 2015 at 1:05
  • $\begingroup$ @Purushothaman I fixed that typo a while ago. It is supposed to read $2i\hbar xp-2i\hbar px=-2\hbar^2$. $\endgroup$
    – Ryan Unger
    Apr 3, 2015 at 1:07

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