This is problem $1.2$ of Molecular Quantum Mechanics by Atkins, 4th edition. I'm given the momentum operator




and I need to find $x$ in this particular representation. A suitable solution, inspired by the formalism of the quantum mechanical harmonic oscillator is:

$$x=-i\sqrt{\frac{m\hbar}{2}}(A-B) $$

Checking it:

$$ [x,p]=-i\sqrt{\frac{m\hbar}{2}}\sqrt{\frac{\hbar}{2m}}\left( [A+B,A-B]\right)=-\frac{i\hbar}{2}\left( [A,A]-[A,B]+[B,A]-[B,B]\right)=\frac{i\hbar}{2}2[A,B]=i\hbar$$

How to derive (perhaps algebraically) another particular solution for $x$ without having the harmonic oscillator hint?

  • 7
    $\begingroup$ Have you tried some arbitrary linear combination $x=aA+bB$? Requiring $[x,p]=i\hbar$ then requires $a-b=i\sqrt{2m\hbar}$. If you then want a specific solution to come out uniquely you'd need a second condition. (For the harmonic oscillator, you also ask that $x$ be hermitian for $B=A^\dagger$.) $\endgroup$ – Emilio Pisanty Mar 18 '13 at 22:48
  • $\begingroup$ Hi Emilio thanks for the comment. I was thinking (and maybe is asking too much) in not forcing $x$ to be an specific combination of $A$ and $B$, say, a linear one. The specific solution in my question is because we can always add an arbitrary function of $p$ to $x$ and the CCR $[x,p]=i\hbar$ will still hold. So inserting $x$ and $p$ into the CCR and then derive some general form for $x$ given $[A,B]=1$. Also, thanks for the hermitian requeriment reminder $\endgroup$ – Jorge Lavín Mar 18 '13 at 22:52
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    $\begingroup$ @Nivalth Do you know the commutation rule between x and the A and B? $\endgroup$ – JKL Mar 19 '13 at 23:08
  • $\begingroup$ @JKL No, it is not give but it should be something like $[x,A+B]=Ci\hbar$ for some $C$ $\endgroup$ – Jorge Lavín Mar 20 '13 at 14:52
  • $\begingroup$ @Nivalth, in the above comment, C is easily determined by knowing the canonical [x,P]. $\endgroup$ – sujeet Mar 24 '13 at 21:38

As was said in the comments, my attempt to a solution is to decompose $\hat{x}$ as a linear combination $\alpha \hat{A} + \beta \hat{B}$. This seems to be general, because if $\hat{p}$ is a linear combination of $\hat{A}$ and $\hat{B}$, then the position will be too, since it's the inverse Fourier transform of $\hat{p}$ and such transformation is linear. I'd them use $[\alpha \hat{A} + \beta \hat{B},\hat{p}] = i \hbar$ and find that (as Emilio said) $\alpha - \beta = i\sqrt{2\pi m}$. It is, with the current information we have, impossible to determine individually $\alpha$ and $\beta$, but to provide the answer you're looking for we can choose $\alpha = -i\sqrt{\frac{\pi m}{2}} = -\beta$.

Sorry I didn't write a comment. I've just start using SE and have not enough rep points. Consider this attempt as a comment and erase it, if you find better.


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