Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 5 down vote favorite
1

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

$$p=\sqrt{\frac{\hbar}{2m}}(A+B)$$

with

$$[A,B]=1$$

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?

share|improve this question
4  
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 $B=A^\dagger$.) – Emilio Pisanty Mar 18 at 22:48
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 – Nivalth Mar 18 at 22:52
@Nivalth Do you know the commutation rule between x and the A and B? – JKL Mar 19 at 23:08
@JKL No, it is not give but it should be something like $[x,A+B]=Ci\hbar$ for some $C$ – Nivalth Mar 20 at 14:52
@Nivalth, in the above comment, C is easily determined by knowing the canonical [x,P]. – sujeet Mar 24 at 21:38

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.