Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I understand from some studies in mathematics, that the generator of translations is given by the operator $\frac{d}{dx}$.

Similarly, I know from quantum mechanics that the momentum operator is $-i\hbar\frac{d}{dx}$.

Therefore, we can see that the momentum operator is the generator of translations, multiplied by $-i\hbar$.

I however, am interested in whether an argument can be made along the lines of "since $\frac{d}{dx}$ is the generator of translations, then the momentum operator must be proportional to $\frac{d}{dx}$". If you could outline such an argument, I believe this will help me understand the physical connection between the generator of translations and the momentum operator in quantum mechanics.

share|improve this question
    
Did you mean to say "rotations", or was that a typo? –  twistor59 Nov 25 '12 at 15:55
    
typo, i meant to say translations. Thanks –  Mew Nov 25 '12 at 15:59
    
Well, if you think about it, Taylor expansion about $p_{0}$ is really $\left.\exp(i\Delta p\partial_{x})f(p)\right|_{p=p_{0}}=f(p_{0}+\Delta p)$ for some constant $\Delta p$, with $\hbar=1$. That's the general idea here –  Alex Nelson Nov 25 '12 at 17:32

1 Answer 1

up vote 5 down vote accepted

In the position representation, the matrix elements (wavefunction) of a momentum eigenstate are $$\langle x | p\rangle = \psi_p(x) = e^{ipx}$$ The wavefunction shifted by a constant finite translation $a$ is $$\psi(x+a)$$ Now the momentum operator is the thing which, acting on the momentum eigenstates, returns the value of the momentum in these states, this is clearly $-i\frac{d}{dx}$.

For our momentum eigenstate, if I spatially shift it by an infinitesimal amount $\epsilon$, it becomes $$\psi(x+\epsilon) = e^{ip(x+\epsilon)} = e^{ip\epsilon}e^{ipx} = (1+i\epsilon p + ...)e^{ipx}$$ i.e. the shift modifies it by an expansion in its momentum value. But if I Taylor expand $\psi(x+\epsilon)$, I get $$\psi(x+\epsilon)= \psi(x)+\epsilon \frac{d}{dx} \psi(x)+... = \psi(x)+i\epsilon(-i\frac{d}{dx})\psi(x)+... $$ So this is consistent since the infinitesimal spatial shift operator $-i\frac{d}{dx}$ is precisely the operator which is pulling out the momentum eigenvalue.

share|improve this answer
    
Thanks this is the best explanation I've seen. –  Mew Nov 26 '12 at 1:22

Your Answer

 
discard

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

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