# Momentum as Generator of Translations

+ 4 like - 0 dislike
1844 views

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.

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user Mew
Did you mean to say "rotations", or was that a typo?

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user twistor59
typo, i meant to say translations. Thanks

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user Mew
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

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user Alex Nelson

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user twistor59
answered Nov 25, 2012 by (2,500 points)
Thanks this is the best explanation I've seen.

This post imported from StackExchange Physics at 2014-03-22 17:10 (UCT), posted by SE-user Mew

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOv$\varnothing$rflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.