# Momentum as Generator of Translations

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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
asked Nov 25, 2012
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

## 1 Answer

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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

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