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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.

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    $\begingroup$ 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 $\endgroup$ – Alex Nelson Nov 25 '12 at 17:32
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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.

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  • $\begingroup$ Though the above is a simplified version. This wiki has some good foundations to the above. en.wikipedia.org/wiki/… $\endgroup$ – Sai krishna Deep Aug 15 '15 at 10:21
  • $\begingroup$ This answer shows that the assumption $\langle x \vert p \rangle = e^{ixp}$ is equivalent to the statement that $p$ is the generator of translation. (You can start from either one and derive the other one.) But when I read the question, it seems to me like the OP is asking for something much more fundamental than that. $\endgroup$ – sasquires Oct 10 '18 at 21:02

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