I have been able to prove this relation by using a certain method, but it uses the fact that $$\hat p=-i\hbar\frac\partial{\partial x},\tag{1}$$ which is a relation I have avoided so far, so I wish to prove it without using that ($\hat p$ has simply been defined as the generator of spatial translation). Is it possible to prove that $$[\hat p,F(\hat x)]=-i\hbar\frac{\partial F(\hat x)}{\partial x}.\tag{2}$$ without using this relation? I can't see how to get even the first step.
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3$\begingroup$ What is your definition, in mathematical terms, of $\hat p$? is the (defining) identity $[\hat x,\hat p]=i\hbar$ allowed? $\endgroup$– AccidentalFourierTransformCommented Oct 19, 2017 at 17:50
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1$\begingroup$ Imagine the term $\hat{x}^n$ in the Taylor expansion of F. Can you find, recursively, it s commutator with $\hat p$? $\endgroup$– Cosmas ZachosCommented Oct 19, 2017 at 17:51
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$\begingroup$ @AccidentalFourierTransform yes that is allowed. $\endgroup$– John DoeCommented Oct 19, 2017 at 17:59
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/87038/2451 , physics.stackexchange.com/q/78222/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Oct 19, 2017 at 18:13
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$\begingroup$ @Qmechanic Yes, I used this link to make one of the steps in my comment on the accepted answer :) $\endgroup$– John DoeCommented Oct 19, 2017 at 18:16
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2 Answers
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Write $F(\hat{x})$ as a power series with arbitrary coefficients. Use linearity to express $[\hat{p}, F(\hat{x})]$ as a linear combination of commutators of the form $[\hat{p}, \hat{x}^n]$. You can calculate the latter commutator using induction and the identities $[A, BC] = B [A, C] + [A, B]C$ and $[\hat{x}, \hat{p}] = i \hbar$. Then recombine the power series into a single expression.
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2$\begingroup$ I did it: $$F(x)=\sum_{j=0}^\infty f_j x^j;$$$$\begin{align} [p,F(x)]&=\sum_{j=0}^\infty f_j[p,x^j]\\ &=-i\hbar\sum_{j=0}^\infty f_j jx^{j-1}\\ &=-i\hbar\sum_{j=0}^\infty f_j \frac\partial{\partial x} x^j\\ &=-i\hbar \frac\partial{\partial x} \sum_{j=0}^\infty f_jx^j\\ &=-i\hbar \frac{\partial F(x)}{\partial x}\end{align}$$ Thanks! $\endgroup$– John DoeCommented Oct 19, 2017 at 18:10
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1$\begingroup$ How did you found that $[p,x^j] =-i \hbar jx^{j-1}$? $\endgroup$ Commented Aug 6, 2018 at 3:13
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1$\begingroup$ I have found the answer to my previous question here. $\endgroup$ Commented Aug 6, 2018 at 3:31
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You can define the momentum operator as the generator of translations. Abstractly, the translation operator $T_a = \exp(-ia\hat{p})$ acts on eigenstates of position via $T_a|x\rangle = |x+a\rangle$. Hence, if $\langle x|\psi\rangle = \psi(x)$ then $\langle x| \exp(ia\hat{p})|\psi\rangle = \psi(x+a)$.
Expanding the exponential and the result in powers of $a$ tells us the action of the generator: $$\langle x|(I+ia\hat{p})|\psi\rangle = \psi(x+a) = \psi(x)+a\frac{\partial\psi}{\partial x}$$ $$\Rightarrow \langle x| \hat{p}|\psi\rangle = -i\frac{\partial\psi}{\partial x}$$
In other words the $\hat{p}$ operator acting on a state with wavefunction $\psi$ yields a state with wavefunction $-i\frac{\partial\psi}{\partial x}$.
answered Oct 19, 2017 at 17:59
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1$\begingroup$ The OP wishes to prove without using the fact that $p$ generates translations... $\endgroup$ Commented Oct 19, 2017 at 20:28
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$\begingroup$ @ZeroTheHero I think you may have misread what I wrote - we have defined $p$ in this way. I just didn't want to use the fact that $p=-i\hbar \frac\partial{\partial x}$. $\endgroup$– John DoeCommented Oct 19, 2017 at 22:52
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$\begingroup$ I have edited my answer to make it slightly clearer the distinction between the abstract generator of translations and the differential operator $-i\partial_x$. In any case, the statement that $[x,p]=i\hbar$ is exactly equivalent to the statement that $p$ generates translations. $\endgroup$ Commented Oct 20, 2017 at 16:09