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.
Asked
Active
Viewed 1,513 times
1
-
3What is your definition, in mathematical terms, of $\hat p$? is the (defining) identity $[\hat x,\hat p]=i\hbar$ allowed? – AccidentalFourierTransform Oct 19 '17 at 17:50
-
1Imagine the term $\hat{x}^n$ in the Taylor expansion of F. Can you find, recursively, it s commutator with $\hat p$? – Cosmas Zachos Oct 19 '17 at 17:51
-
@AccidentalFourierTransform yes that is allowed. – John Doe Oct 19 '17 at 17:59
-
1Possible duplicates: https://physics.stackexchange.com/q/87038/2451 , https://physics.stackexchange.com/q/78222/2451 and links therein. – Qmechanic Oct 19 '17 at 18:13
-
@Qmechanic Yes, I used this link to make one of the steps in my comment on the accepted answer :) – John Doe Oct 19 '17 at 18:16
2 Answers
6
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.
tparker
- 47,418
-
2I 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! – John Doe Oct 19 '17 at 18:10
-
1
-
1
3
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 $-i\frac{\partial\psi}{\partial x}$.
Luke Pritchett
- 7,195
-
1The OP wishes to prove without using the fact that $p$ generates translations... – ZeroTheHero Oct 19 '17 at 20:28
-
@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 $p=-i\hbar \frac\partial{\partial x}$. – John Doe Oct 19 '17 at 22:52
-
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. – Luke Pritchett Oct 20 '17 at 16:09