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See the following relation:

$$i\hbar \frac{\partial \hat{F}}{\partial \hat {x}} = [\hat{F}, \hat{p}],$$

assuming that $\hat{x}$ and $ \hat{p}$ are the position and momentum matrices and $\hat{F}(\hat{x}, \hat{p})$ an arbitrary function of these in Heisenberg matrix mechanics.

Is it possible to prove this equation using only the canonical commutation relation $$[\hat{x}, \hat{p}] = i\hbar \hat{I}$$ and matrix algebra? (I know how to prove it using the momentum operator definition $$\hat{p} = -i\hbar\frac{\partial}{\partial x}$$ but I would like to know if there is another way)

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Consider the commutation relation $[\partial_x, x] = ?$. Adding a test function we would have $$\partial_x (x \Psi) - x \partial_x \Psi = \Psi + x \partial_x \Psi - x \partial_x \Psi = \Psi,$$hence $[\partial_x, x] = 1.$ The canonical commutation relation is therefore suggestive that $\hat p = -i\hbar \partial_x + f(\hat x)$ directly.

One can also imagine that we write out some sort of multinomial expansion, $$\hat F = F(\hat x, \hat p) = \sum_{mn} F_{mn} ~\hat x^m ~\hat p^n.$$ The commutator here is,$$[\hat F, \hat p] = \sum_{mn} F_{mn}~\big(\hat x^m ~\hat p^{n+1} - \hat p~\hat x^m~\hat p^n\big)= \sum_{mn} F_{mn}~[\hat x^m, \hat p]~\hat p^n.$$ Working out this commutator is quite straightforward: $$ \begin{align} [\hat x^m, \hat p] &= \hat x^m~\hat p - \hat p~\hat x^m\\ &=\hat x~\hat x^{m-1}~\hat p - [\hat p, \hat x]~\hat x^{m-1} - \hat x~\hat p~\hat x^{m-1} \\ &= \hat x [\hat x^{m-1},\hat p] - [\hat p, \hat x]\hat x^{m-1}. \end{align}$$We can thereby prove by induction,$$[\hat x^m, p] = i\hbar ~m~\hat x^{m-1}.$$Plugging this in above gives $$ i\hbar~\sum_{mn} F_{mn} ~m~\hat x^{m-1}~\hat p^n = i\hbar\frac{\partial}{\partial \hat x} \sum_{mn} F_{mn}~\hat x^m~\hat p^n = i\hbar\frac{\partial\hat F}{\partial \hat x}.$$

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  • $\begingroup$ Thank you very much! And I guess, you can extend that to series in x and p as well! $\endgroup$
    – John Doe
    Sep 11 '17 at 22:10
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In classical Hamiltonian mechanics, you would have

$$\frac{\partial F}{\partial x} = \{F,p\}$$

where $\{\ \cdot\ ,\ \cdot\ \}$ are the Poisson brackets. Apply canonical quantisation à la Dirac and you're done.

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  • $\begingroup$ Yes, you are totally right, but can you prove it using only the ccr? (I am reading a book about quantum mechanics which said that from the ccr the relation can be proven) $\endgroup$
    – John Doe
    Sep 11 '17 at 19:54
  • $\begingroup$ Well the ccr comes from Dirac quantisation, so indirectly you are also using them. Another way of proving this that I can think of is to somehow start with the knowledge that $U_\delta=e^{i\delta p}$ translates by $\delta$ (i.e. $p$ is the generator of translations). Then $U_\delta^* F(x) U_\delta = F(x+\delta)$. By taking the limit $\delta \to 0$ and equating both sides at each order, you will recover the sought result. $\endgroup$
    – Phoenix87
    Sep 11 '17 at 20:03

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