0
$\begingroup$

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)

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

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 $$ i\hbar~\sum_{mn} F_{mn} ~m~\hat x^{m-1}~\hat p^n = i\hbar\frac{\partial}{\partial \hat x} \sum_{mn} F_{mn}~\hat x^m~\hat p^n = i\hbar\frac{\partial\hat F}{\partial \hat x}.$$

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you very much! And I guess, you can extend that to series in x and p as well! $\endgroup$
    – John Doe
    Sep 11, 2017 at 22:10
1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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, 2017 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, 2017 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.