I want to evaluate $\left[x,\frac{\partial}{\partial x}\right]$ using a Poisson bracket. Can this be done? I have heard that the commutator bracket is $i\hbar$ times the Poisson bracket.I tried to do this , I know A is incorrect, but how please someone explain. I tried to do this , I know A is incorrect, but how please someone explain.

  • $\begingroup$ When you write $\left[x,\frac{\partial}{\partial x}\right]$ are you talking about the QM commutator of position and momentum, or just between these two functions - I ask this because in the question you also have $i\hbar$? $\endgroup$
    – joseph h
    Commented Dec 22, 2020 at 3:59
  • $\begingroup$ I am talking generally, just these two functions. $\endgroup$ Commented Dec 22, 2020 at 4:04
  • $\begingroup$ I think what you are trying to do is impossible. You can only take the Poisson brackets of functions of the state variables $x$ and $p$, not of operators like $\partial/\partial x$. $\endgroup$
    – Buzz
    Commented Dec 22, 2020 at 4:43
  • $\begingroup$ I have just left an answer to what you attempted above. And why it's incorrect. $\endgroup$
    – joseph h
    Commented Dec 22, 2020 at 4:57

1 Answer 1


I'm going to address the attempt you made and point out your error. Why do you have a term with $\large\frac{\partial}{\partial p_x}$? This term is not necessary.

Let's evaluate the commutator by letting it act on a function $f$ so that

$$\left[x,\frac{\partial}{\partial x} \right]f = (x \frac{\partial}{\partial x}f -\frac{\partial}{\partial x}[xf]) $$

Note that the last quantity on the RHS is the derivative of a product, so you must use the product rule. Now

$$\left[x,\frac{\partial}{\partial x} \right]f = (x \frac{\partial f}{\partial x} -\frac{\partial}{\partial x}[xf]) = x \frac{\partial f}{\partial x} - \frac{\partial x}{\partial x}f - \frac{\partial f}{\partial x}x $$

Since $\large \frac{\partial x}{\partial x} = 1$ we are left with

$$\left[x,\frac{\partial}{\partial x} \right]f = -f $$


$$\left[x,\frac{\partial}{\partial x} \right] = -1 $$

There is a way to also do this from building the QM analogue of the Poisson bracket since in your question you include the term $i\hbar$. The Poisson bracket is used in classical mechanics especially in the Hamiltonian formulation.

When you say that

I've heard that the commutator bracket is $i\hbar$ times the Poisson bracket.

you must mean the canonical commutator of position-momentum where

$$x \rightarrow \hat x \\\ \text{and} \\\\ p \rightarrow -i\hbar \frac{\partial}{\partial x} $$

The general relationship between a commutator and the Poisson bracket is given by the relation

$$[\hat a,\hat b]= i \hbar\{a,b\}$$

where the object $\{a,b\}$ is the Poisson bracket. So in the case of the position-momentum commutator,

$$\left[\hat x,\hat p \right] = i \hbar \{ \hat x,\hat p \}$$

But since

$$[\hat{x},\hat{p}]= i \hbar{\bf 1}$$

this would imply that

$$\{q,p\} = \bf 1$$

This is consistent with the result above (but there may be an error with the sign). Please double check. Since the title of your question maybe implies that you want a relation between the two, you can also write

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


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