2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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

2
$\begingroup$

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

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

leaving

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


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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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