Commutativity Equation Of Hamilton and Position Operators

[buraqenigma]

How can we show $$\left[\hat{H},\hat{x}\right]=\frac{-i\hbar}{m} \hat{p_{x}}$$ ?

 

[buraqenigma]

is this regular ?

$$\left[\hat{p_{x}}^2,\hat{x}\right]=\left[\hat{p_{x}},\hat{x}\right]\hat{p_{x}}+\hat{p_{x}}\left[\hat{x},\hat{p_{x}}\right]$$ ( from leibniz rule)

 

[cristo]

How can we show $$\left[\hat{H},\hat{x}\right]=\frac{-i\hbar}{m} \hat{p_{x}}$$ ?

Well, do you know the form of the Hamiltonian?

 

[genneth]

$$[p^2,x] = ppx - xpp = p[p,x]-pxp - [x,p]p + pxp$$

 

[cristo]

$$\left[\hat{p_{x}}^2,\hat{x}\right]=\left[\hat{p_{x}},\hat{x}\right]\hat{p_{x}}+\hat{p_{x}}\left[\hat{x},\hat{p_{x}}\right]$$ ( from leibniz rule)

Yes, that is correct.

 

[buraqenigma]

if we know $$\left[\hat{p}_x,\hat{x}\right]=-i\hbar$$ we can show this equation. Thanks my friends for your helps.

 

[dextercioby]

So how does the Hamiltonian look like ?

 

[Reshma]

if we know $$\left[\hat{p}_x,\hat{x}\right]=-i\hbar$$ we can show this equation. Thanks my friends for your helps.

Yes, but you will have to start with the Hamiltonian operator first,
$$\hat H = {\hat {p}^2 \over {2m}} + V(\vec r)}$$
and arrive at your result.