# Physical Chemistry: Math

1. Oct 11, 2007

### igaffed

hello all,

I am doing some homework and need some verification on the math:

[Px, Px^2]f= PxPx^2f - Px^2Pxf

Px= -iħd/dx
Px2= -i^2ħ^2d^2/dx^2

-iħd/dx(-i^2ħ^2d^2/dx^2f) - (-i^2ħ^2d^2/dx^2)( -iħd/dxf)

i^3ħ^3d/dx(d^2f/dx^2) - i^3ħ^3 d^2/dx^2(df/dx)

the number following the "^" sign just means raised to that power. Would someone verify if this is correct please? thank you.

2. Oct 13, 2007

### Gokul43201

Staff Emeritus
When you write Px and Px^2, do you mean $P_x$ and $P_x^2$?

1. If the above is true, then there's a very simple value for the commutator that can be got without having to write the momentum operator in the position representation ($-i \hbar \partial/\partial x$). The last line of your calculation simplifies to this final answer.

2. $$\frac{\partial}{ \partial x} \frac{\partial f}{ \partial x} = \frac{\partial ^2f}{ \partial x^2 }$$

and so on...