- #1
kcirick
- 54
- 0
Question:
If g(p) can be Taylor expanded in polynomials, then prove that:
[tex] \left[x, g\left(p\right)\right] = i\hbar \frac{dg}{dp} [/tex]
To start, I multiply the wave function [itex] \Psi [/itex] and expand the commutator:
[tex] \left( xg\left(p\right)-g\left(p\right)x \right)\Psi [/tex]
then expand g(p) using Taylor:
[tex] x\left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)\Psi + \left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)x\Psi [/tex]
Then I'm stuck... I'm nowhere near getting [itex]i\hbar[/itex] or [itex]\frac{dg}{dp}[/itex]
Any help would be appreciated. Thank you!
-Rick
If g(p) can be Taylor expanded in polynomials, then prove that:
[tex] \left[x, g\left(p\right)\right] = i\hbar \frac{dg}{dp} [/tex]
To start, I multiply the wave function [itex] \Psi [/itex] and expand the commutator:
[tex] \left( xg\left(p\right)-g\left(p\right)x \right)\Psi [/tex]
then expand g(p) using Taylor:
[tex] x\left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)\Psi + \left(g\left(a\right) + g'\left(a\right)\left(p-a\right)\right)x\Psi [/tex]
Then I'm stuck... I'm nowhere near getting [itex]i\hbar[/itex] or [itex]\frac{dg}{dp}[/itex]
Any help would be appreciated. Thank you!
-Rick