Average momentum squared of Psi(100) of hydrogen atom

[phys_chemist]

Homework Statement

Calculate <p²> for ψ₁₀₀ of the hydrogen atom

Homework Equations

ψ₁₀₀ = 1/(√pi) (1/a₀)^3/2 e^-r/a<sub>0

0</sub>∫^∞ r ⁿ e^{-B r}dr = n!/Bⁿ⁺¹

p² = -hbar ∇² = -hbar² (r² d²/dr² +2 r d/dr) (ψ does not depend on ø or θ)

The Attempt at a Solution

<p²> = ∫ψ*(p²ψ)dV
∫dV = 4pi₀∫^∞r²dr

<p²> = ∫1/(√pi) (1/a₀)^3/2 e^-r/a₀ (-hbar² (r² d²/dr² +2 r d/dr) 1/(√pi) (1/a₀)^3/2 e^-r/a₀)dV

<p²> = -4 hbar²/ a₀³ ∫r² e^-r/a₀ ( (r/a₀)² - 2r/a₀ ) e^-r/a₀ dr

<p²> = -4 hbar²/ a₀³ ∫e^-2r/a₀ ( r⁴/a₀² - 2r³/a₀ ) drThe problem that I am running into is that I am calculating the integral to be 0:

B= 2/a₀ n₁= 4 n₂=3

∫e^-2r/a₀ ( r⁴/a₀² - 2r³/a₀ ) dr = (4!/ (a₀² (2/a₀)⁵) - (2*3!/ (a₀ (2/a₀)⁴ = (24a₀³/32 - 12a₀³/16) =0

Am I doing something wrong, because I don't think <p²> would be 0 (that would indicate that the momentum is always 0)?

 

[Simon Bridge]

First thing I noticed:

p² = -hbar ∇2 = -hbar² (r² d²/dr² +2 r d/dr)

Going by the laplace operator in spherical-polar coordinates:
re: http://en.wikipedia.org/wiki/Laplace_operator#Three_dimensions $$\nabla^2\psi(r)=\frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial}{\partial r}\psi(r)$$

 

[phys_chemist]

First thing I noticed:

Going by the laplace operator in spherical-polar coordinates:
re: http://en.wikipedia.org/wiki/Laplace_operator#Three_dimensions $$\nabla^2\psi(r)=\frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial}{\partial r}\psi(r)$$

Thank you. I left out the 1/r^2 in the laplace operator. When I redid the problem after fixing the operator, the answer I am now getting makes sense.

 

[Simon Bridge]

Easy to do - there are so many examples where you don't need it.