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.

 

[paranormal]

Your calculation is correct, but there is a subtle error in your reasoning. The momentum operator is a differential operator, not a constant. This means that it acts on the wavefunction to produce a new wavefunction, which is then integrated over to calculate the expectation value. In other words, you cannot simply pull the momentum operator out of the integral and treat it as a constant.

The correct calculation would be:

<p2> = ∫ψ*(p2ψ)dV
= ∫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
= -4 hbar2/ a03 ∫r2 e-r/a0 ( (r/a0)2 - 2r/a0 ) e-r/a0 dr
= -4 hbar2/ a03 ∫e-2r/a0 ( r4/a02 - 2r3/a0 ) dr

Now, we can use integration by parts to evaluate the integral:

= -4 hbar2/ a03 [ e-2r/a0 (r4/a02 - 2r3/a0) |∞0 - ∫e-2r/a0 (-4r3/a03 + 6r2/a02) dr ]
= -4 hbar2/ a03 [ 0 - (0 - (4*3!/ (a03 (2/a0)4) + 6*2!/ (a02 (2/a0)3)) ]
= -4 hbar2/ a03 [ 0 - (0 - (6a03/32 + 12a03/16)) ]
= 3 hbar2/ a03

So, the expectation value of p2 for ψ100 of the hydrogen atom is 3 hbar2/ a03.