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Kinetic Energy

  • #1
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ok. this is an easy enough thing to prove in one dimension but my question seems to be 3 dimensional and it's causing me some hassle:

show the expectation value of the kinetic energy in a bound state described by the spherically symmetric wavefunction [itex]\psi_T(r)[/itex] may be written

[itex] \langle \hat{T} \rangle = \frac{\hbar^2}{2m} \int \int \int \vline \frac{\partial \psi_T}{\partial r} \vline^2 d^3r[/itex]

so far i have

[itex] \langle \hat{T} \rangle = -\frac{\hbar^2}{2m} \int_V \psi^* \nabla^2 \psi d^3r[/itex]

usually one uses integration by parts here (at least in the 1D case) but that got me lost and so i tried expanding the laplacian in spherical polars but then the terms didn't cancel properly. any ideas?

thanks.
 

Answers and Replies

  • #2
gabbagabbahey
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Combine your two methods. First expand the Laplacian in spherical polars (use the fact the [itex]\psi[/itex] is spherically symmetric), then integrate by parts

EDIT: Better yet, use the vector product rule [itex]\mathbf{\nabla}\cdot(f\textbf{A})=f(\mathbf{\nabla}\cdot\textbf{A})+\textbf{A}\cdot(\mathbf{\nabla}f)[/tex] to integrate by parts and avoid expanding the Laplacian altogether.
 
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  • #3
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ok so i get (for the integral):

[itex] \int_V \psi^* \frac{1}{r^2} \frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial r} ) d^3r[/itex]

so do i set [itex]u=\frac{\psi^*}{r^2} , v=\frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial r} ) [/itex]?
 
  • #4
gabbagabbahey
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ok so i get (for the integral):

[itex] \int_V \psi^* \frac{1}{r^2} \frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial r} ) d^3r[/itex]

so do i set [itex]u=\frac{\psi^*}{r^2} , v=\frac{\partial}{\partial r} ( r^2 \frac{\partial \psi}{\partial r} ) [/itex]?
Not quite. Remember, [itex]d^3r=r^2\sin\theta drd\theta d\phi[/itex], and [itex]\psi[/itex] is spherically symmetric, so it depends only on [itex]r[/itex] and you can replace the partial derivatives with ordinary ones.

You may also want to look at the edited version of my previous post to see an easier method.
 
  • #5
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ok. yeah this is pretty neat. i'm sure i've seen it before as well but i'm not sure how to justify one of the terms vanishing (obviously it's due to it being in a bound state but where do i say this)

for example, i have got to:

[itex]-\frac{\hbar^2}{2m} \int \int \int \left( \vec{\nabla} \cdot \left( \psi^* \vec{\nabla} \psi \right) - \left( \vec{\nabla} \psi \right) \cdot \left( \vec{\nabla} \psi^* \right) \right) d^3r[/itex]

so the second term gives us what we want and we need to get rid of the first term.

presumably i do the integral first and THEN say it goes to zero. so it would be like:

[itex]\int_V \vec{\nabla} \cdot \left( \psi^* \vec{\nabla} \psi \right) dV = \int_S \left( \psi^* \vec{\nabla} \psi \right) \cdot \vec{n} dS[/itex]
what do i say at this point to justify this going to zero?

thanks.
 
  • #6
gabbagabbahey
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Well, the volume integral is over all space, so the surface integral is at infinity (Basically take your surface to be a thin spherical shell of radius [itex]r[/itex] and then take the limit as [itex]r\to\infty[/itex]) What must happen to the wavefunction as [itex]r\to\infty[/itex]?
 
  • #7
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Well, the volume integral is over all space, so the surface integral is at infinity (Basically take your surface to be a thin spherical shell of radius [itex]r[/itex] and then take the limit as [itex]r\to\infty[/itex]) What must happen to the wavefunction as [itex]r\to\infty[/itex]?
ok well grad psi is parallel to the normal so will it not just be

[itex]4 \pi \psi^*(\infty) \vec{\nabla} \psi(\infty)=0[/itex] since [itex]\psi(\infty)=0[/itex]
 
  • #8
gabbagabbahey
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ok well grad psi is parallel to the normal so will it not just be

[itex]4 \pi \psi^*(\infty) \vec{\nabla} \psi(\infty)=0[/itex] since [itex]\psi(\infty)=0[/itex]
Does [itex]4\pi[/itex] have units of area?:wink:

The idea is that, for a bound state, [itex]\psi(r)[/itex] (and hence also [itex]\psi^*[/itex]) must fall off at least as fast as [itex]1/r[/itex] and so [itex]\mathbf{\nabla}\psi[/itex] will fall off at least as fast as [itex]1/r^2[/itex]. When you take the limit as [itex]r\to\infty[/itex] of the surface integral then, you find it will be zero (since it will fall off at least as fast as [itex]1/r[/itex])
 
  • #9
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(i)how do we know [itex]\psi(r)[/itex] falls of as 1/r?

(ii) what was wrong with what i did? i don't understand the problem since i just did the theta and phi integrals on their own since the integrad only depended on r.

(iii) so is there a way to write this down mathematically or would you just have got to the surface integral that i had and then used the "wordy" explanation that you had above?

thanks.
 
  • #10
vela
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(i)how do we know [itex]\psi(r)[/itex] falls of as 1/r?
It has to fall off at least as fast as 1/r otherwise the wavefunction isn't normalizable.
(ii) what was wrong with what i did? i don't understand the problem since i just did the theta and phi integrals on their own since the integrad only depended on r.
You looked at only part of the integrand. Besides the angular part, the dS contributes a factor of r2, so it's not enough just to say the part you wrote goes to zero as r goes to infinity because if it doesn't do it fast enough, the integral still won't converge.
 

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