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3D harmonic oscillator ground state

by StatusX
Tags: ground, harmonic, oscillator, state
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StatusX
#1
Oct29-06, 03:50 PM
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I've been told (in class, online) that the ground state of the 3D quantum harmonic oscillator, ie:

[tex]\hat H = -\frac{\hbar^2}{2m} \nabla^2 + \frac{1}{2} m \omega^2 r^2 [/tex]

is the state you get by separating variables and picking the ground state in each coordinate, ie:

[tex]\psi(x,y,z) = A e^{-\alpha(x^2+y^2+z^2)}[/tex]

where [itex]\alpha = m \omega/2\hbar[/itex], and this state has energy [itex]3\hbar \omega /2[/itex] (the sum of that from each coordinate).

On the other hand, going back to the schrodinger equation, assume a spherically symmetric solution (ie, l=0) and make the transformation [itex]\chi(r)= r \psi(r)[/itex]. Then the schrodinger equation becomes:

[tex]-\frac{\hbar^2}{2m} \frac{d^2 \chi}{dr^2} + \frac{1}{2} m \omega^2 r^2 \chi =E \chi[/tex]

which is the 1D SHO equation, and so we have the solution:

[tex]\chi(r) = B e^{-\alpha r^2} [/tex]

or:

[tex]\psi(r) = \frac{B}{r} e^{-\alpha r^2} [/tex]

with an energy [itex]\hbar \omega /2[/itex]. In fact, the 1st excited state of the 1D harmonic oscillator, with energy [itex]3\hbar \omega /2[/itex], is [itex]\chi(r)=A r e^{-\alpha r^2} [/itex], and this is just the solution found above by separating variables. So that wasn't the ground state, this new one (which isn't in separated variables form) is. Is this a correct analysis? Is there a general way to determine if a given state is the ground state of a given hamiltonian?
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StatusX
#2
Oct30-06, 11:33 AM
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I just lost credit on a HW for this. Now this is personal. Any ideas?
StatMechGuy
#3
Oct30-06, 11:49 AM
P: 223
I'm not really sure on where your second form of the Schrodinger equation is coming from, but that's just my not knowing laplacians off the top of my head. Also, check on what the actual ground state is for your second equation, because the ground state absolutely is [tex]3/2 \hbar \omega [/tex]

StatusX
#4
Oct30-06, 12:14 PM
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3D harmonic oscillator ground state

You can verify it's an eigenstate directly:

[tex]\nabla^2 \psi = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2 \frac{\partial \psi}{\partial r} \right) + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial \psi}{\partial \theta} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial^2 \psi}{\partial \phi^2} = \frac{1}{r^2}\frac{d}{dr} \left( r^2 \frac{d \psi}{dr} \right)[/tex]

[tex]\psi = \frac{1}{r} e^{-\alpha r^2}[/tex]

[tex] \frac{d \psi}{dr}= \frac{-1}{r^2} e^{-\alpha r^2} + \frac{1}{r} (-2 \alpha r )e^{-\alpha r^2} = -\frac{1}{r^2} ( 1+2 \alpha r^2 ) e^{-\alpha r^2} [/tex]

[tex]\frac{d}{dr} \left( r^2 \frac{d \psi}{dr} \right) = \frac{d}{dr} \left( -(1+2 \alpha r^2 ) e^{-\alpha r^2} \right) = -(4 \alpha r ) e^{-\alpha r^2} - (1+2 \alpha r^2) (-2 \alpha r) e^{-\alpha r^2} = (4 \alpha^2 r^3 - 2 \alpha r ) e^{-\alpha r^2} [/tex]

So:
[tex]\nabla^2 \psi = \frac{1}{r} (4 \alpha^2 r^2 - 2 \alpha) e^{-\alpha r^2} = ( 4 \alpha^2 r^2 - 2 \alpha) \psi [/tex]

[tex] \hat H \psi = -\frac{\hbar^2}{2m} \nabla^2 \psi + \frac{1}{2} m \omega^2 r^2 \psi =\left(-\frac{\hbar^2}{2m} ( 4 \alpha^2 r^2 - 2 \alpha) + \frac{1}{2} m \omega^2 r^2 \right) \psi[/tex]

and with [itex]\alpha = m \omega/2 \hbar [/itex], this reduces to:

[tex] \hat H \psi = \left[ \left( 4\left(-\frac{\hbar^2}{2m} \right) \frac{m^2 \omega^2}{4 \hbar^2}+ \frac{1}{2}m \omega^2 \right) r^2 - 2 \left(-\frac{\hbar^2}{2m}\right) \left(\frac{m \omega}{2 \hbar} \right) \right] \psi = \frac{1}{2} \hbar \omega \psi [/tex]


I think all that algrebra is right. I'm guessing the problem is some subtlety with the divergence at the origin. I'll keep thinking about it.
StatMechGuy
#5
Oct30-06, 02:22 PM
P: 223
My guess would be to check that you have the correct value for your [tex]\alpha[/tex]
dicerandom
#6
Oct30-06, 03:02 PM
P: 308
By assuming that l=0 you've constrained the particle to move only radially and therefore reduced the problem to a purely 1D motion which is why you're getting the 1D result for the ground state energy.
StatusX
#7
Oct30-06, 03:55 PM
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Quote Quote by StatMechGuy
My guess would be to check that you have the correct value for your [tex]\alpha[/tex]
What do you mean? It works, so what could be wrong with it?

Quote Quote by dicerandom
By assuming that l=0 you've constrained the particle to move only radially and therefore reduced the problem to a purely 1D motion which is why you're getting the 1D result for the ground state energy.
But its still a 3D state that has a lower energy than the state that's usually called the ground state of the 3D SHO. Plus, that state has l=0 too (as I mentioned, it corresponds to the first excited state of the equivalent 1D oscillator)
dicerandom
#8
Oct30-06, 05:34 PM
P: 308
Quote Quote by StatusX
But its still a 3D state that has a lower energy than the state that's usually called the ground state of the 3D SHO. Plus, that state has l=0 too (as I mentioned, it corresponds to the first excited state of the equivalent 1D oscillator)
OK, I've been scratching my head over this for a while now and I think I've figured it out. I noticed one problem, in that when you did your first simplification of the SE your term for the radial derivative is incorrect, it should be the same as the first term in the Laplacian you posted. That doesn't change the fact that your [itex]\psi[/itex] with the 1/r dependence satisfies the correct SE though (which you computed, and I duplicated, so either it's right or we're both wrong ).

I believe that the way out of this mess is the fact that the second [itex]\psi[/itex], due to its 1/r dependence, is not square integrable and so it is not a member of the Hilbert space and is not a valid wavefunction.
quasar987
#9
Oct30-06, 10:20 PM
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[tex]\int \psi^2dV = \int B^2\frac{e^{-2\alpha r^2}}{r^2}r^2\sin\theta d\theta\d\phi dr=4\pi B^2\int e^{-2\alpha r^2}dr[/tex]

no?
dicerandom
#10
Oct31-06, 12:18 AM
P: 308
Quote Quote by quasar987
[tex]\int \psi^2dV = \int B^2\frac{e^{-2\alpha r^2}}{r^2}r^2\sin\theta d\theta\d\phi dr=4\pi B^2\int e^{-2\alpha r^2}dr[/tex]

no?
D'oh! I got caught up with the 1D idea and forgot to use the spherical differential volume element. Nevermind, back to the drawing board.
StatusX
#11
Oct31-06, 05:51 PM
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P: 2,566
Ok, I figure it out. The problem is that while:

[tex]\frac{1}{r^2}\frac{d}{dr}\left( r^2 \frac{d}{dr} \left( \frac{1}{r} \right) \right) = 0[/tex]

for all [itex]r \neq 0[/itex], we can't just say the Laplacian is zero because of the behavior near the origin. In fact, from electrostatics we know:

[tex]\nabla^2 \frac{1}{r} = - 4 \pi \delta^3(\vec r) [/tex]

So adding this extra term in, you get:

[tex]\hat H \psi = \frac{1}{2} \hbar \omega \psi + \frac{\hbar^2}{2m} 4 \pi r \delta^3(\vec r) \psi [/tex]

So [itex]\psi[/tex] is not an eigenstate, and in fact:

[tex]<\psi | \hat H| \psi> = <\psi | \frac{1}{2} \hbar \omega + \frac{\hbar^2}{2m} 4 \pi r \delta^3(\vec r)| \psi> = \frac{1}{2} \hbar \omega + \frac{4 \pi \hbar^2}{2m} \int_{R^3} \psi^2 r \delta^3(\vec r) d^3 \vec r = \infty[/tex]

because [itex] r \psi^2 \propto 1/r[/itex] near the origin.

Still, this example worries me, because how can we be sure that there are no states with energy less than the [itex]3 \hbar \omega/2[/itex] state? I can see this is the lowest state in separated variables form, but why can't there be a lower state not in this form?
dicerandom
#12
Oct31-06, 06:38 PM
P: 308
Quote Quote by StatusX
Still, this example worries me, because how can we be sure that there are no states with energy less than the [itex]3 \hbar \omega/2[/itex] state? I can see this is the lowest state in separated variables form, but why can't there be a lower state not in this form?
You can prove it using the raising and lowering operators and the Dirac formalism without ever having to write down a wave function. Wikipedia covers it in a section of their Quantum harmonic oscillator article, for more detail there's also some lecture notes (PDF) available from MIT's OpenCourseWare for their 1st quarter QM. The relevant section of the lecture notes starts on page 7 of that PDF.


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