3D harmonic oscillator ground state

In summary, the ground state of the 3D quantum harmonic oscillator is the state you get by separating variables and picking the ground state in each coordinate.
  • #1
StatusX
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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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  • #2
I just lost credit on a HW for this. Now this is personal. Any ideas?
 
  • #3
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]
 
  • #4
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.
 
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  • #5
My guess would be to check that you have the correct value for your [tex]\alpha[/tex]
 
  • #6
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.
 
  • #7
StatMechGuy said:
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?

dicerandom said:
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)
 
  • #8
StatusX said:
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 :wink: ).

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.
 
  • #9
[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?
 
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  • #10
quasar987 said:
[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.
 
  • #11
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?
 
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  • #12
StatusX said:
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 http://ocw.mit.edu/NR/rdonlyres/9991BFAD-67AB-458F-B009-FFBC36F665A7/0/lecture18.pdf (PDF) available from MIT's OpenCourseWare for their http://ocw.mit.edu/OcwWeb/Physics/8-04Spring-2006/LectureNotes/index.htm. The relevant section of the lecture notes starts on page 7 of that PDF.
 
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Related to 3D harmonic oscillator ground state

1. What is a 3D harmonic oscillator ground state?

The 3D harmonic oscillator ground state refers to the lowest energy state of a particle that is confined in a three-dimensional harmonic potential well. This potential well can be thought of as a "bowl" or "trough" that the particle is trapped in and can oscillate back and forth within.

2. How is the ground state of a 3D harmonic oscillator calculated?

The ground state of a 3D harmonic oscillator can be calculated using the Schrödinger equation, which is a mathematical equation that describes the behavior of quantum particles. The solution to this equation yields the energy levels and corresponding wave functions, with the ground state being the lowest energy level.

3. What is the significance of the ground state in a 3D harmonic oscillator?

The ground state is significant because it represents the most stable and lowest energy state of the system. This means that the particle is in its most probable state and will not spontaneously transition to a higher energy state without external influence.

4. How does the ground state of a 3D harmonic oscillator differ from higher energy states?

The ground state of a 3D harmonic oscillator has the lowest energy and is associated with the most compact and stable wave function. Higher energy states have more nodes and are less likely to be occupied by the particle, as they require more energy for the particle to transition to.

5. How does the ground state of a 3D harmonic oscillator relate to real-life systems?

The 3D harmonic oscillator is a simplified model that can be used to describe many physical systems, such as atoms, molecules, and solid materials. While these systems may have more complex potentials, the concept of a ground state remains relevant as the lowest energy and most stable state of the system.

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