
#1
Oct2906, 03:50 PM

HW Helper
P: 2,566

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? 



#2
Oct3006, 11:33 AM

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P: 2,566

I just lost credit on a HW for this. Now this is personal. Any ideas?




#3
Oct3006, 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]




#4
Oct3006, 12:14 PM

HW Helper
P: 2,566

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. 



#5
Oct3006, 02:22 PM

P: 223

My guess would be to check that you have the correct value for your [tex]\alpha[/tex]




#6
Oct3006, 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.




#7
Oct3006, 03:55 PM

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#8
Oct3006, 05:34 PM

P: 308

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
Oct3006, 10:20 PM

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PF Gold
P: 4,768

[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? 



#10
Oct3106, 12:18 AM

P: 308





#11
Oct3106, 05:51 PM

HW Helper
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? 



#12
Oct3106, 06:38 PM

P: 308




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