What is the Normalized Ground State Energy of a 3-D Harmonic Oscillator?

[pprie003]

Homework Statement

What is the normalized ground state energy for the 3-D Harmonic Oscillator

Homework Equations

V(r) = 1/2m(w^2)(r^2)

The Attempt at a Solution

I started with the wave fn in spherical coordinates, and have tried using sep of variables, but keep getting stuck when trying to simplify.

 

[blue_leaf77]

Welcome in PF!
Ok concerning your first problem, have you tried what you can end up with when you view the problem in Cartesian coordinates instead?

 

[theodoros.mihos]

See this

 

[vela]

Homework Statement

What is the normalized ground state energy for the 3-D Harmonic Oscillator

Homework Equations

V(r) = 1/2m(w^2)(r^2)

The Attempt at a Solution

I started with the wave fn in spherical coordinates, and have tried using sep of variables, but keep getting stuck when trying to simplify.

That'll work, but we can't see what the actual problem is if you don't show any work.

 

[pprie003]

Here is what I got:
$\frac {- \hbar ^2} {2m} \, \Big[\frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂}{∂r} \Big) \, + \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂} {∂θ} \Big) + \frac {1}{r^2 \sin^2θ} \, \frac {∂^2}{∂φ^2} \Big]ψ + \frac {1}{2} m \,ω^2 \, r^2ψ \ = Eψ$
Let ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)

 

[blue_leaf77]

Ok if you insist on working in spherical coordinate. Then the next step is substituting the separated wavefunction right, after which you will get separate differential equations each of which is only in $r,\theta,$ and $\phi$. How do these equations look like?

 

[vela]

Here is what I got:
$\frac {- \hbar ^2} {2m} \, \Big[\frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂}{∂r} \Big) \, + \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂} {∂θ} \Big) + \frac {1}{r^2 \sin^2θ} \, \frac {∂^2}{∂φ^2} \Big]ψ + \frac {1}{2} m \,ω^2 \, r^2ψ \ = Eψ$
Let ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)

This is hardly showing your work. Or if this is all you did, it doesn't really qualify yet as an attempt. If you're stuck at this point, you should go back and read your textbook, look at similar examples, and make some more headway before posting again. If you have a specific question, come back and ask that.

 

[mfb]

I started with the wave fn in spherical coordinates, and have tried using sep of variables, but keep getting stuck when trying to simplify.

Please show that, otherwise we can't tell where you got stuck which makes it impossible to help.

Using cartesian coordinates is easier here.

 

[pprie003]

Sorry I left out info, I was having a hard time writing the code, but here is the rest of the work (the problem tells me to work in spherical coordinates):
$\frac {- \hbar ^2} {2m} \, \Big[\frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂}{∂r} \Big) \, + \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂} {∂θ} \Big) + \frac {1}{r^2 \sin^2θ} \, \frac {∂^2}{∂φ^2} \Big]ψ + \frac {1}{2} m \,ω^2 \, r^2ψ \ = Eψ$
Let ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)
$ΘΦ \Big[\frac {- \hbar ^2} {2m} \, \frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂R}{∂r} \Big) + \frac {1} {2} m \, w^2 \, r^2 \, R \, \Big] + RΦ \Big[ \frac {- \hbar ^2} {2m} \, \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂Θ} {∂θ} \Big) \, + \frac {1} {2} m \, w^2 \, r^2 \, Θ \Big] + RΘ \, \Big[ \frac {- \hbar ^2} {2m} \, \frac {1}{r^2 \sin^2θ} \, \frac {∂^2Φ}{∂φ^2} \, + \frac {1} {2} m \, w^2 \, r^2 \, Φ\Big] = ERΘΦ$

Next Divide by ψ:

$\frac {1} {R} \, \Big[\frac {- \hbar ^2} {2m} \, \frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂R}{∂r} \Big) + \frac {1} {2} m \, w^2 \, r^2 \, R \, \Big] + \frac {1} {Θ} \, \Big[ \frac {- \hbar ^2} {2m} \, \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂Θ} {∂θ} \Big) \, + \frac {1} {2} m \, w^2 \, r^2 \, Θ \Big] + \frac {1} {Φ} \Big[ \frac {- \hbar ^2} {2m} \, \frac {1}{r^2 \sin^2θ} \, \frac {∂^2Φ}{∂φ^2} \, + \frac {1} {2} m \, w^2 \, r^2 \, Φ\Big] = E$

It is at this point that I get lost. When I did this same problem in cartesian coordinates it separated nicely. I'm not sure what to do next. Obviously I have read and re-read my textbook (which I believe is not suitable for an intro class) with no such luck. Some things I have thought about (from READING) is substituting the square of the angular momentum operator and its eigen value, but not entirely sure how to approach this. I'm not just seeking the correct solution, but rather how I can tie what I've read to the advice from someone else. I do appreciate the help.

 

[pprie003]

Welcome in PF!
Ok concerning your first problem, have you tried what you can end up with when you view the problem in Cartesian coordinates instead?

Yes I did this problem, which separated nicely into 3 1-D Harmonic Oscillators, and was able to find the energy and first excited state wave fn as well as its degeneracy.

 

[vela]

You have too many terms from the potential.

Sorry I left out info, I was having a hard time writing the code, but here is the rest of the work (the problem tells me to work in spherical coordinates):
$\frac {- \hbar ^2} {2m} \, \Big[\frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂}{∂r} \Big) \, + \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂} {∂θ} \Big) + \frac {1}{r^2 \sin^2θ} \, \frac {∂^2}{∂φ^2} \Big]ψ + \frac {1}{2} m \,ω^2 \, r^2ψ \ = Eψ$
Let ψ(r,θ,φ) = R(r)Θ(θ)Φ(φ)
$ΘΦ \Big[\frac {- \hbar ^2} {2m} \, \frac {1} {r^2} \, \frac{∂}{∂r} \ \Big(r^2 \frac {∂R}{∂r} \Big) + \frac {1} {2} m \, w^2 \, r^2 \, R \, \Big] + RΦ \Big[ \frac {- \hbar ^2} {2m} \, \frac {1} {r^2 \sinθ} \frac{∂}{∂θ} \, \Big( \sinθ \frac {∂Θ} {∂θ} \Big) \, + \frac {1} {2} m \, w^2 \, r^2 \, Θ \Big] + RΘ \, \Big[ \frac {- \hbar ^2} {2m} \, \frac {1}{r^2 \sin^2θ} \, \frac {∂^2Φ}{∂φ^2} \, + \frac {1} {2} m \, w^2 \, r^2 \, Φ\Big] = ERΘΦ$

When you applied the Laplacian to the wave function, you got
$$-\frac{\hbar^2}{2m}\left[\Theta\Phi \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 R') +R\Phi\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\ \Theta') + R\Theta\frac{1}{r^2\sin^2\theta}\Phi''\right].$$ That part was fine. Now you add in the potential term to get
$$-\frac{\hbar^2}{2m}\left[\Theta\Phi \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 R') +R\Phi\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\ \Theta') + R\Theta\frac{1}{r^2\sin^2\theta}\Phi''\right] + \frac 12 m\omega^2 r^2 R\Theta\Phi$$ for the lefthand side of the equation.

When you divide by $R\Theta\Phi$, you'll be left with the term $\frac 12 m\omega^2 r^2$. It only depends on $r$, so group it in with the expression for the radial function.

 

[blue_leaf77]

First of all the term $\frac{1}{2}m\omega^2 r^2$ should appear only once.

 

[pprie003]

You have too many terms from the potential.

When you applied the Laplacian to the wave function, you got
$$-\frac{\hbar^2}{2m}\left[\Theta\Phi \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 R') +R\Phi\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\ \Theta') + R\Theta\frac{1}{r^2\sin^2\theta}\Phi''\right].$$ That part was fine. Now you add in the potential term to get
$$-\frac{\hbar^2}{2m}\left[\Theta\Phi \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 R') +R\Phi\frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta\ \Theta') + R\Theta\frac{1}{r^2\sin^2\theta}\Phi''\right] + \frac 12 m\omega^2 r^2 R\Theta\Phi$$ for the lefthand side of the equation.

When you divide by $R\Theta\Phi$, you'll be left with the term $\frac 12 m\omega^2 r^2$. It only depends on $r$, so group it in with the expression for the radial function.

I don't know why I did all that extra stuff. Still, won't this leave the $\frac{1}{r^2}$ in the angular terms? Should I factor out that term, then substitute the square of angular momentum operator?

 

[vela]

You can get rid of that $r$ dependence in those terms by multiplying both sides of the equation by $r^2$.