Solving Spherical Harmonics Homework

[says]

Homework Statement

The spherical harmonic, Y_m,l(θ,φ) is given by:
Y_2,3(θ,φ) = √((105/32π))*sin²θcosθe^2iφ

1) Use the ladder operator, L₊ = +ħe^iφ(∂/∂θ+icotθ∂/∂φ) to evaluate L₊Y_2,3(θ,φ)

2) Use the result in 1) to calculate Y_3,3(θ,φ)

Homework Equations

L₊Y_m,l(θ,φ)=A_m,lY_m+1,l(θ,φ)
A_m,l=ħ√l(l+1)-m(m+1)

Y_m,l(θ,φ) = (-1)^m √[((2l+1)/4π) ((l-m)!/(l+m)!)] P_m,lcosθe^imφ

The Attempt at a Solution

1) For m=2, l=3

A_2,3 = ħ√3(3+1)-2(2+1) = ħ√6

∴L₊Y_2,3(θ,φ)=A_2,3Y_2+1,3(θ,φ)

=ħ√6 Y_3,3(θ,φ)

2) Y_3,3(θ,φ) = (-1)³ √[((2(3)+1)/4π) ((3-3)!/(3+3)!)] P_3,3cosθe^i3φ

= -1 √[(7/4π)(1/720)]P_3,3cosθe^i3φ

I don't think I've fully understood the question because I haven't really used the result in 1) to calculate 2)...

 

[Orodruin]

You are not supposed to leave an associated Legendre function hanging. You are supposed to use your result from (1) to find an explicit expression for $Y_{33}$.

You also have not solved (1) because you are supposed to evaluate $L_+ Y_{23}$ explicitly, not express it in terms of $Y_{33}$.

 

[says]

I've used the raising operator, L₊ and operated on Y_2,3.

The only other way I can think of evaluating (1) would be by L²Y_2,3 = constantY_2,3

where L² is the angular momentum operator, which can be made up of L₊L_- raising operators.

The spherical harmonics are eigenfunctions of the operator L², and the constant is an eigenvalue, equal to ħ² l(l+1)

 

[Orodruin]

I've used the raising operator, L+ and operated on Y2,3.

No, you have not. You are supposed to operate on the actual expression given. Then in (2) you are supposed to compute $Y_{33}$ using the relations you have already given.

You have the raising operator represented as a derivative operator and you have the expression for $Y_{23}$, all you have to do is apply it.

 

[says]

ahhh ok

L₊Y_m,l(θ,φ) = +ħe^iφ(∂/∂θ+icotθ∂/∂φ)√((105/32π))*sin²θcosθe^2iφ

Can I rearrange that equation so it looks like this (below), grouping the partial derivatives?

√((105/32π))ħe^iφ (∂/∂θ [sin²θ cosθ] + icotθ ∂/∂φ [e^2iφ])

 

[Orodruin]

No. That is not how derivatives work.

 

[says]

Ok, I'm not entirely sure how to evaluate that first equation in post #5 then.

 

[Orodruin]

Just perform the derivatives. I am sorry, but I cannot see where the difficulty in this lies. You just need to use the distributive property of the linear differential operators.

 

[says]

Ok, I made an attempt with that differential.

(1)
L₊Y₂₃ = +ħe^iφ [ ∂/∂θ + icotθ ∂/∂φ ] (√(105/32π) sin²θ cosθ e^2iφ)

L₊Y₂₃ =+ħe^iφ√(105/32π) [ (sin2θ cosθ - sin³θ)e^2iφ - (2cotθ sin²θ cosθ)e^2iφ

L₊Y₂₃ =+ħe^3iφ√(105/32π) [ (sin2θ cosθ - sin³θ) - (2cotθ sin²θ cosθ)

(2) If question (1) is correct then (2) should be:

A₂₃Y₃₃ = Answer from question (1)

A₂₃ = ħ√6

Y₃₃ = (e^3iφ√(105/32π) [ (sin2θ cosθ - sin³θ) - (2 cotθ sin²θ cosθ) ] / √6

Y₃₃ = √(35π/8) e^3iφ [ (sin2θ cosθ - sin³θ) - (2 cotθ sin²θ cosθ)]

 

[says]

Is there any way to check these answers? This is similar to Griffiths problem 4.23

 

[Orodruin]

Your answer could use some cleaning up. For example, how can $\cot(\theta) \sin^2(\theta)\cos(\theta)$ be written on a simpler form? I also strongly suggest against having a $\sin(2\theta)$.

Once you have it you can just compare to the usual expression for the spherical harmonics in terms of the associated Legendre functions.

 

[says]

simplifying those two terms:

2cotθsin²θcosθ
= 2(cosθ/sinθ)sin²θcosθ
=2cos²θsinθ

sin2θcosθ-sin³θ
=2sinθcos²θ-sin³θ∴ L₊Y₂₃= +ħe^3iφ√(105/32π) [ 2sinθcos²θ - sin³θ - 2cos²θsinθ]
= +ħe^3iφ√(105/32π) [-sin³θ]
= -ħe^3iφ√(105/32π) [sin³θ]

∴ Question (2)

Y₃₃ = - e^3iφ (√(35π)/8) (sin³θ)

 

[Orodruin]

So how does that compare with what you would find with the usual expression for $Y_{33}$?

 

[says]

https://research.csiro.au/static/dcm/HydrogenAtomWaveFunction.htm

It seems like the denominator inside my square root is incorrect!

 

[Orodruin]

Is the 8 inside your square root or not? It is unclear from your notation.

You will probably find it easier to get through with notation if you use $\LaTeX$.

 

[says]

Sorry, yes my 8 is inside the square root

 

[Orodruin]

Then you made an arithmetic error https://www.wolframalpha.com/input/?i=105/(32*6)

Also, you factor of $\pi$ has somehow jumped up to the nominator.

 

[says]

oh dear...

Thank you for your help, @Orodruin much appreciated!

 

[Orodruin]

It should be mentioned that different people use different normalisation conventions for the spherical harmonics. In quantum mechanics we are very used to normalising states and typically normalise our orthogonal functions to have norm one. However, in different disciplines, other conventions are used, see https://en.wikipedia.org/wiki/Spherical_harmonics#Conventions

 

[DrClaude]

in different disciplines, other conventions are used

On that topic, may I add that I have never seen the notation $Y_{m,l}$, which made me first think that the OP had the operation of $L_+$ wrong. The only conventions I am aware of are $Y_{l,m}$ and $Y_l^m$.

 

[says]

Yes, I was writing them the other way around because I didn't know how to do that formatting. I'll try it for next time!

 

[Orodruin]

On that topic, may I add that I have never seen the notation $Y_{m,l}$, which made me first think that the OP had the operation of $L_+$ wrong. The only conventions I am aware of are $Y_{l,m}$ and $Y_l^m$.

Me neither, I used the OP's notation just to avoid in-thread confusion.

 

[says]

In my post #12 I have a minus sign (in the -sin³θ), which I've then taken out to the front.

I've looked at the Griffiths problem 4.23, which is similar to this problem: http://physicspages.com/pdf/Griffiths%20QM/Griffiths%20Problems%2004.23.pdf

I don't think I've missed another minus sign to cancel it out though, and the answer in the other link doesn't have a minus sign: https://research.csiro.au/static/dcm/HydrogenAtomWaveFunction.htm

 

[Orodruin]

Just as with normalisation, there are different sogn conventions. You really should check against the definitions in your original source, not against a random internet page.