Spherical harmonics and P operator

In summary, the author provides a summary of the content. The operator P is defined and it is shown that under parity (r, \theta, \phi) it changes to (r,\pi-\theta, \phi\pm\pi). The spherical harmonics are a product of the associated Legendre function P_{l}^{m}(x) and the azimuthal exponential. Under parity (r, \theta, \phi) changes to (r,\pi-\theta, \phi\pm\pi
  • #1
paweld
255
0
Let's define operator P:
[tex]
P \phi(\vec{r})=\phi(-\vec{r})
[/tex]

Does anyone know simple and elegant prove that [tex]P|lm\rangle = (-1)^l |lm\rangle[/tex]
([tex]|lm\rangle[/tex] is spherical harmonic).
 
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  • #2
paweld said:
Let's define operator P:
[tex]
P \phi(\vec{r})=\phi(-\vec{r})
[/tex]

Does anyone know simple and elegant prove that [tex]P|lm\rangle = (-1)^l |lm\rangle[/tex]
([tex]|lm\rangle[/tex] is spherical harmonic).

I don't think this is true.

Take:
[tex]
|1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\theta)e^{i\phi}
[/tex]
[tex]
P|1 1\rangle = (3/(8\pi))^{(1/2)}sin(\theta)e^{-i\phi}
[/tex]

The [tex]\phi[/tex] component screws up the relation.
 
  • #3
The spherical harmonics are a product of the associated Legendre function [tex] P_{l}^{m}(x)[/tex] and the azimuthal exponential. Under parity [tex](r, \theta, \phi) [/tex] changes to [tex](r,\pi-\theta, \phi\pm\pi)[/tex], which means that [tex]cos \theta[/tex] changes to [tex]-cos \theta[/tex]

From Rodrigos formula [tex]P_{l}^{m}(x) \mbox{ }\alpha \mbox{ } (1-x^2)^{\frac{m}{2}} \frac{d^{l+m}}{dx^{l+m}}(1-x^2)^l [/tex]

where x is [tex]cos \theta[/tex]. Under x-->-x this changes by [tex](-1)^{m+l}[/tex]. However, the azimuthal exponential changes by: [tex]e^{i m (\phi \pm \pi)}=(-1)^m e^{i m \phi}[/tex].

Multiplying the azimuthal and polar parts together, the result changes by [tex](-1)^{l+m}(-1)^m=(-1)^l[/tex]
 
  • #4
ygolo said:
I don't think this is true.

Take:
[tex]
|1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\theta)e^{i\phi}
[/tex]
[tex]
P|1 1\rangle = (3/(8\pi))^{(1/2)}sin(\theta)e^{-i\phi}
[/tex]

The [tex]\phi[/tex] component screws up the relation.

In spherical coordinates P acts as follows:
[tex]
P \psi(r,\theta,\phi)=\psi(r,\pi-\theta,\phi+\pi)
[/tex]
(Am I right?)
So:
[tex]
P|1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\pi-\theta)e^{-i(\phi+\pi)} =- |1 1\rangle
[/tex]
 
  • #5
paweld said:
In spherical coordinates P acts as follows:
[tex]
P \psi(r,\theta,\phi)=\psi(r,\pi-\theta,\phi+\pi)
[/tex]
(Am I right?)
So:
[tex]
P|1 1\rangle = -(3/(8\pi))^{(1/2)}sin(\pi-\theta)e^{-i(\phi+\pi)} =- |1 1\rangle
[/tex]

Ah, OK. Thanks for the correction. I was going outside the domain of [tex]\theta[/tex] and [tex]\phi[/tex] and the [tex]\phi[/tex] transformation was just wrong.
 
Last edited:
  • #6
RedX said:
The spherical harmonics are a product of the associated Legendre function [tex] P_{l}^{m}(x)[/tex] and the azimuthal exponential. Under parity [tex](r, \theta, \phi) [/tex] changes to [tex](r,\pi-\theta, \phi\pm\pi)[/tex], which means that [tex]cos \theta[/tex] changes to [tex]-cos \theta[/tex]

From Rodrigos formula [tex]P_{l}^{m}(x) \mbox{ }\alpha \mbox{ } (1-x^2)^{\frac{m}{2}} \frac{d^{l+m}}{dx^{l+m}}(1-x^2)^l [/tex]

where x is [tex]cos \theta[/tex]. Under x-->-x this changes by [tex](-1)^{m+l}[/tex]. However, the azimuthal exponential changes by: [tex]e^{i m (\phi \pm \pi)}=(-1)^m e^{i m \phi}[/tex].

Multiplying the azimuthal and polar parts together, the result changes by [tex](-1)^{l+m}(-1)^m=(-1)^l[/tex]

Maybe somone knows more "physical" prove? I mean prove which doesn't use of explicite formlua for spherical harmonics.
 
  • #7
RedX said:
The spherical harmonics are a product of the associated Legendre function [tex] P_{l}^{m}(x)[/tex] and the azimuthal exponential. Under parity [tex](r, \theta, \phi) [/tex] changes to [tex](r,\pi-\theta, \phi\pm\pi)[/tex], which means that [tex]cos \theta[/tex] changes to [tex]-cos \theta[/tex]

From Rodrigos formula [tex]P_{l}^{m}(x) \mbox{ }\alpha \mbox{ } (1-x^2)^{\frac{m}{2}} \frac{d^{l+m}}{dx^{l+m}}(1-x^2)^l [/tex]

where x is [tex]cos \theta[/tex]. Under x-->-x this changes by [tex](-1)^{m+l}[/tex]. However, the azimuthal exponential changes by: [tex]e^{i m (\phi \pm \pi)}=(-1)^m e^{i m \phi}[/tex].

Multiplying the azimuthal and polar parts together, the result changes by [tex](-1)^{l+m}(-1)^m=(-1)^l[/tex]

Maybe somone knows more "physical" prove? I mean prove which doesn't use explicite formlua for spherical harmonics.
 
  • #8
paweld said:
Maybe somone knows more "physical" prove? I mean prove which doesn't use explicite formlua for spherical harmonics.

Parity and rotations commute (why?). Since parity and rotations commute, by Schur's lemma that means all states of total angular momentum [tex]\ell [/tex] have the same parity. So parity does not depend on the quantum number 'm', but only '[tex]\ell [/tex]' (if you're not familiar with Schur's lemma then the fact that rotations commute with parity implies that angular momentum commutes with parity, so the raising and lowering operators that are built from the angular momentum operators commute with parity - therefore states with different m but same [tex]\ell [/tex] have the same parity). An easy test for parity is to test the m=0 or m=[tex]\ell [/tex] spherical harmonic. The m=0 spherical harmonic coincides exactly with the Legendre function, which is odd if [tex]\ell[/tex] is odd and even if [tex]\ell[/tex] is even.

The important point is that you can test m=anything to determine the parity for all [tex]\ell[/tex]. The spherical harmonics for m=[tex]\ell[/tex] takes on the simple form: [tex] sin^{\ell} \theta e^{i\ell \phi}[/tex] and that's all you have to test for parity.
 
  • #9
All spherical harmonics with m ne 0 can be obtained by rotation and linear combination from the one with m=0 so it is sufficient to look at the function with m=0 (thats the Schur's lemma stuff of RedX). The harmonic with m=0 and l can be obtained from z^l by orthogonalization to the functions with l'<l. As under parity z->-z, the function transforms as z^l ->(-z)^l.
 

Related to Spherical harmonics and P operator

What are spherical harmonics?

Spherical harmonics are mathematical functions that are used to represent the angular part of solutions to certain partial differential equations. They are commonly used in physics and mathematics, particularly in problems involving spherical symmetry.

How are spherical harmonics related to the P operator?

The P operator, also known as the parity operator, is a mathematical operator that changes the sign of a function when its coordinates are reflected through the origin. Spherical harmonics are eigenfunctions of the P operator, meaning they have a well-defined parity and can be used to classify functions based on their symmetry properties.

What is the significance of spherical harmonics in quantum mechanics?

In quantum mechanics, spherical harmonics play a crucial role in the description of the angular momentum of particles. They are used to represent the angular part of wavefunctions, which determine the probability distribution of a particle's position and momentum. Spherical harmonics are also used to classify atomic orbitals and to calculate transition probabilities between energy levels.

How are spherical harmonics visualized?

Spherical harmonics can be visualized as a series of concentric spherical shells, with each shell representing a different order of the function. The amplitude of the function increases towards the poles and decreases towards the equator, creating a complex pattern of nodal lines. This visualization helps to understand the symmetry properties of spherical harmonics and their role in physical systems.

What are some applications of spherical harmonics?

Spherical harmonics have many applications in physics and mathematics, including the analysis of electromagnetic fields, the study of heat conduction in spherical objects, and the solution of boundary value problems. They are also used in computer graphics to create realistic lighting and shading effects, and in geodesy to model the Earth's gravitational field.

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