28
votes
Accepted
How should we think about Spherical Harmonics?
How should we think about Spherical Harmonics?
In short: In the same way that you think about plane waves.
Spherical harmonics (just like plane waves) are basic, essential tools. As such, they are ...
- 135k
17
votes
Angular orientation of exact solution of the Hydrogen Schrödinger Equation
Yes, you are right. If there is no external interaction that breaks the rotational symmetry, we are committed to think that the probability of finding an electron around the nucleus cannot depend on ...
- 78.1k
16
votes
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Why do we need the Condon-Shortley phase in spherical harmonics?
You don't need it: it's a sign convention and the only thing you need to do with it is to be consistent. (In particular, this means always checking that the sign and normalization conventions for $Y_{...
- 135k
11
votes
Irreducible form of Spherical tensor operators
The formula handling has been well addressed by ZeroTheHero, but I think there's a bit more to be said about your second question,
Also is $Y_{l}^{m}$ defined as irreducible since we can write ...
- 135k
11
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How should we think about Spherical Harmonics?
How familiar are you with Fourier analysis?
For instance, if you want a complete basis of functions defined on the unit square, you would pick:
$$ |n, m\rangle \equiv e^{2\pi i nx}e^{2\pi i my}$$
with ...
- 39.6k
10
votes
How should we think about Spherical Harmonics?
The Spherical Harmonics are a complete orthonormal basis
for the space of the functions defined on the sphere.
This statement means:
You can decompose any arbitrary function $f(\theta,\phi)$
defined ...
- 41k
9
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How to calculate the angular momentum states of isotropic quantum harmonic oscillator?
As you have probably already worked out, the two eigenspaces that you're interested in, $N=2$ and $N=3$, are formed by a direct sum of subspaces of different angular momentum; thus, the $N=2$ ...
- 135k
9
votes
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Can you expand a real scalar field $\phi(t,\mathbf{x})$ in terms of spherical harmonics?
You can expand any object in any basis you want, this is precisely the point of bases. Whether a given basis is useful or not is an entirely different question. In flat space $\mathbb R^n$ the basis ...
9
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Angular orientation of exact solution of the Hydrogen Schrödinger Equation
We can find a nice visual representation of the Spherical Harmonics in this wiki article. You can see that in fact, only $l=0$ is spherically symmetric.
We know that:
The ground state is the only ...
- 3,369
8
votes
What is the magnetic quantum number for $p_x$, $p_y$, $p_z$?
The $\text{p}_x$, $\text{p}_y$, $\text{p}_z$ orbitals are "real" $\ell=1$ wavefunctions.
The actual angular wavefunctions (i.e., simultaneous eigenfunctions of $\hat{{L}}^2$ and $\hat{L}_z$) ...
- 564
7
votes
Irreducible form of Spherical tensor operators
To understand a tensor operator one first needs to be reminded that it's an operator. Whereas an (angular momentum) state $\vert \ell m\rangle$ transform as a column vector under rotation:
\begin{...
- 47.9k
7
votes
Accepted
Is it a coincidence or a convention that the spherical harmonics $Y_{\ell m}(\theta, \phi)$ are eigenstates of $L_z$?
There is no global $z$ axis that permeates the universe. As such, since the choice of $z$ axis is necessarily arbitrary for every specific situation, there is nothing special about the choice of $L_z$ ...
- 135k
7
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Systematic expansion of $e^{i\vec{k}\cdot\vec{r}}$ in atomic physics in terms of Legendre polynomials and identifying different $l$ terms
The expression you found in Sakurai's textbook is the correct one and can be further expanded in spherical harmonics, if needed.
The difference with electrostatics is due to the different position ...
7
votes
Accepted
Quantum mechanics angular momentum spherical tensor components
There are several approaches you can take for converting between cartesian and spherical tensors. You can start with the definition of spherical vectors:
$$ \hat e^{\pm} = \frac 1 2 \mp[\hat x \pm i\...
- 39.6k
7
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How should we think about Spherical Harmonics?
As knowing your current knowledge I would like to a very basic interpretation of the following:
$P_{l,0}$ called Legendre's polynomials, are a complete orthonormal basis on the circle.
The other one ...
- 12.1k
6
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Why is $\langle x \rangle =0$ for the ground state hydrogen atom?
In spherical coordinates $r$ is the distance from the origin and can only ever be positive.
So the expectation value $\langle r \rangle$ has to be positive.
However in rectangular Cartesian ...
- 99.9k
6
votes
In a spherically symmetric central potential why do we look for eigenfunctions of the angular momentum operator?
Well, to put it shortly, it's just because $L^2$ and $L_z$ are two observables that have no $r$ dependence. Since the kinetic terms of $H$ can be written as functions of $r$ and $L^2$, and the ...
- 2,317
6
votes
Coordinate-free general solution to the wave equation
In fancy "coordinate-free" language, the solutions to the wave equation $(d^* d + d d^*) \phi = 0$ are called harmonic $0$-forms. Hodge's theorem states that for certain compact manifolds, harmonic ...
- 105k
6
votes
Accepted
Physically unacceptable solutions for the QM angular equation
We are in principle trying to solve the angular TISE problem$^1$
$$ \vec{\bf L}^2Y~=~\hbar^2\ell(\ell+1)Y, \qquad {\bf L}_zY~=~\hbar m Y, $$
on the unit 2-sphere $\mathbb{S}^2$. However, we are using ...
- 213k
6
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Can a spherical function with only $r$-dependence be an eigenstate of $L^2$ and $L_z$?
These are differential operators, so a function of $r$ only is effectively a constant, corresponding to the zero eigenvalues of both $L^2$ and $L_z$.
- 65.1k
5
votes
Accepted
Why are circular unit vectors often defined as $\hat{\mathbf e}_\pm = \mp (\hat{\mathbf e}_x \pm i \hat{\mathbf e}_y)/\sqrt{2}$
One answer, connecting he antisymmetry of the wedge product and the (Lie) commutator, is through the Wigner-Eckart theorem.
Let
$$
\hat T_{10}=\hat L_z\, \qquad \hat T_{1\pm 1}=\mp \frac{1}{\sqrt{2}}
...
- 47.9k
5
votes
Quantization of electrons' angular momentum in atoms and molecules
This kind of depends on exactly what it is you're talking about.
In multi-electron atoms, the total angular momentum $\mathbf J = \mathbf L+\mathbf S$, which includes the orbital and spin angular ...
- 135k
5
votes
Accepted
Covariant derivative of spherical harmonics
Since $f = f(\theta,\phi) = Y^l_m(\theta,\phi)$ is a scalar function, $\nabla_a f$ is a covariant vector, and so it's covariant derivative is that of the covariant derivative of a covariant vector, ...
- 4,101
5
votes
What are spherical tensors?
According to the Peter-Weyl theorem, if $G$ is a compact topological group, then every strongly continuous unitary representation $G\ni g \mapsto U_g$ in a fixed Hilbert space ${\cal H}$ is the ...
- 78.1k
5
votes
Accepted
Series Solution of Laplace Equation in Spherical Coordinates
You need both solutions in general. Your potential is not uniquely defined by being harmonic, you need to append boundary conditions. If your domain is a ball (includes the origin), this excludes the $...
- 15.1k
5
votes
Accepted
What is the relation between irreducible representation of $SO(3)$ and spherical harmonics?
At fixed $l$, the $2l+1$ spherical harmonics:
$$ Y_l^m(\theta, \phi) \ \ \ m \in [-l, -l+1, \cdots, l-1, l] $$
are a $2l+1$ dimensional irreducible$^1$ representation of $SO(3)$, with Casimir ...
- 39.6k
5
votes
What is the relation between irreducible representation of $SO(3)$ and spherical harmonics?
You could do worse than simply going to your text, e.g., Modern Quantum Mechanics (3rd Ed) by J J Sakurai & Jim Napolitano, and reviewing the definition,
$$\langle \hat {\mathbf n}|\ell, m\rangle= ...
- 66.3k
5
votes
Accepted
Can the irrotational vortex be described using spherical harmonics?
For some physical intuition, if a vector field is conservative (i.e. can be written as the gradient of a potential), then you can think of the potential as a "landscape" such that the vector ...
- 71.5k
4
votes
Accepted
Choosing $A_l=0$ to guarantee bounded potential in infinity
It would only equal $e^{-r}$ for $\gamma=0$. For $\gamma = \pi$ you get $P_l(-1) = (-1)^l$, changing your answer to $e^r$ in that direction.
- 22.8k
4
votes
Monopole Spherical Harmonics
This is not a direct answer to the question asked, but rather a different approach to the problem. It's much easier to derive the monopole harmonics from a geometric consideration of the Hopf map ${...
- 56.6k
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