Rotation of Spherical Harmonic

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Rotation of a vector is easy and straight forward, just apply a rotation matrix on the vector. But rotating a function may be tricky, we need to transform the coordinate one by one.

r \rightarrow r' = R\cdot r

\displaystyle f(r)  \rightarrow f(r')

When rotating spherical harmonic, the thing becomes get. We can treat the spherical harmonic as a eigen state of angular momentum operator. The problem will becomes easy.

Recall that, the rotation operator for a eigen-state is

\displaystyle R(\alpha, \theta, \phi) = e^{-i\alpha J_z} e^{-i\theta J_y} e^{-i \phi J_z},

The matrix element is the Wigner D-matrix,

D^j_{m'm} = \left<jm'|R|jm\right>

The spherical harmonic is Y_{lm}(\Omega)

Thus, the rotation of spherical harmonic is

\displaystyle R(Y_{lm}) = \sum_{m=-j}^{j} Y_{lm'} D^j_{m'm}


Y'_{l} = Y_{l} \cdot D_{l}(R)


The above picture is Y_{20} rotated by \theta = 45 \deg.

We can see, the spherical harmonic formed a close set under Wigner D-matrix, that combination of them are themselves.

The derivation is follow. Notice that

Y_{lm} = \langle \hat{r} | lm\rangle

\langle \hat{r}| U(R) = \langle R(\hat{r}) |


Y_{lm}(R(\hat{r})) = \langle R(\hat{r}) |lm\rangle = \langle \hat{r} |U(R) |lm \rangle

We can use the identity

\sum_{m'} |lm'\rangle \langle lm' | = 1

\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} \langle \hat{r} | lm' \rangle \langle lm' |U(R) |lm \rangle


\displaystyle Y_{lm}(R(\hat{r})) = \sum_{m'} Y_{lm'}(\hat{r}) D_{m'm}^{l}(R)


Y'_{l} = Y_{l} \cdot D_{l}(R)

It is easy to show that

(D_{m'm}^{l}(R))^{\dagger} = D_{mm'}^{l*}(R) = D_{m'm}^l(R^{-1})


D_{l}^{\dagger} = D_{l}^{T*}

Note that

D_{l} ^{\dagger} \cdot D_{l} = 1 =  D_{l} \cdot D_{l} ^{\dagger}



Wigner-Eckart theorem

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The mathematical form of the theorem is, given a tensor operator of rank k, T^{(k)}, The expectation value on the eigen-state \left|j,m\right> of total angular momentum J is,

\left<j m\right|T_q^{(k)} \left|j' m'\right> = \left<j' m' k q| j m \right> \left<j||T^{(k)}||j'\right>

where, \left<j||T^{(k)}||j\right> is reduced matrix element. The power of the theorem is that, once the reduced matrix element is calculated for the system for a particular (may be the simplest) case, all other matrix element can be calculated.

The theorem works only in spherical symmetry. The state are eigen-state of total angular momentum. We can imagine, when the system rotated, there is something unchanged (which is the reduced matrix element). The quantum numbers m, m' define some particular direction of the state, and these “direction” will cause an additional factor, which is the Clebsch-Gordan coefficient.

Another application is the Replacement theorem.

If any 2 spherical tensors A^{(k)}, B^{(k)} of rank-k, using the theorem, we have,

\displaystyle \left<j m|A^{(k)}|j' m' \right> = \frac{\left<j||A^{(k)}||j'\right>}{\left<j||B^{(k)}||j'\right>} \left<j m|B^{(k)}|j' m' \right>

This can prove the Projection theorem, which is about rank-1 tensor.

L , J are orbital and total angular momentum respectively. The projection of L on  J is

L\cdot J = L_z J_z - L_+ J_- - L_-J_+

The expectation value with same state \left|j m\right> ,

\left< L\cdot J\right> = \left< L_z J_z\right> - \left< L_+ J_-\right> - \left<L_- J_+\right>

using Wigner-Eckart theorem, the right side becomes,

\left< L \cdot J \right> = c_j \left<j||L||j\right>

where the coefficient c_j only depends on j as the dot-product is a scalar, which is isotropic. similarly,

\left< J \cdot J \right> = c_j \left<j||J||j\right> ,

Using the Replacement theorem,

\displaystyle \left< L \right> = \frac{\left<j||L||j\right>}{\left<j||J||j\right>} \left<J \right>

Thus, we have,

\displaystyle \left< L \right> = \frac{\left< L\cdot J \right>}{\left<J\cdot J\right>} \left<J \right>

as the state is arbitrary,

\displaystyle L = \frac{L\cdot J}{J\cdot J} J

this is same as the classical vector projection.



Forced damped oscillation

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I come across with this problem today. It said, when the driving force is changing, what is the peak driving frequency for maximum oscillation amplitude? It is like, asking what is the resonance frequency, right? so I though it should be the natural frequency. But no, the answer is something else.

I remember that “all forced oscillation is moving with the driving frequency”. And I don’t know that the damping will change the oscillation frequency at all! I though it is just a damping factor(term) with the oscillation!

When I search the web, many people use

\displaystyle x(t) = A \exp(-\lambda t) \cos(\omega t + \phi) ,

, where \omega is the natural frequency, as a solution, and calculated \lambda and \phi . That may be the impression that all motion has a simple frequency. But in fact, the phase factor is \omega dependent.

Here i am going to go through it again.

The motion of equation take the from

\displaystyle \frac{d^2x}{dt^2} + \omega^2 x + R \frac{dx}{dt} = A \cos(f t) ,

where R is the damping constant, and A is the magnitude of the driving force, and f is the driving frequency.

A general solution is a homogeneous solution, which is the driving term is zero, or A=0, plus a particular solution.

The homogeneous solution is

\displaystyle x(t) = C_1 e^{-\frac{R}{2}t} \cos( \omega_D t + C_2),


 \displaystyle \omega_D^2 = \omega^2 - \frac{R^2}{4}

and C_1, C_2 are fixed by the boundary condition. So, the damping will change the oscillation frequency. When consider the frequency is related to the energy. The damping draws the system energy to the environment, not just the oscillation amplitude, but also the frequency changed. And the frequency becomes smaller.

The particular solution is

\displaystyle x(t) = \frac{A}{f^4 + f^2 (R^2-2\omega^2) + \omega^4} ( (\omega^2 - f^2) \cos(ft) + fR \sin(ft))

We can see, when the driving force is large, the particular solution dominated, and it does not depends on any boundary condition. The oscillation frequency is f. However, look at the denominator, it can be arranged as

\displaystyle (f^2 - (\omega^2 - \frac{R^2}{2}))^2 + R^2 \omega^2 + \frac{R^4}{4}

The peak is at \omega_F = \sqrt {\omega^2 - \frac{R^2}{2}} ! not w_D, not the natural frequency!




Now, lets us have two systems, P and Q.

two systems.PNG

Suppose, we already know that the damping of Q is large than P. And, by measuring, the damping frequencies are the same for P and Q too! Thus

\omega_D^2 (P) = \omega(P)^2 - R_P^2/4 = \omega(Q)^2 - R_Q^2/4 = \omega_D^2(Q)

means that the natural frequency of  Q is also larger than that of P.

R_P  < R_Q \iff \omega(P) < \omega(Q)

So, lets us set P is the group of lines on the left, which share a common natural frequency, and Q is the group of lines on the right, which share a common natural frequency too. The same color for same damping constant.

Thus, The system of P and Q, in which the damping constant of P is smaller than that of Q, we may guess something like left blue line and right orange line.

two systems_2.PNG

However, when we really fulfill the constrain, this is like this:

two systems_3.PNG

The blue curved never crossed with other curves. And it looks like figure 2. But there is difference. Let overlap two plots.

two systems_4.PNG

Suppose the curves can be crossed. That means, at some particular frequency, the system P and Q oscillating at same amplitude, but the damping of Q is larger than that of P, or the energy loss of Q is larger than that of P, given the same driving force! That is impossible!

Although the frequency of Q is larger, but the drop of amplitude is much faster, unless the damping of Q is very large. but in that case, the frequency spectrum does not have any peak at all.







Exchange energy

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In QM, when treating a system with identical particles, the exchange term raised because the system must be the same after particle exchanged.

I am reluctant to use the term “exchange interaction” or “exchange force”, “exchange potential”, because there is no interaction, no force carrier, no potential at all.

Suppose the Hamiltonian for single fermion is

H\psi = \epsilon \psi

The 2-particle system, the Schrodinger equation is

H_T \Psi = H_1 + H_2 + H'= E \Psi

where H' is the interaction between the 2 particles.

Now, guessing the total wave function to be

\Psi = a \psi_1(r_1) \psi_2(r_2) + b \psi_1(r_2)\psi_2(r_1) = a \Psi_n + b\Psi_e

The first term is “normal”, the second term is “exchanged”. substitute to the equation

 a H_T \Psi_n + b H_T \Psi_e = a E \Psi_n + b E \Psi_e

multiply both side with \Psi_1 or \Psi_2, and integrate r_1, r_2, we have

\begin{pmatrix} J & K \\ K & J \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = E \begin{pmatrix} a \\ b \end{pmatrix}


J = \int \psi_1^*(r_1) \psi_2^*(r_2) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2

K = \int \psi_1^*(r_2) \psi_2^*(r_1) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2

Here, we assumed H' is symmetric against r_1, r_2. This is a fair assumption as H' is the mutual interaction, and H_1, H_2 are independent.

The J is the energy from the interaction of the particles.

The K is the energy due to EXCHANGE of the particles.

As we can see, the energy J = \epsilon_1 + \epsilon_2 + J'.

The H_T in K can be reduced to H' when the eigen wave function \psi_1 , \psi_2 are in difference orbits. When \psi_1 = \psi_2 , K = \epsilon_1 + \epsilon_2 + K'.

Thus the individual energy can be subtracted, yield,

\begin{pmatrix} J' & K' \\ K' & J' \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \Delta E \begin{pmatrix} a \\ b \end{pmatrix}

The solution is

\Delta E = J' + K', (a,b) = (1,1)

This is the symmetric state. For fermion, the spin-part is anti-symmetric. This is the spin-singlet state.

\Delta E = J' - K', (a,b) = (1,-1)

This is the anti-symmetric state. For fermion, the spin-part is symmetric, or spin-triplet state.

It is worth to notice that, in the symmetric state, the energy is higher because the two particles can be as close as possible, that create large energy. In the opposite, in the anti-symmetric state, the two particle never contact each other, thus, the interaction energy reduced.

We can summarized.

The energy J' is the shift of energy due to the mutual interaction.

But the system of identical particle subjects to the exchange symmetry. The exchanged wave function is also a state, that the total wave function has to include the exchanged wave function. This exchanged wave function creates the exchange term or exchange energy K'.

In the system of Fermion, the exchange energy is related to Pauli exclusion principle.

Hartree-Fock for 2-body ground state

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Long time ago, we talked about the mean-field calculation, an touched Hartree-Fock method. In that time, we explained excited-state approach. Now, we explain another approach by variation of the wave functions. This approach is inevitable in atomic physics, because the potential is fixed.

The Hamiltonian is

H = H_1 + H_2 + V_{12}

Since the spin component is anti-parallel, the space part of the total wave function is

\Psi = \phi_1(r_1) \phi_2(r_2)

The Schrodinger equation is

H\Psi = E\Psi

Integrate both side with \phi_1(r_1)

\int dr_1 \phi_1(r_1) H \Psi = \int dr_1 \phi_1(r_1) E \Psi

using the normalization and define \left<x| 1\right> = \phi_1(r_1)

\left<1|H|1\right> \phi_2(r_2) = E \phi_2(r_2)


\left<2|H|2\right> \phi_1(r_1) = E \phi_1 (r_1)

expand H = H_1 + H_2 + V_{12}

(H_2 + \left<1|V_{12}|1\right>) \phi_2(r_2) = (E - \left<1|H_1|1\right>) \phi_2(r_2)

(H_1 + \left<2|V_{12}|2\right>) \phi_1(r_1) = (E - \left<2|H_2|2\right>) \phi_1(r_1)

Now, we have 2 equations,  an initial guess of \Psi = \phi_1(r_1) \phi_2(r_2) ,

The difficulty is that, the \left<1|V_{12}|1\right> contains 2 variables.