Exchange energy

June 15, 2017

GoLuckyRyan Basic , , , Leave a comment

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}

where

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.

some work done

July 9, 2011

GoLuckyRyan Lab Log, review, theory , Leave a comment

recently, i am working on 3 things, both are calculation.

the 1st thing is a concrete foundation of microwave inducted optical nuclear polarization theory, which is the theory that my experiment is current using. i found some papers talk on the theoretical treatment on this subject. however, they are all start from some un-stated assumptions, for example, they are started with the Hamiltonian of hyperfine interaction, without state clear where the term comes from. i guess it is very elementary.  nevertheless, different papers may use different Hamiltonians. moreover, some of the mathematics techniques or steps were skipped and directly jump to result. Thus. that is my motivation on building a concrete theory, based on dipole -magnetic field interaction.

when i finished the hyperfine interaction, i have to apply a magnetic field and it cause a Zeemen splitting of energy level. and the Zeeman effect depends on the strength of the magnetic field. they are called strong field, intermedia field and weak field. in different field strength, the origina Hamiltonian may or may not be treated as a perturbation. Thus, i have to know the magnitude of the hyperfine structure. at the point, i come up with in-consistance value that cannot smoothy translate from weak field to intermedia field and strong field.

So, i restudy the atomic theory on Hydrogen atom. Start from Bohr model, fine structure to hyperfine structure and calculated the energy level numerically, so that i can get a full and complete picture on this matter. then i can now compare the magnetic field strength. it turns out that it is not that clear when is weak or strong, because it is depends on the total vale of the Hamiltonian, not the coefficient. Thus, for higher orbital angular momentum, a very small magnetic field can be treated as strong field, coz it already large enough to break the coupling. i am still working on the hyperfine Zeeman effect.

Meanwhile, my professor asked me to calculate the filling factor of the sample and NMR coil. the NMR signal strength depends on the geometry of the  coil and sample. and this geometric factor is called filling factor. it is a good name, i thing, since the maximum value is that the sample completely filled the coil. In order to calculated this, i parallel calculated the magnetic field generated by a cylindrical uniformly magnetized rod and the magnetic induction due to multiple from a single coil. the far field of the rod is done and basically, it is dipole, but the near field is complicated, the multiple terms appear and they are convergence slowly.

that’s why i don’t have post for 2 weeks.

Quick reminder for Perturbation theory

June 26, 2011

GoLuckyRyan Basic, theory , Leave a comment

when calculate the energy from Hamiltonian. usually, we only know the exact form from a simple Hamiltonian but not the exact form of the Hamiltonian. Thus, we have to apply a perturbation to find the approximate energy.

the formula is :

if

H_0 \left|n^0\right>= E_n^0\left|n^0\right>

is the exactly solvable Hamiltonian, energy and eigenstate for the n-h state, then the REAL energy for the total Hamiltonian

H= H_0 +V

is

E_n = E_n^0 + V_{nn} + \sum_{k\neq n}{\frac{|V_{nk}|^2}{E_n-E_k}}+....

where

V_{nk} = \left<n^0|V|k^0\right>

so far as I encounter, the exact state is not quite important.

Detail walk through.

Set the perturbation strength as

H = H_0 + \lambda V,

where \lambda is the switch for the perturbation.

Suppose the solution for n-th level is

\displaystyle \left|n \right> = \sum_{r}^{\infty} \lambda^r \left|n^{(r)}\right> ,

with energy

\displaystyle E_n = \sum_{r}^{\infty} \lambda^r E_n^{(r)} .

Put in the equation,

\displaystyle (H_0 + \lambda V) (\sum_{r}^{\infty} \lambda^{r} \left| n^{(r)} \right>) = (\sum_r^{\infty} \lambda^r E_n^{(r)}) (\sum_r^{\infty} \lambda^r \left|n^{(r)}\right>)

expand and collect \lambda^r

 \displaystyle H_0 \left| n^{(r)}\right> + V \left| n^{(r-1)}\right> = \sum_l^r E_n^{(k)} \left| n^{(r-k)} \right>

we can see, for r = 0, this is the original, unperturbed solution.

For r =1 , this is

\displaystyle H_0 \left| n^{(1)} \right> + V \left| n^{(0)} \right> = E_n^{(0)} \left| n^{(1)} \right> + E_n^{(1)} \left| n^{(0)} \right>

To get E_n^{(1)} , we apply \left< n^{(0)} \right| from the left, the first terms of left and right side will cancel out.

\left< n^{(0)} | V | n^{(0)} \right> = E_n^{(1)}

For higher order, one can calculate using the lower orders. for example,

E_n^{(2)} = - \left< n^{(1)} | (H_0 - E_n^{(0)} )| n^{(1)} \right>

The state \left| n^{(1)} \right> can be constructed using eigen states,

\displaystyle \left| n^{(1)} \right> = \sum_{m} a_m \left| m \right>

put in the equation, using orthogonal relation,

\displaystyle a_m = \frac{ \left<m|V|n\right>}{E_n - E_m}

For Coulomb Potential, here is some common expected values:

\left< n,l,m| \frac{a}{r} | n,l,m \right> = \frac{1}{n^2}

\left< n,l,m| \frac{a^2}{r^2} | n,l,m \right> = \frac{1}{(l+1)n^3}

\left< n,l,m| \frac{a^3}{r^3} | n,l,m \right> = \frac{1}{l(l+1/2)(l+1)n^3}

\left< n,l,m| \frac{r}{a} | n,l,m \right> = \frac{1}{2} \left(3n^2-l(l+1)\right)

\left< n,l,m| \frac{r^2}{a^2} | n,l,m \right> = \frac{1}{2} \left( n^2 ( 5n^2 -3l(l+1)+1) \right)

NMR (nuclear magnetic resonance)

January 27, 2011

GoLuckyRyan Basic, no math , , , , , , , , , , Leave a comment

NMR is a technique to detect the state of nuclear spin. a similar technique on electron spin is call ESR ( electron spin resonance)

The principle of NMR is simple.

  1. apply a B-field, and the spin will align with it due to interaction with surrounding and precessing along the B-field with Larmor frequency, and go to Boltzmann equilibrium. the time for the spin align with the field is call T1, longitudinal relaxation time.
  2. Then, we send a pule perpendicular to the B-field, it usually a radio frequency pulse. the frequency is determined by the resonance frequency, which is same as the Larmor frequency. the function of this pulse is from the B-field of it and this perpendicular B-field with perturb the spin and flip it 90 degrees.
  3. when the spin are rotate at 90 degrees with the static B-field, it will generate a strong enough signal around the coil. ( which is the same coil to generate the pule ) and this signal is called NMR signal.
  4. since the spins will be affected by its environment, and experience a slightly different precession frequency. when the time goes, they will not aligned well, some precess faster, some slower. thus, the transverse magnetization will lost and look as if it decay. the time for this is called T2, transverse relaxation time.

by analyzing the T1 and T2 and also Larmor frequency, we can known the spin, the magnetization, the structure of the sample, the chemical element, the chemical formula, and alot many others thing by different kinds of techniques.

For nuclear physics, the use of NMR is for understand the nuclear spin. for example, the polarization of the spin.