slater determinant - university of crete · slater determinant •antisymmetric case: •initial...

Slater determinant• Antisymmetric case:

• Initial state(no specific symmetry) is the product of the diagonal elements

The symmetric case cannot be written as handily, but all we have to do is replace a – with a + everywhere in the deteminant expansion

Does this mean everytime we need to compute something we need to do the calculations withthis HUGE

wavefunction?

Fortunately, not!

occupation formalism

It suffices to decleare how many(NOT which ones) electrons are in each single-paticle state ψ(χi), e.g.

1s22s2p4

The Slater determinant produces the full ψ.

Second quantizationThe Slater determinant is not very

convenient for 2 reasons:

Α)To compute inner products(or mean values of operators) we need to compute a very large number of such inner products(a typical number of electrons in an atom is of the order of 100)

Β)If we have interactions(as in perturbation theory) we have linear combinations of such determinants

Can we make our life easier?

• Recall harmonic oscillator: creation and destruction operators

• Describe a state as follows

• means Ni particles(e.g. Electrons) in the ith state

We can describe transitions via the creation and destruction operators

• Thus

• An obvious check is that we cannot destroy an electron in state j(e.g. 5p) if there is no electron there to start with.

Orthogonality

• Creation operator

• Obviously a transition is represented by the product of a destruction and a creation operator

How do we represent operators?It suffices to produce the matrix elements <i|O|j>

if Ο is a one-body operator Ο(x1), e.g. the position or momentum of a particle

Ο=Σij <i|O|j>a+i aj where i,j are one-electron states , e.g. 1s,2p,…

Such an operator can cause a transition from j to any i(even i=j)

We also have two-body operators, such as the interaction between two particles which depends on their positions: V(x1,x2) we have an integral over x1,x2 in the inner product

V=(1/2)Σijkm <ik|O|jm>a+i a+

kaj am

i.e. Such an operator can cause a transition and in one interaction the states of two particles can change

We do NOT deal with determinants because we are interested not in the wavefunction itself, but in the results of physical processes.

Spin: Rules

• Can take either integer or half-integer values-but only one category depending on the nature of the particle(fermion/boson)

• Independent of the particle motion(like an internal degree of freedom)

• For a given particle, constant value(e.g. ½ for electrons, protons, neutrons,1 for photons, 0 for pions and kaons)

• There is no classical description for the internal degree of freedom associated with spin

For spin to be relevant to our observations, it needs to have some interaction

• Strong and weak nuclear force depends on spin

• Electromagnetic interactions: Each particle with spin and charge has a magnetic moment

Lande factor, g ≈ 2

Addition of angular momenta• What happens in a system with a number of

angular momenta?• e.g.• Classical example: Solar system: We have

rotation around the sun plus internal rotation. The total angular momentum, is conserved, not necessarily the individual ones

• Same here: angular momentum of an isolated system is conserved, but what we have is subsystems that are noninteracting in some approximation

Addition of angular momenta• A very general issue. main idea: It's the total,

not the individual angular momentum that is conserved(e.g. The solar system and planets example)

• Might refer to:• - addition of different angular momenta on the

SAME body(e.g spin and orbital ang. Momenta on the same electron)

• -addition of angular momenta from different constituents of a body(e.g. A many-electron wavefunction)

Again basis change• We had a basis consisting of the products

of the eigenfunctions of J21,Jz1 and of

J22,Jz2

Now J21,J2

2 and J2=(J1+J

2)2 ,Jz

Clebsch-Gordan(alrebraic coefficient)

These two bases are in the same vector space

Any vector/function of the vector space can be written as a linear combination of any of the two bases

Hence the basis vectors of one basis can be written as linear combinations of the basis vectors of the other basis

The coefficients of the linear combinations(the inner products) are precisely the Clebsch-Gordan coefficients

Definition and Notations

In bibliography we find the following notations:

Possible values of JTriangular inequality:

Easier way: Μ=m1+m2=>Mmax= j1-j2 <=Jmax

|Mmin|=|m1+m2|min=|j1-j2|>= Jmin

Dimension of the vector space(=number of basis vectors) is independent of the basis(in 3d, we can choose cartesian, spherical or cylindric basis, but need 3 vectors in any

case). Hence the dimensions of the vector space=(2j1+1)

(2j2+1)=∑J=|j1-j2|

j1+j2 (2J+1)

maximum Μ=J1+J2

Jmax=J1+J2

if m=J1+J2-1, two ways: m1=J1-1,m2=J2, η m1=J1,m2=J

2-1

etc

How many eigenstates do we have?

• In the initial basis we had (2j1+1)(2j2+1)

• Basis change does not change the vector space dimensionality. Hence in the new basis we have

∑j=|j1-j2| j1+j2(2j+1) =(2j1+1)(2j2+1)

Practically the addition of angular momenta is very important

• In many cases(L-S or Russel-Saunders coupling) the orbital angular momentum and spin are two completely independent systems, so we compute the total L and total S. These two systems are connected with weak forces that are spin-dependent, and give eigenstates with specified total angular momentum.

L-S Coupling-For light atoms (Z<~30)

-orbital angular momenta interact to give a total orbital angular momentum. Spins also interact to give a total spin. Finally total orbital angular momentum and total spin interact to give a total angular momentum

-Lifts the degeneracy for parallel and antiparallel spins-thus explains spectra to a large extent

-True for weak magnetic fields

At the other end(Ζ>30) the spin-related forces are especially strong

• Interactions between spin and orbital angular momentum are more important than interactions between orbital-orbital or spin-spin

• We can ignore interactions between particles, but not spin. Every particle(electron) is described by a definite TOTAL angular momentum and the eigenstates of the Hamiltonian arise by adding the total angular momenta of electrons(j-j coupling)

• Note that in general it DOES matter which type of addition is valid. Mathematically of course these are just different bases and we can always describe a vector in any basis as a linear combination of the basis vectors of another base of the same vector space.

More specifically, the Hamiltonian has a term of the form

(a∑iLi+b ∑iSi)2 or [∑i(aLi+bSi)2]

• Since L,S are vectors and the Hamiltonian is not, so it must contain some sort of inner product

• The first extreme case is L-S, i.e.• • Η’=

• And the second one j-j: In the first we add together the orbital angular momenta of all electrons i of the system to get the total orbital angula rmomentum and also all the spins together to get the total spin and finally we add the total orbital angular momentum with the total spin. In jj coupling we add orbital angular momentum and spin for each electron and at the end we add the total angular momentum of each electron

• There are also other different forms, such as intermediate coupling JK

Technically, the issue is finding the base change coefficients

• From independent angular momenta to total

• Note that for the largest m=j1+j2, the coefficient is 1 since only φj1m1 χj2m2 contribute

• If we act with the operator (J(1)-+J(2)

-)((1) and 2 refer to the two 2 angular momenta)

How do we compute them?• Algebraically. We do not do integrals.

There is a closed-form formula, but it is too complicated

• For complex situations, there are tables-nowadays this is done by a computer routine

• It is important to understand how this works:

Recursion relations

But J±=J1± + J2± ,hence acting with J1± + J2± on the expansion of |JM>:

In the last line we set m1->m1± 1, m2->m2± 1 in the sum(in any event the total m of each side is the same in the CG). Comparing:

ℏ

Use of recursion relation for construction

For + and maximum=J:

0=C+(j1,m1−1)(j1m1−1j2m2 |JJ) +C+ (j2,m2−1) (j1m1j2m2 −1|JJ).

take <j1 j1 j2 J-j1|JJ> positive

Special cases

(j1j1j2j2 | (j1 + j2)(j1 + j2)) = 1.

J+ gives 0 when acting on this state

How do we move to other J<j1+j2? Orthogonality +normalization, see problems

Examplee.g. L=J1=1,S=j2=1/2

<1 1 1/2 ½|3/2 3/2>=1

If there is only one way to have this result, then the coefficient is absolutely 1(+1 or -1: + for the maximum)

In other words:

| j1=1 m1=1 j2=½ m2=½>=|j1=1 j2=½ J=3/2 M=3/2>

How do we construct the rest?

Use of reduction

• Μ =m1+m2. (2j1+1) values for m1, (2j2+1) values for m2=>(2j1+1)(2j2+1) values for Μ=> may be represented as a(matrix (2j1+1)(2j2+1), e.g. j1=3,j2=1/2

m1=-3 m1=-2 m1=-1 m1=0 m1=1 m1=2 m1=3

m2=-1/2

m2=1/2 CG=1

A more general case: j1=3,j2=2

Each dot a combination |j1m1>|j2m2>

Blue dots: m1+m2=2

For specified J, use recursion connect the red dots

As in the previous example the top right corner gives CG=1. The next (2,2),(3,1) give the CGs for J=5 and two different CG for J=4

For J<j1+j2, each dot gives a CG

Green dots connected via J-:

In that case the matrix loses its ends

e.g.J=3<3+2. So (2J+1)=7 states

Largest M for:

(m1,m

2)=(1,2),(2,1),(3,0)

The lowering operator gives 4 elements with M=2(next parallel line)

Start with largest m1 If we use

the recusrsions(e.g. Red dots), the blue dot here is connected to (2,1) and the dot right above it, which is however 0 because it gives an Μ>j1+J2

Symmetries

Laser foundamentals• One category of particles(bosons) that includes

photons like company!• They prefer to be all in the same state• consequently:• Α) photon emission is easier when there is

already a photon present • Β) The emitted photon much prefers to have the

IDENTICAL characteristics(color, direction, polarization etc) as the existing photon=> monochromaticity, directionality

Basic ideas• (without math) • 3 ingredients: • Pumping mechanism provides energy and achieves population

inversion• Resonance chamber with mirrors at the end• Active Medium

Active medium is the material that produces and amplifies the radiation(and gives the laser its name, e.g. ruby, Nd-Yag, gas etc laser)

This is what we exploit to create an avalanche effect

• We have a perfectly reflecting mirror at the one end and a semitransparent(= lets some, but not all radiation through and reflects the remaining radiation) mirror at the other end.

• The photons that do not cross traverse the tube and cause more emissions with the same radiation characteristics(frequency, polarization, diraction), which in turn add up to the photons already traversing the tube. Every time a part of them escapes as a laser beam from the semitransparent mirror.

Laser not in operation

Atoms in the ground state

mirrotSemitransparent mirror(reflects a part and lets the rest through)

Through an electric discharge or radiation, we exhite the atoms of the active medium in the tube

Excited atoms

Some methods of excitation• α) gas discharge, e.g. CO2 laser and Argon Ions• Excitation via molecular/ionic collisions with electrons. This can be a

2-step process: creation of free electrons and collisions with molecules/ions

• β) Optical pumping, e.g. dye laser excited by argon ion laser. Other ways of excitation: laser, lamps(flash-lamps arc-lamps) etc.

• γ)electrical excitation• Current for semiconductor lasers• (for excimer laser), high voltage electron beam

Some atoms deexcite by emitting photons

Light emission

Some of the emitted photons move parallel to the tube axis, some of thise reflect off the semitransparent mirrors at the end.

The photons traversing these paths induce other excited atoms to emmit photons with the SAME characteristics(color, directionality etc)

Which is just what the laser is: Monochromatic radiation with the

same phase and direction

Classical fields

Maxwell equations:

Stimulated emission and absorption

• Hamiltonian: Atom+EM field

• EM field described via Ε(x,t), B(x,t) or Α(x,t), φ

q=-e electonic charge

Atomic hamiltonian(e.g. hydrogen)

The gauge problem

• The Ε,Β fields are invariant under a gauge transformation with χ any function

Consequently we are free to choose gauge

Lorenz gauge:∇.Α+c-1 ∂φ/∂t=0(Gauss)- ∇.Α+c-2 ∂φ/∂t=0(SI)

Things are simplest for the Coulomb gauge: φ=0, ∇.Α=0 –separates equations for φ, Α. The wave equation depends on the transverse current

The perturbation in the Coulomb gauge

H1 was:

becomes(P.A=A.P): Η1=(q/2Μc)[2A.P]=-eA.P/mc

small

Monochromatic perturbation: Α=2Αοε cos(ω n.r/c –ωt)

Unit polarization vector

Unit propagation vector

Golden Rule

Form for harmonic (H’~exp(iωt)) perturbation:

W =(2π/ ℏ)ρ(Ε) |H’mk|2

/c2

Dipole approximation:λ=2πc/ω >> atomic dimensions,

So we need the momentum matrix

• [p2,x]=-2i ℏ p

• <f|p|j>=(i/2 )ℏ <f|[p2,x]|j>=(2mi/2ℏ)<f|[Ho,x]|j>

• Since the rest of Ho does not involve

momentum and commutes with x:

• <f|[Ho,x]|j>=<f|H

ox-xH

o|j>=(E

f-E

j)<f|x|j>

• <f|p|j>=[im(Ef-E

j)/ ℏ]<f|x|j>=im(ω

fj)<f|x|j>

slater determinant - university of crete · slater determinant •antisymmetric case: •initial...

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