Electric Dipole Moment

Definition of Electric Dipole Moment

The Energy of a system of charges is defined as:

ε=∫ρΦ dV
where ρ is the charge density and Φ the potential both functions of coordinates.Choosing a particular origin and expanding the potential in a Taylor series about the origin:

ε=∫ρ{Φ(r₀)+x_i&part_i&Phi(r₀)+½ x_ix_j&part_i&part_j&Phi(r₀)+...}

ε=∫ρΦ(r₀)+d_iE_i+d_ijE_iE_j+...

where d_i is the Electric Dipole Moment and d_ij is the Induced Dipole Moment and E the electric field.

For discrete charges the dipole moment is

d=&Sigma_ir_ie_i

where i denotes the sum over the charges.

EDM for a system
Let P is the Parity operator.If P is conserved then the energy eigenstates are also
parity eigenstates.Under a parity transformation

&lang m|P⁺PdP⁺P|m &rang =-&lang m|d|m &rang
.
So the dipole moment must vanish.A non-vanishing dipole moment signals violation of parity.


Let us consider non-degenerate energy eigenstates characterised by only by its angular momentum operator J.

Under the action of time reversal,

T d T^-1 =d

T J T^-1 =-J

implying the EDM has to vanish.

So, the presense of an electric dipole moment would violate both time reversal and parity.

However, a system possessing non-degenerate eigenstates which are parity eigenstates as well can have a non vanishing dipole moment.This is because the energy shift is quadratic in &Delta² where &Delta is &lang+|zE_z|- &rang and therefore quadratic in E.So it is an induced dipole moment and doesnt imply the violation of time reversal.


When the system is in a degenerate state ( for example the ground state of a water molecule), the energy shift can be linear in E, but then it does not imply violation of time reversal invariance.This is usually called the permanent dipole moment.

Projection Operators

Permutation Operator: The Permutation Operator is defined by its action on a N-particle wavefunction as follows:
p_ijψ(...,i,...,j,...)= ψ(...,j,...,i,...) ( 1)
In general, p_ij²= e^2iφ
The order of the permutation group of order N is N!.Further, if the hilbert space of each single particle state is d,dimension of the total hilbert space formed is d^N.The particles are assumed to be indistinguishable.

Completely Symmetric & Completely Antisymmetric States and Projection Operators:
Notation: |i1,...,iα,...,iN> is a multiparticle state which means that particle no α is in a state denoted by |iα>.If the |i>'s form a complete orthonormal system in the space of N-particle system,the completely symmetric and completely antisymmetric basis states is defined by:
S_±|i1,...,iα,...,iN>=(1/√ N!) Σ_p(±)^pp|i1,...,iα,...,iN> (2)
By definition,it is clear that S_± are 1-dimensional subspaces.This we can also prove by writing down the projection operators onto these states.The trace of the projection perators is equal to the dimensionality of the hilbert space onto which they project as can be seen from the example of a three dimensional vector.From this elementary consideration, the following properties of the projection operators can be formulated:
1.Tr P < D,where D is the dimensionality of the hilbert space.
2.Tr P is always integral.
3.If Tr P =0 then it projects on to the null space.
4.If P_iP_j=&delta_ij and Tr P_i=1, then i can lie between 1 and D.
5.Iff 1 ≤ i ≤ D, then Σ _i P_i=I

Now, let us give some theorems and their proofs.
Theorem 1: The trace of a projection operator is equal to the dimension of the Hilbert space over which it projects.
Proof:Let the basis of an d dimensional Hilbert space be denoted as {|i 〉 }.The projection operator P_r which projects over a r dimensional subspace can then be written as Σ _k=1 ^r |k 〉 〈 k|.Then Tr P= Σ_l=1 ^D 〈 l| P |l 〉 = Σ_l=1 ^D 〈l Σ_k=1 ^r |k 〉 〈 k|l 〉 =Σ_k=1 ^r Σ_l=1 ^D 〈 l|k 〉〈 k|l 〉=Σ_k=1 ^r1 = r, since
〈 k|l 〉=δ_kl.

Theorem 2:The Projection Operator on to the completely symmetric and the completely antisymmetric states are given by
P_±=(1/√ N!)S_± (2)
Proof:From our definition of the S_± operators we know that there can only be a single symmetric or anti-symmetric state.So let us write
P₊=c₊ S₊.We need to find c₊ such that P₊²=P₊
Now, P₊²|i1,...,iα,...,iN&rang =(1/N!)c₊²&Sigma_qpqp|i1,...,iα,...,iN&rang
=(1/N!)c₊²&Sigma_phh|i1,...,iα,...,iN&rang
=c₊²&Sigma_hh|i1,...,iα,...,iN&rang
(since if pq=h, then p=hq^(-1) and has N! distinct elements.)
=c₊²(√ N!)S₊|i1,...,iα,...,iN&rang
=c₊(√ N!)P₊|i1,...,iα,...,iN&rang
which implies c₊=(1/√ N!)
Thus:
P_±=(1/N!) &Sigma _p(±)^pp|i1,...,iα,...,iN&rang

Theorem 3:Tr P_±=1
Proof: We shall prove this theorem only for the case of P₊.The other one involves more complicated algebra, but gives the same result.
Tr P₊=&Sigma{|i 〉 } &lang i1,...,iα,...,iN|(1/N!)&Sigma_pp|i1,...,iα,...,iN&rang
Only p=1 gives non zero contribution and there are N! such identities are generated due to the sum on p and hence we have the result Tr P₊=1

noethers theorem

1. we proved the noether's theorem( for every continuous symmetry with end points fixed characterized by m parameters, there are exactly m conserved quantities characterised by Q_k, where Q_k=(delta L /delta q_dot)*(delta q / delta alpha_k), alpha ks are parameters.) this we did by writing the transformed q_i's as functions of q_j and alpha_k, effecting the variation of the action,setting it to zero for a symmetry and using the lagrange equations.
2.next, we also considered the cases where the end points vary and noted that such a variation of end points causes the lagrangian to remain otherwise unchanged.this variation causes a variation of the coordiantes at the boundary, but along the trajectory.the limits of integration change.for the action to remain unchanged, the change of lagrangian due to this must equal the change of lagrangian due to the coordinate variation.a short calcualtion then yielded the conservation of energy.
3.we also saw that the alpha_k 's form a lie algebra
4.we derived the energy momentum tensor of the field.we noted that the tensor was not unique and so certain tensors(anti-symmetric in the last two indices) can be added to it to symmetrize it.
5.we derived its form for the classical electromagnetic field, checked that the tensor was symmetric and traceless.
6.it was also noted that this tensor cannot always be diagonalised; since under such a transformation the metric (which is not an identity metric, it is a minkowski metric) must also be preserved.

further points of discussion

1.we discussed the possibilty of whether all canonical transformations can be represented by rotations of phase space and reasoned that in general it cannot.the reasoning follows the lines of symplectic structure of the poisson brackets which implies that canonical transformations cannot be
in general represented by rotations of phase space.
2.the second pair of maxwell equations
3. we saw that quantum mechanics is a field theory in some sense.this is because we can always obtain the klein-gordon equation (which describes quantum mechanics) from an action which can be said to describe a classical field theory in 0+1 dimension.
4.some discussions on path integrals

interesting points of fields in lower dimensions

there are some points about electrodynamics you can rather easily see from the field tensor.

in 1+1 dim, the field tensor is a 2X2 matrix with only one independent component.this component is the 01 component:which is one space and one time.so this is an electric field.note that E is a polar vector and so is the 01 component; hence the electric field is a scalar in 1+1 dimension.

all components exhausted, there is no B in 1+1 dimension!!

this also follows from Biot-savart law; the force on a current due to another current acts in a direction prependicular ot the line joining them, which is not possible in 1 space.

now lets come to 2 +1 dim.the electric field has 2 components.

what about the magnetic field ?

recall biot-savart law.force is given as a cross pdt, it has the structure of Fj=εjkvkB so it turns out that it is a scalar.also since F and v are polar vectors,B must ont change sign under parity and hence it is a true vector.

Bianchi identity: suppose u have an antisymmetric object and u have an operator that acts linearly on the tensor , then the sum of all cyclic permutations of the tensor add up to zero.
Fij=DjAi – DiAj (1)
define tijk=DkFij (2)
then the bianchi identity is
Tijk=tijk+tjki+tkij=0 (3)
as can be shown by substituting 1 & 2 in 3.

Next,using the anti-symmetric properties of Tijk in 3 we see that only 4 non-trivial equations result.knowing that this involves no source terms, we suspect these to be the source free maxwell equations.there are the right number of them!so the bianchi identity are the source free maxwell equations

why do you need a four-vector potential to characterize the field in the action formulation?well,this is because you have E from a scalar potential and B from a vector potential.

note that the convention E= -div (phi) comes from the convention that a positive charge moves in the direction of lower field.

the sturcture of the bianchi identity, ie, of some linear operator acting on an antisymetric structure is the same as for maxwell equations.

ED#

proceeding further, we wrote down the action of the complete field+particle system and then used the variational techniques with respect to different cocordinates to get maxwell equations in four-dimensional form.symmetry arguments for gauge invariance yielded the continuity equation.the properties of the field tensor in 1+1 and 2+1 dim were also deduced.

Into Electrodynamics

With the notational ideas cleared up, i move into the covariant formulation of electrodynamics.But first, a short review of relativistic mechanics.
Using the principle of least action,the lagrangian is determined to be
L=-mc^2*sqrt(1-v^2/c^2)
Then we go ahead to find the expressions for momentum and energy in the case of a free particle.the euler-lagrange equations of motion(ELEOM) give us the condition that the four-velocity must be a constant.
LL makes the comment that the energy in relativistic mechanics is always positive and proportional to the mass, unlike in class mech where it can be either positive or negative.That would be because in class mech u do have the potential energy of interaction of a system of particles.but to do so in rel. mech. one has to explicitly include the momentum and energy of the field produced by the interacting particles.
A short review was also made of the angular momentum tensor, its antisymmetric nature.

Next, we come to the section where the field is viewed as an physical entity that mediates interactions between particles, since relativity allows only contact interaction.A particle imparts the information about its motion to the field surrounding it, which then propagate it far away.Also, since relativiy disallows the concept of rigid bodies, the so called elementary particles must be pointlike.
the next sections are more mathematical.in them we use the four-vector A to characterise the field, obtain the lagrangian, the generalised momentum,the hamiltonian.Then we proceed further to obtain the ELEOM and define the elecric and magnetic fields.interestingly, we see that the work done on the particle is only by the electric field E, since the magnetic field B is always perpendicular to the velocity v.
Equations of motion in classical mechanics are time reversal invariant; the same thing is true in electrodynamics under the transformation t-> -t; E -> E ; B -> -B which in turn implies phi -> phi; A -> -A; phi and A being the components of the four vector A.
then i studied gauge invariance and solutions for E,B and energy in constant electomagnetic fields.

four vectors

Thought it would be wise to recall some of the concepts before proceeding further.So,I started with revising the section on Four Vectors of sec 6 of Landau-Lifshitz(LL).
there is one thing i wanted to clarify.A quantity that is a product of a pseudoscalar and a scalar is a pseudoscalar; the product of a pseudoscalar with another pseudoscalar would be a scalar.I can extend the same argument to vectors and tensors; and then i would claim that (levi-civita_iklm)*(levi-civita_prst) is a true tensor.

hi this is a new post..but theres a problem with this account..