Problem 1.5

The time-averaged potential of a neutral hydrogen atom is given by

where q is the magnitude of the electronic charge, and being the Bohr radius. Find the distribution of charge( both continuous and discrete) that will give this potential and interpret your result physically.

Solution:

We can solve the distribution of charge by solving first the charge density using the Poisson’s equation

We are given a hint to find the discrete distribution of charge, meaning likely a delta function will be in our answer.

Now, we solve Poisson’s equation for ,

where .

But . So,

Using the product rule on the first term, we obtain

Then we distribute the term and we get

And then using product rule, we have

After the terms cancel, we are only left with

Using our delta function equation for the first term,

thus

Physically, this is the point charge of the proton nucleus represented by the

delta function at the center of the atom, surrounded by the negative electron

cloud.

MSPhysics1-MSU-IIT

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Problem 1.10
Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

Proof:

To prove this problem, we are going to use the Green’s  Second Identity which is given by,

Choosing (the scalar potential), and be the integration variable, we have

.       Eq.(1)

Let us solve Eq.(1) term by term. For the first integral,

,   since

,  since if  V contains .

For the second integral,

,  since .

But because there is no charge in the volume that we are integrating(Charge-free) . So the second integral becomes

.

For the third integral,

.

For the fourth integral,

.

But , then

.

Using Divergence Theorem,

,

the fourth integral becomes

.

But and (again, this is true for a charge-free volume! ), then the fourth integral  would be equal to zero, that is,

.

Thus, Eq.(1) is simplified into

.

Hence, the scalar potential is then equal to

.         Eq.(2)

Now, we have proven the mean value theorem.  Eq.(2) says that the potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

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Use Gauss’s theorem and to prove the following:

(a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.)

Solution:

Suppose that the field were initially nonzero. Since this is a conductor, any charges in the interior would move in response to the field. After a time, this process stops since the moving charges produce currents which dissipate energy. The final configuration would then have charges arranged so that the interior is zero. Recall that in equilibrium, the electric field inside a conductor is zero. Since everywhere inside the conductor, then from Gauss’s Law, , the charge density everywhere in the interior. Therefore, every point inside a conductor has zero charge, and any excess charge can only reside on the surface of the conductor.

(b)  A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it.

Solution:

Part 1. Consider the charge exterior to the conductor which produces an electric field, as shown in the figure.

The electric field in the conductor is zero, with induced charge densities on the exterior and interior surfaces of the conductor.

i) Imagine moving a charge on the interior surface from point A to point B along path 2 which goes throughout the conductor itself. Since in the conductor, along this path.

ii) Move the same charge from A to B along path 1, in the interior cavity of the conductor. Since the electrostatic field is conservative, along its path.

This must be true also for any path in the interior. So generally, in the interior. Therefore the conductor shields its interior from field due to charge placed outside.

Part 2. Consider a positive charge placed inside a hollow conductor as shown in the figure.

The charge induces a charge density in the interior surface of the conductor in such a way that the electric field in the interior of the conductor is zero. Assuming that  the conductor is charge neutral, this means that there is an induced charge density on the exterior surface of total charge Q. If we apply Gauss’s Law to the Gaussian surface G surrounding the conductor, the total charge enclosed is still . Therefore, there is an electric field outside the conductor.

(c) The electric field at the surface of a conductor is normal to the surface and has a magnitude , where is the charge density per unit area on the surface.

Solution:

Note that in equilibrium, the field at exterior surface must be normal to the surface, so that the tangential component is zero. The magnitude of the field is derived using Gauss’s Law with a Gaussian pillbox which cuts through the surface. The electric field is zero on the conducting side of the pillbox. So,

, with the area on the surface.

Rearranging, we get

.

Define as the charge per unit area ,

.

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Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.

To prove this formula, we use the following:

Where: and

Using the equation above:

We can factor out in the first term to give:

Note that for the second term, the permutation of indices are odd, rearranging them to ijk will give the negative:

Thus,

Prove: ___________________________________________________________

Proof: First, we define the following vectors as:

; and

Now, 

if we let i=k, then . Furthermore,

Now, the derivative of orthonormal basis , that is, and the derivative of a coordinate X, . Also, , thus

=

=

=

=

It is noted that . Therefore,

=

This is further equivalent to the ratio of the component of perpendicular to , that is

=

since .

To show their equivalence, we use the BAC-CAB Rule in the definition of . So,

=

=

Thus, we conclude that

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MSPhysics1

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We are going to prove the following vector identity using Levi-Civita symbol:

Solution:

Let    ,     ,   ,   .

Then,

By definition:

We have to let m=n so that,

Levi-Civita symbol can be expressed in terms of Kronecker delta given by:

Thus,

Show that
(a)
and
(b)

Solution:
Let

and

(a)

since

and

Thus,

and

(b)

since

and

Therefore,

About Me:
I am Yusof-Den Jamasali, an MS Physics student of Mindanao State University – Iligan Institute of Technology. I like painting, and drawing. And loves to eat avocado and carrot.

In quantum mechanics, relativistic correction to the energy levels of a system is used when it is introduced by a little disturbance we often recognized as . Fine structure is an example of this where the splitting of spectral lines of atoms is due to its first-order relativistic corrections. Here is an example of finding the first-order relativistic corrections of a given system.

Our task is to find the lowest-order relativistic corrections to the energy levels of the one-dimensional harmonic oscillator.

Note: Our reference through all these is Jackson’s book of Quantum Mechanics

Start:

We begin by eq’n 6.52 ,

where

and

So,

with

Substituting this, we get

                    (1)

We now introduce the ladder operators. That is,

Using these, we could then derive basing on Eq’n 2.69,

Note that only equal numbers of  raising and lowering operators will survive.

By eq’n 2.66,

Going back to (1) to get ,

Thus, the lowest-order relativistic correction of one-dimensional harmonic oscillator is

In the May 15, 1935 issue of Physical Review , Albert Einstein co-authored a paper with his two postdoctoral research associates at the Institute for Advanced Study, Boris Podolsky and Nathan Rosen. The paper, known as EPR, became a centerpiece in debates, challenging the validity of Quantum Theory.

The paper features a striking case where two quantum systems interact in such a way as to link both their spatial coordinates in a certain direction and also their linear momenta (in the same direction). As a result of this “entanglement”, determining either position or momentum for one system would fix (respectively) the position or the momentum of the other.  In quantum mechanics, in the case of two physical quantities described by non- commuting operators, the knowledge of one prevents the knowledge of the other (Heisenberg Uncertainty Principle). Thus, the paper asserts that, either (1) the quantum- mechanical description of reality given by the wave function is not complete or (2) when the operators corresponding to two physical quantities, like position and momentum, do not commute the two quantities can not have simultaneous reality. The authors affirm that one or another of these assertions must hold, giving rise to these two premises: (1) if quantum mechanics were complete (first option failed) then the second option would hold, that is, incompatible quantities cannot have real values simultaneously but (2) that if quantum mechanics were complete, then incompatible quantities (in particular position and momentum) could indeed have simultaneous, real values. They conclude that quantum mechanics is incomplete. The conclusion certainly follows since otherwise if the theory were complete one would have a contradiction. To establish the premises, the authors discuss the idea of a complete theory, offering only a necessary condition. In order for a theory to be complete, every element of the physical reality must have a counterpart in the physical theory, further requiring the criterion: If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of reality corresponding to that quantity. This is the famous EPR Criterion of Reality.

To realize their assertions further, the authors provided a thought experiment wherein the two quantum systems (one system is labeled Albert while the other system is named Neils) interact in such a way that  conservation of relative position and momentum hold following their interaction. The paper constructs an explicit wave function for the combined system that satisfies both conservation principles. The critical point in the paper centers on the two assumptions made by the authors, namely, separability and locality.  The first assumption states that at the time when measurements will be performed on Albert’s system there is some reality that pertains to Niels’ system alone. In effect, they assume that Niels’ system maintains its separate identity even though it is correlated with Albert’s. The second assumption supposes that no real change can take place in Niels’ system as a consequence of a measurement made on Albert’s system. Locality implies that the prediction of the position of Niels’ system does not involve any change in the reality of Niels’ system. Since the prediction does not disturb Neils’ system, all the pieces are in place to apply the Criterion of Reality. Hence, the authors concluded that Niels’ system can have real values or elements of reality for both position and momentum simultaneously. The negation of the first premise leads to the negation of the only alternative.

Following the result of the thought experiment, separability, locality as well as the application of the Criterion of Reality, EPR concludes that quantum- mechanical description of a physical reality given by the wave functions is not complete.

Reference:

A. EINSTEIN, N. ROSEN and B. PODOLSKY, Phys. Rev. 47, 777 (1935).

In quantum mechanics, the presence of spin-orbit coupling gives rise to the Hamiltonian that will no longer commute with , and , so the spin and orbital momenta are not separately conserved.

In order to understand this concept better, a commutation problem for orbital angular momentum , spin , and spin-orbital momentum is presented here.

• Consider the fundamental commutation relations for angular momentum. The individual components of the spin do not commute with each other. That is,

;         ;            ;

• The commutation relation for the ‘instrinsic’ angular momentum is much like a ‘carbon copy’ to that of the ‘extrinsic’ angular momentum

;         ;            ;

But the spin-obit Hamiltonian does commute with , and the total angular momentum which is

From the given fundamental commutation relations above, we then now seek the commutators of the following commutation relations:

(a.)           (b.)           (c.)

(d.)           (e.)             (f.)  

As a hint here, and satisfy the fundamental commutation relations for angular momentum, but and commute with each other.

(a.)

same goes for the other two-components, so

(b.) is identical only with

(c.)

(d.) commutes with all the components of (and ), so

(e.) Likewise,

(f.)

where

The first, second and third terms vanish from the results of (d.) and (e.). Thus,

This means that the quantities , and are conserved. That is, the eigenstates of and are not “good” states to use in perturbation theory, but the eigenstates of , and are.

The problem presented here is based on the problem 6.16 from D.J. Griffiths “Introduction to Quantum Mechanics”.

About the author: Marichu T. Miscala is currently taking up her masters in Mindanao State University – Iligan Institute of Technology. She is most interested in pursuing a career in advanced research.