The Student Room Group

Scroll to see replies

Reply 20
Worzo
Well there's the deeper meanings and the basics, true. That's about 5% of a physics degree. Most of it is taken up with applying the basic things that you've learnt, and that's a whole lot harder and more intellectual than you might think. What I was trying to say was that you don't have to ponder the big questions in order to ponder hard questions.


Nothing! You may well be, as far as I know, hence my (!) after I said it. Someone taking on an OU physics degree is pretty hardcore, as far as I'm concerned - you must be on the ball.


You'll be happy to know that my colleague thought you were pretty on the ball too, he suspected you may be looking at a good pass.

I know what you mean though, I think anyone taking on a physics degree better be pretty on the ball, or your probably going to be flummoxed, but then as you say I am prepared, and forewarned is fore armed, let's face it anyone who thinks physics is going to be easy is an idiot and probably shouldn't be taking physics, I don't think your the first to warn me of this, but I thank you for it anyway.

I chose physics because I love the ideas behind it, even classical mechanics has a mathematical elegance, quantum is weird enough to make you go bwah! I think that's probably one of the most important things to be motivated, you might not enjoy all of it, but you have to be keen.

Again thanks for the advice, coming from someone who knows first hand, any advice is more than welcome.
I studied theoretical physics back in the 1970s. I'm still as passionate about it today.

Be careful. Physics is addictive.
Reply 22
Actually after talking to some people about it I'm feeling somewhat less passionate about it than I was. But I'm sure it'll pass.

In the interest of well just about anyone who cares why things are the way they are here is an excerpt copied as best I can from the original work.


Fundamentals of modern physics by R.M.Eisenberg 1967.pg.128-131.


6, The Wilson-Somerfeld Quantization Rules

The success of the Bohr Theory, as measured by it's agreement with experiment, was certainly very striking. But it only accentuated the mysterious nature of the postulates on which it was based. One of the biggest of these mysteries was the question of the relation between Bohr's quantiztion of the angular momentum of an electron moving a circular orbit and Planck's quantization of the total energy of an entity, such as an electron, executing simple harmonic motion. In 1916 some light was shed upon this by Wilson and Sommerfield, who enunciated a set of rules for the quantization of any physical system for which the coordinates are periodic functions of time. These rules included both the Planck and the Bohr quantization of special cases. They were also of considerable use in broadening the range of applicability to quantum theory. These rules can be stated as followed:

For any system in which the coordinates are periodic functions of time, there exists a quantum condition for each coordinate. These quantum conditions are.

Unparseable latex formula:

$\oint P_qdq=n_qh$

(5-20)

Where q is one of the co-ordinates, Pq is the momentum assosciated with the coordinate,nq is the quantum number which takes the integral values, and \oint means the integration is taken over the co-ordinate q

There are two tables here, showing a wave function and roFigure(5.10)

showing the time dependence of coordinates I'll try and reproduce them in mathcad or something at some point, if I can.

The means of these rules can be best illustrated in terms of some specific examples.

Consider a particle of mass m moving with constant angular velocity ω\omega in a circular orbit radius ro. The position of the particle can be specified by the polar coordinates and θ\theta. The behaviour of these two coordinates is shown in fig 5-10(apologies not present)as functions of time. They are both periodic functions of time, if we consider r=ro to be a limiting case of this behaviour. The momentum associated with the angular coordinate θ\theta is the angular momentum L=mr2dθdtL=mr^2 \frac{d\theta}{dt}. The momentum associated with the radial coordinater t is the radial momentum pr=mdrdtp_r=m\frac{dr}{dt}. In the present case r=ro,drdt=0\frac{dr}{dt}=0 anddθdt=ω\frac{d\theta}{dt}=\omega a constant. Thus L=mr02ωL=mr_0^2\omega and Pr = 0. We do not need to apply equation (5-20) to the radial coordinate in the limiting case in which the coordinate is a constant. The application of the equation oto the angular coordinate θ\theta is easy to carry through for the present example. We have q=θq=\theta and Pq=LP_q=L, a constant. Write nq=nn_q=n; then

Unparseable latex formula:

$\oint P_q dq=\oint Ld\theta=L\oint d\theta =L\int_0^{2\pi}=2\pi=2\pi L $



So the condition

Unparseable latex formula:

$\oint P_q dq=n_qh




Here ends part one, more maths to follow when I can be arsed to type the rest out.:smile:


Anyway provided By DR EPABailey from my work. To answer his own question, and something he thought those who are learned in the ways of physics would like to know?


[Part 2]Those who dislike bare quantum equations should look away.

Unparseable latex formula:

$2\pi L=nh$



Unparseable latex formula:

$L=nh/2\pi\equiv n\hbar



Diagram Simple:- sin wave between -x0 and x0 with t on the x axis

figure5-11. The time dependence coordinate of a simple harmonic oscillator.

Position of the particle can be specified by the single linear coordinatex. The behavious of this coordinate with time is illustrated in (5.11) and can be expressed.

Unparseable latex formula:

$x=x_0 sin 2\pi vt=x_0 sin \omega t, \omega\equiv 2\piv$



Where x0 is the amplitude of oscillation. The momentum of the particle is

Unparseable latex formula:

$p=m\frac{dx}{dt}=mx_0\omega cos \omega t$



to evaluate this integral it is convenient to express
Unparseable latex formula:

$cos \omegat$

in terms of x:

Unparseable latex formula:

$x^2=x_0^2 sin^2\omega t =x_0^2(1-cos^2\omega t)$



Then

Unparseable latex formula:

$cos^2\omega t=\frac{x_0^2-x^2}{x_0^2}, cos\omega t=\frac{\sqrt {x_0^2-x^2}}{x_0}$



and

Unparseable latex formula:

$\oint p dx=mx_0\omega\oint \frac{\sqrt {x_0^2-x^2}}{x_0} dx=m\omega... 4\int_0^{x_0} \sqrt{x_0^2-x^2}dx$



The last step depends upon the fact that

Unparseable latex formula:

$\oint = \int_0^{x_0} + \int_{x_0}^0 + \int_0^{-x_0} + \int_{-x_0}^{0} = 4\int_{0}^{x_0}$



Because the integrand is an even function of x, this gives

Unparseable latex formula:

\oint p dx=m\omega 4\left[ \frac{x\sqrt x_0^2-x^2}{2} +\frac{x_0^2}{2} sin^{-1}\frac{x}{x_0}\right]_{0}^{x_0}$



Unparseable latex formula:

$\oint pdx=m\omega 2(x_0^2sin^{-1} 1-x_0^2sin^{-1} 0)



Unparseable latex formula:

$pdx=mx_0^2\omega\pi



We would like to write this in terms of the total enrgy E of the particle; it is easy to do if we recall that the total energy of a harmonic oscillator is equal its kinetic energy when x=0. So

E=12m(dxdt)t=02=12m(x0ωcosωt)t=02=12]mx02ω2E=\frac{1}{2}m\left(\frac{dx}{dt}\right)_{t=0}^{2}=\frac{1}{2}m \left( x_0\omega cos\omega t \right)_{t=0}^{2}=\frac{1}{2}]mx_0{^2}\omega^2

Since x = 0 when t = 0. Thus we have.

Unparseable latex formula:

$\oint pdx=\frac{1}{2}mx_0{^2}}\omega^2 \frac{2\pi}{\omega}=\frac{E}{v}$



and

Ev=nqhnh\frac{E}{v}=n_qh\equiv nh,

E=nhvE=nhv

Which is identical with Plancks quantization law.

7. Sommerfield's Relativistic theory

One of the important applications of the Wilson-Sommerfield quantization rules was to the case of a hydrogen atom in which it was assumed that the electron could move in elliptical orbits. This was done by Sommerfield in order to explain the fine structure of the hydrogen spectrum. The fine structure is the splitting of the spectral lines, into several distinct components,which is found in all atomic spectra. It can be observed using only by using equipment of very high resolution since separation, in terms of wave number, between adjacent components of single order of the spectral line 10-4 times the separation between adjacent lines. According to the Bohr theory, this must mean that what we had thought was a single energy state of the hydrogen atom actually consists of several states which are very close together in energy.



I bet that was fun:smile:, one answer. But it's pre dates Schrödingers operator approach. There apparently is a Bohm model as well, which is not as good as this approach, according to him.
Reply 23
Ok this might need a bit of explanation to the laymen.

To be honest I don't follow most of the steps, my integration isn't that good, so any corrections would be welcome.

Integration is the area under the curve on a graph so:-



Simple sinosoidal wave. expressed by sinθsin\theta

In this case the area under the peaks and troughs of the wave is the Integral. Essentially what the above equation does is place the particles momentum and thus it's energy in terms of \hbar or planks constant by using our initial equation for the momentum of the partical between -xo and xo or the amplitude of wave at maximum value and minimum.

And then solve the initial equation in terms of the simple harmonic oscillator for an electron, the answer given correlates with the energy of the wave, which in term agrees with the equation e=hnv and explains why unlike other waves an electron or a photon or whatever has the equation e=hnv

where h is planks constant.

instead of e=Av^2

Where A is a constant as in a water wave or guitar string or whatever.
Reply 24
Sidhe
Ok this might need a bit of explanation to the laymen.

To be honest I don't follow most of the steps, my integration isn't that good, so any corrections would be welcome.

Seems right to me. Quantising the angular momentum and relating this to the linear momentum under a simple harmonic oscillator will indeed reveal energies quantised in units of ν\hbar\nu. A more sophisticated approach solves the Schrodinger equation for a simple harmonic oscillator potential using operator methods, and the energy eigenvalues reveal the same relationship. They also show a positive energy ground state, which the Sommerfeld approach does not, because a zero-energy quantum state has no wavefunction, and thus does not exist.
Reply 25
Worzo
Seems right to me. Quantising the angular momentum and relating this to the linear momentum under a simple harmonic oscillator will indeed reveal energies quantised in units of ν\hbar\nu. A more sophisticated approach solves the Schrodinger equation for a simple harmonic oscillator potential using operator methods, and the energy eigenvalues reveal the same relationship. They also show a positive energy ground state, which the Sommerfeld approach does not, because a zero-energy quantum state has no wavefunction, and thus does not exist.


Ah good, I'm glad I followed it correctly. thanks.

He sent it to me by post from work(which was nice of him) Probably above my level of study, but what the heck, it's interesting anyway. I figured one or two people might appreciate a bit of quantum mechanics history.

Latest