# Particle Physics - Terms I don't Understand

1. Aug 16, 2004

### Nima

Hello, I've been doing some extra reading in Physics and there are some certain aspects that I don't understand. I'd appreciate some people explaining to me what they mean.

1.) Wave Function
2.) Intrinsic Angular momentum
3.) Relativistic
4.) The Coupling Constant
5.) Spin is parallel/anti-parallel to momentum
6.) Charge Multiplets
7.) Supermultiplets
8.) Isospin
9.) What are the properties associated with each of the defined colour charges? (red, blue, green) i.e.) "If this particle has a colour charge of red = 1, green = 0, blue = 0, it therefore has the following properties..."
10.) What is the Pauli exclusion principle.

Thanks in advance. :)

2. Aug 16, 2004

### zefram_c

Hmm. You've pretty much asked us to explain all of modern physics at the introductory level - a task which no one here is up to due to time constraints. Perhaps there are tutorials on this site. Other than that, keep reading! (What have you been reading, anyway?)

3. Aug 16, 2004

### Nima

It's so frustrating, because I have a big physics book and I have read the particle physics section, and I understand a lot of it but certain bits I don't understand due to not understaind key terms.

I also don't like it when they say things like:

"Dirac postulated that all negative-energy states were filled with electrons. They would, therefore, exert no force on anything and thus would not be observable."

Why won't they explain things, rather than just stating facts such as the above? Could someone explain why, which is what Physics should be about? I'm completely sick and irritated at the stupid fact stating in Physics books, it seems like nobody really understands this stuff, or the books keep it a secret.

4. Aug 16, 2004

### jcsd

Yep, but trying to understand what the Dirac sea is before you understand what the wavefunction is like trying to run before you can walk.

basically there is no way to restrict to the Dirac equation to postive-energy states (due to the relativistic nature of the equation tho' this was not a problem classically, but QM allows sponateous transistions between states) so Dirac postulated that the negative-energy states were already filled and thus transistion is prevented by the Pauli exclusion principle.

5. Aug 16, 2004

### Nima

OK, but could you explain the statement I quoted?

I don't understand why no net force is exerted and thus wouldn't be observable.

6. Aug 16, 2004

### chroot

Staff Emeritus
Nima,

Which big physics book are you reading?

- Warren

7. Aug 16, 2004

### Nima

Modern Physics - Fourth Edition

Authors: Paul A. Tipler and Ralph A. Llewellyn

8. Aug 16, 2004

### jcsd

Fom what I understand is the fact that you have an infinite amount of these negative electrons means that there actions cancel each other out leaving no net effect.

9. Aug 16, 2004

### Nima

Sorry, but what do you mean by "their actions cancel out"?

10. Aug 16, 2004

### jcsd

Though it's worth noting that the idea of the Dirac field has supplanted the idea of the Dirac sea.

11. Aug 16, 2004

### jcsd

What I mean is that the net-force they exert would be exactly zero, due to the fact there are an infinite number of them.

12. Aug 16, 2004

### Nima

Do you mean that the electrons exert forces equal in all directions, and hence the net force that acts on any objects will be zero?

If so, how does that relate to "and thus would not be observable"?

Why wouldn't the electrons in negative energy states be observable?

Last edited: Aug 16, 2004
13. Aug 16, 2004

### jcsd

Yes I do, but I lay no claim to being an expert.

They shouldn't usually be noticeable (but if they didn't have some effect they'd be no piint in postulating their existance), but when an electron goes from a postive to a negative-energy state, a vacancy is created. This vacancy behaves like a positron (i.e. in the model, which was the model that first predicted their existance, positrons are vacant negative-energy states).

I believe that there some problems with the model, for example, it's clear that the individual electrons in the negtaive-enrgy staes have charge (this is a key feature of modelling the positron), yet the infinite set of these 'negative' electrons have no charge or for that matter gravitational mass. But as I said this is not the model that is currently used.

14. Aug 16, 2004

### Nima

Yes, but I still don't know why they wouldn't be noticeable.

15. Aug 16, 2004

### jcsd

I can't say that I'm 100% on this myself, I can see why they might not be noticeable, but I can't see why they should defintely not be noticeable in all situations (excepting situations where a vacancy exists).

16. Aug 16, 2004

### zefram_c

They're not noticeable because - in the absence of gravity effects which we don't know how to include - there's no way to detect the fact that they're there; other than knocking one of them into a positive energy state and leaving a hole behind. Not that it matters anyway; it's much better to use the field theory picture where there are no electrons in the ground state. But QFT is probably too much for someone who doesn't know the meaning of the wave equation (no offense intended; no one was born knowing it). I recommend you keep reading - you will need at least basic QM and SR before you can study the Dirac field.

17. Aug 16, 2004

### jtolliver

The wave function(usually denoted ψ) contains all the information about the state of a particle. The probability of the particle being in a region is the integral of |ψ(x,t)|2 over that region. Also any observable is a linear operator(such as the derivative or just multiplication by x). The state has a well defined value of an observable Q if Qψ(x,t)=qψ(x,t) for some number(not operator) q, and in that case q is the value for the observable. The wave function for a superposition of states is just a linear combination of the wave functions for individual states.
In quantum mechanics the angular momentum operator($i\vec{r}\times \vec{\nabla}}$ in the absence of intrinsic angular momentum) is the generator of rotations. What that means is that when we rotate by a small angle &epsilon;(for simplicity i will assume the rotation is about the z axis) the new wave function is $\psi + i \epsilon J_z \psi$, where Jz is the operator corresponding to the z component of the angular momentum. Intrinsic angular momentum is caused by the wave function mapping position and time into something other than a scalar, for instance in the case of spin 1 particles the wavefunction is a vector. When we rotate something with intrinsic angular momentum we not only have to put the wave function in terms of the new coordinates(which gives the orbital angular momentum) but we also have to rotate the resulting vector(which gives the intrinsic angular momentum).
Relativistic basically just means something has to do with relativity.
The coupling constant is just a coefficient of a term in the lagrangian(dont ask me to explain what the lagrangian is). It is probably simpler to understand what it means by looking at the field equations. For a classical(non-quantum) particle in an electric field $E=\frac{p^2}{2m}+qV$, where E is energy, p is momentum, V is electric potential, m is mass, and q is charge. The coupling constants in this case are 1/m and q. q measures the strength of the electromagnetic interaction(1/m could be seen as measuring the strength of some odd self interaction caused by the particles being annihilated and created in the same spot at the same time, but no one cares). In general coupling constants measure the strength of some interaction
Spin(intrinsic angular momentum) is a vector just like momentum. It turns out that for massless particles the spin must point in the same direction as the momentum, or in the opposite direction(it cant be perpendicular to the momentum). That has to do with the fact that massless particles move at the speed of light in all frames of reference, and thus are never at rest.
Besides rotations and transalations and stuff like that there are also internal transformations. Isospin has three components just like angular momentum and transforms as a vector under isospin transformations, which are like rotations except they don't rotate space. Isospin and Hypercharge are the generators of electroweak guage transformations, and thus are the charges associated with the electroweak force. The electric charge is the hypercharge plus the third component of the isospin.
That question is like asking what properties a vector pointing in the x direction. In this case the wavefunction has 3 components just like a vector(because it is a superposition of the 3 'basis' states corresponding to each color, and a linear combination of the basis vectors is just a vector), and the components are mixed in the same way a rotation mixes them(actually it is somewhat different, because these vectors can be complex, so the rotation group is SU(3) not SO(3).) Just like you can't tell whether a vector points in the x direction, without having a vector known to point in the x direction or vectors known to point in the y and z directions, you can't determine the color charge of a particle, you can only compare it to other colors. The strong force(which corresponds to color rotations in the same way gravity corresponds to ordinary rotations and lorentz boosts) causes particles to tend to combine to form color neutral combinations
It says 2 fermions(such as electrons) cannot be in the same state. The pauli exclusion principle is what causes the electrons in an atom to not all just go into a ground state(if they did chemistry would be very boring).

Thanks in advance. :)[/QUOTE]

18. Aug 17, 2004

### LURCH

Because observation requires force. In a way, observation is a complex chain of events resulting in an impact on our senses. When you look at the wall in front of you, light from a bulb bounces off the wall, making impact (transfering energy; exerting a force) on your retina. If the wall did not have any effect on the light from the bulb, you could not see it. If it also had no effect on the tiisue of your hand, you could not feel it. If it had no effect on anything, you could not observe it. We observe things by the effect they have on other things (including ourselves). As a simple matter of thermodynamics, all these "effects" involve forces. No forces; no effects. No effects; no observation.

19. Sep 24, 2004

### Roberth

Hello,

I have suffered from the same problems that you describe. I think that quantum physicists and theoretical physicists in general often do not (or even cannot) make a distinction between Mathematics and Physics (… sometimes this is difficult). I believe that this is one of the reasons why people have such great difficulties to conceptually understand the subject.

So, I will try to give answers in a way that I would have hoped for.

I will start out very profound and hope you will not be offended if I say certain things that are very basic and that you know already. It will hopefully provide a framework where the complicated things will make easier sense to you.

1.) Wave Function

In 1924, Louis de Broglie suggested that all particles can be considered waves.

He gave a derivation in his PhD dissertation where the major formula was
lambda=h/p
where lambda is the wavelength of the particle, h is Planck´s constant and p is the momentum of the particle. The momentum in turn is p=m*v where m is the mass of the particle and v is the speed of the particle. The formula lambda=h/p
is correct for slow particles. For fast ones you must use the 'relativistic formula' since according to special relativity, the faster a body moves, the more mass it gains. I will not go into that here, but see my answer to 3) below.

As you know, when you use mathematics you think in "functions". In high school you learn 'y is a function of x'. For these functions, you can take the derivative dy/dx as the first derivative and d^2y/dx^2 as the second and so on.

In Physics, the most important derivatives are the first and the 2nd derivative and in particular the 2nd derivative. You may ask, why is that? The reason is related to Newton's law "Force = mass * acceleration". Typically you have a given force in physics and you want to describe the movement of bodies in this force field.

A consequence of this "I have a force and now I try to predict the movement of bodies in its influence" is that you end up with 2nd order differential equations. One such equation is the classical wave equation which describes the movement of a wave on a string

delta^2/delta x^2 [y] - 1/v^2* delta^2/delta t^2 [t] = 0

x is the x coordinate, t is the time, v is a constant and gives the tension of the string and the mass per unit length and

y is the wave function!!

A general solution of this wave equation is

y=f(x-v*t)+g(x+v*t)

where y is the wave function.

Here, I believe is an example why physics (and particularly modern physics) is so difficult to understand. All of a sudden, we use a mathematical term rather than a physical term.

The wave equation is a second order, partial, differential equation and the solutions to such equations are functions, rather than variables.

Erwin Schrödinger tried to find a model of the behavior of the electron in the hydrogen nucleus and because he knew that electrons behave like waves he tried to mathematically model the hydrogen atom as a wave system with a wave equation.

His wave equation was

i*hbar*delta(psi)/delta(t)=-hbar^2/2*m*delta^2(psi)/delta^(x)^2+V*(psi)

Now, if you find a psi that is a solution to this equation then you have a wave function for that equation. The Wave Function Psi is a Solution to this Equation, in the same way as the y above is a solution for the classical wave equation.

The term function comes from the mathematics for second order differential equation and the term wave comes from the physics that the electron has wave character.

2.) Intrinsic Angular momentum

= spin. This is a physical term. The intrinsic (non-orbital momentum) of a particle is called spin. Think of the Earth going around the Sun. The momentum around the Sun is orbital momentum. However, the Earth also turns around its own axis, ie. the Earth spins and has a spin.

The electron rotates (this is not really true, but fine enough as a model) around the atom and therefore creates orbital momentum. However, it also has an intrinsic spin, ie. rotates about itself (again, this is probably not really true but as a model, it is fine).

Spin was discovered in the Stern-Gerlach experiment. Particles are grouped into two types relating to spin. The value of the spin is either half or integer multiples of hbar, ie. 1/2 hbar, 3/2 hbar or 0 hbar (no spin), 1 hbar, 2 hbar. The half multiples are called Fermions, the integer multiples are called Bosons.

3.) Relativistic

='Physics slang'. Results of Einstein's Special Relativity show that mass and energy increase, length contracts and time slows down the faster you move (up to the speed of light). When physicists say relativistic then they usually mean that relativistic effects must be taken into account in a calculation. Most things (but not all) that move less than 10% of the speed of light are not considered as relativistic. The most important mathematical term is the factor k=square_root_of(1-v^2/c^2) where v is the speed of the quantity in question and c is the speed of light. If you use v=0.1 and c=1 (ie. v=10% of the speed of light) then the k term becomes k=square_root_of(1-0.1^2/1^2)=0.995 of the non-relativistic value, so there is almost no difference.

For making quantities smaller, multiply by the k term, for making them larger divide by this term. For example, a 1m rod that moves with 3/5 the speed of light is

1m* square_root_of(1-v^2/c^2) = 0.8m long

4.) The Coupling Constant

Mathematical concept. Describes a number (=constant) that gives the force between two particles where these two particles interact via one of the fundamental four forces (=electromagnetic, strong force, weak force, gravity).

Most often, however, Coupling constant refers to the electromagnetic force between two point charges such as, for example, two electrons.

The quantum mechanical expression of the potential energy between two point charges is

V=((1/(4*pi*epsilon0*hbar*c))*e*e)/r
epsilon0…permittivity of free space, a constant
e…..electron charge 1 = 1.602x10^-19 C
e….electron charge 2 = 1.602x10^-19 C
r…..distance between the charges
hbar… Planck´s constant divided by 2*pi
c….. speed of light

Except for the distance r, every term in the expression is a constant and if you put the numbers in then you get approximately 1/137 = (1/(4*pi*epsilon0*hbar*c))*e*e. This is called the coupling constant of the electromagnetic force. It is also called fine structure constant (same thing).

5.) Spin is parallel/anti-parallel to momentum
Physical and mathematical: parallel means that physically the Spin and angular momentum are aligned in the same direction and mathematically that you add them, just like two vectors. anti-parallel means that physically the Spin and angular momentum are opposite in their direction and mathematically that they must be subtracted, ie. the difference of the vectors taken.

6.) Charge Multiplets
Mathematical concept that describes consequences of isospin in Nuclear Physics. This is fairly advanced. To really understand it, you first must understand eigenfunctions and their relationship to quantum mechanics and spin operators. This is one of the most difficult things in quantum mechanics and takes some time.

I will briefly try to explain. In 1932, Heisenberg (yes, the famous one) suggested that certain groups of particles such as the nucleons, ie. the proton and the neutron, could be considered as the charge state of the same particle. As you probably know, the proton has a positive charge, whereas the neutron is, well, neutral.

Heisenberg suggested that both particles have an isospin of T=1/2 but different 'projections' T3: T3=+1/2 for the proton and T3=-1/2 for the neutron. This is where it becomes difficult. These projections are eigenvalues of an operator, the isospin operator, which acts in the same way as the spin operator.

Formally, the isospin is treated as quantum mechanical angular momentum and calculated with in the same way. Isospin can be used as a quantum number to label isobars (=nuclei with the same atomic number A), where to each nuclear state an isospin T with third component T=1/2(Z-N) is added. Z is the number of protons and N the number of neutrons.

Finally, how is all this related to Charge Multiplets? I show an example.

I use the following notation: CL [37 17] means the nuclei CL (= Chlorine) with mass number A=37 and proton number Z=17. It has 37-17=20 neutrons and hence N= 20.

The following nuclei CL [37 17], A [37 18], K [37 19] and Ca [37 20] have all the same atomic number 37 but different number of protons and neutrons.

Using the formula T=1/2(Z-N), their respective isospins are T3=-3/2, T3=-1/2, T3=+1/2, T3=+3/2.

They form what is called a Charge Multiplet. They have the same mass number but different charges which is expressed as the isospin.

7.) Supermultiplets
Mathematical concept, similar to 6, but related to characteristics of quarks.

8.) Isospin
See 6 above.

10.) What is the Pauli exclusion principle

This is a mathematical concept with physical effects.

It says that no two electrons in the same atom can be in the exactly same state, ie. can have the same quantum numbers n,l,m,s. If n,l,m are the same then their spins must be different. The principle has also a wider application in nuclei where it is partly 'responsible' for certain aspects of the shell model of nuclei and the existence of the so-called magic numbers. Since neutrons and protons are Fermions (spin 1/2 particles) they also obey Pauli´s principle.

Pauli derived it from mathematical considerations which are related to the symmetry requirements for the wave functions.

Cheers,
Roberth

PS: Another reason why people find physics difficult is because it is difficult!

20. Sep 25, 2004

### Roberth

Books

Hi again,

In my explanations above (or below, not quite sure where this will end up) I forgot to mention a few books that you may like if you are seriously interested in Quantum Physics and its applications.

1) Quantum Revolution I - III
subtitle: "Vignettes in Physics"

Author: G. Venkataraman
Publisher: Sangam Books

These are absolutely lovely three books. The author tries, in his own words, to write like Richard Feynman - and he succeeds in my opinion.

These three small booklets (they cost peanuts, about 8\$ at the time I bought them, about 9 years ago) explain quantum mechanics on a freshman level, but unlike many popular books they never become superficial.

You need about the level of 1st year college/university. However, if you have a good 'feeling' for physics and mathematics, you will understand all the conceptual points also with high school math.

In any case, you need a lot less math than your "Modern Physics" (which I know), but it explains quantum mechanics on a far more profound level. Your "Modern Physics" is aimed at 2nd to 3rd year college physics students and this may be the reason why you are struggling with it. It is actually pretty good, in my opinion, but you need the 'mathematics exposure' first.

If you want to be more serious about Nuclear Physics then I recommend
"Particles and Nuclei"
Author: Povth, Rith, Scholz, Zetsche
Publisher: Springer

This is a beginning graduate level book about particles and nuclei, always up-to-date (they bring out a new, updated version every 3 years, or so) but approached from a ´physical point´. It focusses more on experimental facts than on mathematics. You need a good grasp of quantum mechanics for many parts, though. It often uses the Dirac notation (in case you do not know, this is a sort of compact form of dealing with wave functions). You may, however, get the 'jist' of some issues without higher quantum mechanics.

Finally, if you want to get a general feeling of the mathematical level and physics level that is required to be 'a Physicist' then take a look at "Mathematical Methods for Physicists" by Arfken and Weber. Depending on your area of physics, you need about 70%-80% of the contents of that book. For quantum physics, look at "Quantum Mechanics" by Eugen Merzbacher, for example. I do not recommend that you buy any of those, though. They are books for graduates and researches (and may simply depress you )

Roberth

Last edited: Sep 25, 2004