What exactly is supersymmetry?

[Obnoxious]

I read somewhere that this symmetry somehow connects fermions and bosons together, how does it do it? What exactly is supersymmetry?

(sorry if this has been asked already)

[5614]

Supersymmetry is a hypothetical symmetry that links bosons to fermions (and vice versa). With supersymmetric theories all fermion have a "superpartner" which is a boson (and vice versa). Supersymmetry has not been observed in real life but remains a vital part of a few theories of physics, including some extensions to the standard model as well as modern superstring theories.

 

Supersymmetry relates particles of different spins. So matter particles (spin 1/2 (fermions)) are related to force particles (spin 0 or spin 1 (bosons)).

 

referrence to: http://en.wikipedia.org/wiki/Supersymmetry

[Obnoxious]

How are they "related?"

[Severian]

Normally, when one thinks of the usual dimensions of space-time (like 'up-down', 'left-right' etc) one takes for granted that they are bosonic. By that I mean that the operators for the position co-ordinate in a particular direction commute with one another. So that if I ask 'what is the x-coordinate of the particles position?' followed by 'what is the y-coordinate of the particles position?' I will get the same answers as if I had asked for the y-coordinate first and then the x-coordinate.

 

However, quantum mechanics has shown us that assumptions like this are dangerous and some operators do not commute, but anti-commute (they are fermionic), so that if you ask the questions in the opposite order you will get a relative minus sign. Supersymmetry postulates that there are extra dimensions in space time which anti-commute.

 

This sounds silly, but in fact it is a very natural idea. In the '70s Colman and Mandula 'proved' that the maximum symmetry that space-time could have is the Poincarre symmetry of relativity (boosts and translations). It was only the advent of supersymmtry that showed us a flaw in their argument - they had forgotten the possibility of anti-commuting operators (or to be more technical, graded Lie algebras). Now we have 'proven' (until someone else finds something that we have missed) that the maximal symmetry of space-time is the superPioncarre group which consists of the boosts and translations of the Poincarre algebra but also supersymmerty transformations which is basically a rotation in the enlarged space (bosonic + fermionic directions).

 

So what does this mean for particles?

 

Well, if we have extra fermionic directions, a field (in position space) will be not just a function of x and t, but also of this extra direction z (it is normally denoted theta but to save typing I will call it z). This field is called a 'superfield'. The weird thing about z is that it is anti-commuting so the anti-commutator of two such directions z₁z₂ = -z₂z₁ and thus the anti-commutator of z with itself must vanish: zz = -zz = 0.

 

Now, you are all familiar with taylor series where we expand a function f(x) as a power series in x and if x<1 the higher order terms are small because a high power of x is small. If we do the same with a fermionic number z, the expansion of f(z) will only have two terms: f(z)=a+zb since terms like z² (and higher) will vanish. Therefore this expansion is always true (even if z were large). Also notice that since the object f(z) is a bosonic object, then 'a' must be bosonic and 'zb' must be bosonic; but since z is fermionic, b must also be fermionic (add two objects of half-integer spin and you get one with integer spin).

 

Now, a superfield can also be expanded in this way (although it is quite a bit more complicated) and similarly the series will be truncated at a certain order by the anti-commuting of the fermionic direction. One will find that each superfield in 'superspace' can be written as two fields in normal space: one bosonic and one fermionic exactly like 'a' and 'b' earlier.

 

Supersymmetry postulates that every particle we have seen is coming from the expansion of a superfield and is either the bosonic part or the fermionic part. For example, the electron is the fermionic part of the electron-superfield and should have a bosonic partner (usually called the selectron). We are currently looking for these particles in colliders.

 

So when people say that supersymmetry has twice as many particles, this is not true. The different 'particle' (fermionic and bosonic) are just different projections of the same particle in different directions.

 

There are a lot more reasons for believing why supersymmetry is true, but I will leave these for another post.

[Severian]

Did my rather too technical description kill this thread?

[bascule]

Did my rather too technical description kill this thread?

 

Perhaps it merely answered everyone's questions so there was no need for further discussion

[Phi for All]

Did my rather too technical description kill this thread?

Tends to happen when the answer is spot on.

[Johnny5]

You stay technical Severian, I didn't fully understand your answer, but I'm sure others did.

 

I don't see how the commutator fits in really, but Tom Mattson will get around to explaining the commutator, and then I will look at your answer again.

 

Personally, my take on particles is that we are far from understanding them. But i think half of the fun is in trying to, and of course the other half is understanding them.

 

Perhaps I can be slightly more specific. Some particles are very short-lived. Say a lifetime of a nanosecond, others a femtosecond. Now, in the case of a free neutron, i think the average lifetime is around 11 minutes, or maybe 14, i forget. But, and this is my take on it, to say we understand particle physics, would require us to understand these particle lifetimes (transition times). As I recall, there is a long formula, but it doesn't work well.

 

Right now I am still focused on this:

 

Normally' date=' when one thinks of the usual dimensions of space-time (like 'up-down', 'left-right' etc) one takes for granted that they are bosonic. By that I mean that the operators for the position co-ordinate in a particular direction commute with one another. So that if I ask 'what is the x-coordinate of the particles position?' followed by 'what is the y-coordinate of the particles position?' I will get the same answers as if I had asked for the y-coordinate first and then the x-coordinate.

 

However, quantum mechanics has shown us that assumptions like this are dangerous and some operators do not commute, but anti-commute (they are fermionic), so that if you ask the questions in the opposite order you will get a relative minus sign. Supersymmetry postulates that there are extra dimensions in space time which anti-commute.[/quote']

 

You say...

 

"... for granted they are bosonic. By that (you) mean that the operators for the position coordinate in a particular direction commute with one another"

 

I guess, if I had a question for you, that would be it. I don't understand what you mean there.

 

I kind of understand... but not totally. First ask for x coordinate, then ask for y coordinate. Are you talking about some measurement process here?

 

Regards

[Tom Mattson]

Commuting operators are called 'bosonic' because the creation and annihilation operators for bosons must commute. The reason for that is that in experiments it is observed that any number of bosons can occupy the same quantum state, and so the eigenvalues of the bosonic number operator must be allowed to take on any whole number value. This is achieved by using commutating operators to represent the creation and destruction of bosonic particles.

 

Anticommuting operators are called 'fermionic' because their creation and annihilation operators must anticommute. Experimentally it is known that at most one fermion can occupy a quantum state (Pauli Exclusion Principle). So the number operator for fermions can only have eigenvalues of 0 or 1, which is achieved with anticommuting operators.

[Johnny5]

Commuting operators are called 'bosonic' because the creation and annihilation operators for bosons must commute. The reason for that is that in experiments it is observed that any number of bosons can occupy the same quantum state' date=' and so the eigenvalues of the bosonic number operator must be allowed to take on any whole number value. This is achieved by using commutating operators to represent the creation and destruction of bosonic particles.

 

Anticommuting operators are called 'fermionic' because their creation and annihilation operators must anticommute. Experimentally it is known that at most one fermion can occupy a quantum state (Pauli Exclusion Principle). So the number operator for fermions can only have eigenvalues of 0 or 1, which is achieved with anticommuting operators.[/quote']

 

 

This is a good answer Tom.

 

Can you explain the creation and annihilation operators.

 

Commuting operators-- bosonic

Anti-commuting operators-fermionic

 

 

You make it sound like we actually understand the periodic table.

 

That is something I don't believe is possible though.

[Tom Mattson]

Can you explain the creation and annihilation operators.

 

Commuting operators-- bosonic

Anti-commuting operators-fermionic

 

The first 2 pages of the following document explain it succinctly.

 

http://www.cithep.caltech.edu/~fcp/physics/quantumMechanics/secondQuantization/SecondQuantization.pdf

[Kygron]

How much does this involve complex(imaginary) numbers?

[Severian]

Perhaps I can be slightly more specific. Some particles are very short-lived. Say a lifetime of a nanosecond' date=' others a femtosecond. Now, in the case of a free neutron, i think the average lifetime is around 11 minutes, or maybe 14, i forget. But, and this is my take on it, to say we understand particle physics, would require us to understand these particle lifetimes (transition times). As I recall, there is a long formula, but it doesn't work well.

[/quote']

 

I would disagreee with that. We understand (and can predict) these decay widths very well. When the decay is via the strong force it is difficult to theoretically predict widths to high precision (because the force is so strong) but we can always do it to within 10% or so, and understand the source of this error. In fact, every measurement ever made in particle physics agrees with the Standard Model to within the accuracy of the experiment and the theoretical prediction. That is a pretty amazing acheivement I think.

 

 

You say...

 

"... for granted they are bosonic. By that (you) mean that the operators for the position coordinate in a particular direction commute with one another"

 

I guess, if I had a question for you, that would be it. I don't understand what you mean there.

 

I kind of understand... but not totally. First ask for x coordinate, then ask for y coordinate. Are you talking about some measurement process here?

 

OK, in QM and QFT we have operators which tell us the value of an experimental quantity when we act on a physical state. So for example, in the Schroedinger equation the energy operator is i*hbar*d/dt (not sure if LaTeX is working again yet...): taking this operator and acting on the state tells us the energy of the state. Making this 'measurement' actually (usually) changes the state of the system, so you can imagine that if we make two different measurements we might get different answers depending on which order we make the measurements. The classic example of this is position and momentum: measuring a particle's position and then its momentum will give different answers from measuring its momentum and then its position. The momentum and position operators do not commute.

 

Now, when you measure a position, you are in fact making 3 measurements (4 if you you are measuring its position in time as well as space) because you are measuring the x-coordinate, the y-coordinate and the z-coordinate. All of these measurements can be made 'at once' because measuring one has no effect on the others. The measurements commute.

 

The extra 'fermionic' dimensions I mentioned in the previous post are not like this. They 'anticommute' with each other, so if I make the measurements in the opposite order I will find a relative minus sign.

 

The link between fermions/bosons and anticommutation/commutation has already been described by Tom, so I won't go into that....

[Johnny5]

I read the article posted by Dr. Mattson, and not suprisingly understood extradinarily little.

 

I have a book on Quantum mechanics written by Hans C. Ohanian, which is where I first encountered creation/annihilation operators.

 

I think the impediment to me learning how to use them, is that I don't see what their utility is.

[Tom Mattson]

I read the article posted by Dr. Mattson' date='

[/quote']

 

Please don't call me that.

 

If you must know I have an MS, not a PhD. So you may call me Master, if you wish.

 

I have a book on Quantum mechanics written by Hans C. Ohanian, which is where I first encountered creation/annihilation operators.

 

I think the impediment to me learning how to use them, is that I don't see what their utility is.

 

Their utility, aside from their natural, obvious application to the harmonic oscillator in quantum mechanics, is that they are used (in part) to describe quantized fields.

[Johnny5]

Please don't call me that.

 

If you must know I have an MS' date=' not a PhD. [/quote']

 

I think I'll stick with Tom then.

 

 

Having a masters is practically like having a PhD anyways.

 

Their utility, aside from their natural, obvious application to the harmonic oscillator in quantum mechanics, is that they are used (in part) to describe quantized fields.

 

Ok... I'll keep that in mind. I gathered that much from the link you posted. It is highly abstract information though. Without an experiment firmly in mind, it's difficult to understand the point of them.

[Severian]

I have a book on Quantum mechanics written by Hans C. Ohanian' date=' which is where I first encountered creation/annihilation operators.

 

I think the impediment to me learning how to use them, is that I don't see what their utility is.[/quote']

 

In QFT pretty much everything is an operator. We move from classical mechanics to QM by "first quantisation" where we turn things like momentum and position into operators, so that an operator acting on the (still classical) state gives the value of the associated quantity. The problem with QM is that the number of particles is fixed: for example there is no way to create a particle-antiparticle pair in QM unless you put them in by hand.

 

We move to QFT by "second quantisation", quantising the fields themselves. This means that the field [math]\phi(x)[/math] becomes an operator. It is natural to decompose the field into plane wave states as:

 

[math] \hat\phi(x) = \Int \frac{d^4p}{(2 \pi)^4 \left( \hat a_p e^{-ipx} + \hat a_p^{\dagg} e^{ipx} \right) 2 \pi \delta(p^2-m^2) [/math]

 

(I hope LaTeX is finally working!)

 

When one does the math, one finds that the operator [math]\hat a_p[/math] destroys one quantum of momentum p and [math]\hat a_p^{\dagg}[/math] creates one quantum. The annihilation operator can then be reinterpreted as creating an anti-particle.