# The QH QM QA thread.

Discussion in 'Physics & Math' started by QuarkHead, Mar 22, 2008.

1. ### QuarkHeadRemedial Math StudentValued Senior Member

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So here I will ask some really stupid questions about quantum theory.

I'll dribble them out, one by one, and fully expect either derision or elementary responses from our resident experts.

So. I took chemistry in college when dinosaurs roamed the Earth, and this is what I recall.

Obviously the electron is a charged extra-nuclear particle, with, I assume mass. How come mass?

The nucleus, as I was taught, comprised protons (with mass and charge), neutrons (with mass but no charge) and neutrinos (mass, and no charge). They also said said there was bunch of other stuff I wasn't to bother my pretty head about; we'll come to that

So here is my first embarrassing question:

You guys talk about particle "mass" using units of electron volts eV. So how can a particle without charge but with so-called mass be described in these terms?

3. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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The definition of an electron volt is the amount of energy it takes to move an electron through one volt of potential. So it's only a unit of energy.

Also, we work in units where hbar = c = 1. In these units, energy = mass and length = time, and mass = 1/length. So EVERYthing can be measured in electron volts.

5. ### AlphaNumericFully ionizedModerator

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You mean "Why does the electron have mass?"

Pushing aside the philisophical discussion of "Why anything?", the electron has mass, the Z boson has mass, the quarks have mass, the W bosons have mass while the gluons and the photon don't because of how they all couple to the Higgs field.

How mass arose 'naturally' within quantum mechanics was an issue for quite a while because there was no appriory reason why some particles had mass and others didn't or why there was mass at all. Then a few (very) bright sparks realised that the patterns of masses we see, including the massless ones, could be explained by a scalar field (the Higgs field) coupling to them all and then obtaining a vacuum expectation value (VEV).

When you write down the Lagrangian in QFT for the Standard Model, Higgs and all, there's no mass terms. But the Higgs is in a lot of places. It has a special potential which means that rather than prefering to settle at a zero energy, it prefers to settle at a non-zero energy and then move about that value (obviously I'm throwing specifics out the window here). Because it's got this big VEV, this VEV appears in all the equations of motion for other particles, but it does so in just the right way to say "Z and W bosons have mass. Gluons don't. Neither does the photon. Quarks and leptons have mass too".

This process can be done without the Higgs boson, but the mechanism which I just described (the Higgs mechanism) has got to be done by something which has a lot of the properties of the Higgs boson (quite massive, forms a spin 0 state, gets a VEV). The LHC will be make or break because it's able to scan the entire energy range the theory allows the Higgs to be in.

Does that help or hinder?
There's no neutrinos within the nucleus. The protons and neutrons are comprised of quarks. Every now and again one of these quarks will turn into a different quark and spit out a W boson. This W then turns into a lepton (electron, muon or tau or one of their antipartners) and a neutrino (or antineutrino). So while weak decay results in electrons and neutrinos being emitted from the nucleus, they weren't in there hiding, their energy was.
1J is expended moving 1 Coulomb of charge through a potential different of 1 Volt. Alternatively, one electron-volt of energy is expended moving 1 electron's charge through a PD of 1 volt. Thus you can use "electron-volts" instead of "joules". Doesn't matter if there's anything charged or not in your system, you are doing nothing more than saying "Feet" instead of "Metres" or "Miles" instead of "Kilometres".

And since $E=mc^{2}$ you can relate mass and energy. So instead od saying "10kg" you can say it in terms of energy and the convenient units in particle physics are 'electron volts'. It's useful because the typical energies of particles at room temp are about 0.01~1eV, electron energies around the atom are 1~1KeV (ie it takes 13.6eV to get the electron to escape from hydrogen) and then you get into particle masses. An electron is about 500,000eV in mass. This has nothing to do with it's charge, but it's convenient for things like measuring the energy of photons in the photoelectric effect.

7. ### QuarkHeadRemedial Math StudentValued Senior Member

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So, if I read this right, the "mass" of a particle is defined by the amount of energy required to "push it up" to relativistic speed?

Now Alpha: I do thank you for for your response, truly. But you must not assume that I understand terms like Z bosons, W bosons, VEV, Higgs mechanism etc.

I want to proceed very cautiously here, and build from the ground of my very inadequate knowledge of QM.

OK with you?

8. ### Walter L. WagnerCosmic Truth SeekerValued Senior Member

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2,549
QH:

Using Einstein's forumla, E = mc^2, mass and energy are equivalent.

Rather than using mass in grams for an electron, proton, etc., it is instead converted into units of energy, then divided by c^2. Often, it is simply expressed in the units of energy.

Thus, the rest mass of a proton is about 938.2723 MeV/c^2, usually approximated as about .94 GeV, or frequently simply as 1 GeV

The rest mass of an electron is 510.9906 KeV/c^2, usually approximated as about 511 KeV, which is also the energy of the photon given off when an electron and positron annihilate, referred to as the annihilation energy.

Ben's defintion of an eV [electron Volt] is correct, and it is a small amount of energy.

"The electronvolt (symbol eV) is a unit of energy. It is the amount of energy equivalent to that gained by a single unbound electron when it is accelerated through an electrostatic potential difference of one volt, in vacuo. In other words, it is equal to one volt (1 volt = 1 joule per coulomb) multiplied by the (unsigned) charge of a single electron. The one-word spelling is the modern recommendation[1], although the use of the earlier electron volt still exists.

One electronvolt is a very small amount of energy:

1 eV = 1.602 176 53(14)×10−19 J. [2] (or approximately 0.160 aJ)"

from Wikipedia at: http://en.wikipedia.org/wiki/Electronvolt

The values provided are from Nuclides and Isotopes, Fourteenth Edition

Hope this helps!

Last edited: Mar 22, 2008
9. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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In the spirit of keeping this discussion on track, I am going to (try) and moderate this very closely.

I have deleted posts by Reiku and 2inquisitive, because Reiku's post didn't address QH's quesiton, and 2inquisitive was asking Walter something.

10. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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The best way to say it (I think), is that in units where c = 1, the rest mass and the energy are the same thing. Think of it as the energy of the photons if the particle met it's anti-particle.

11. ### James RJust this guy, you know?Staff Member

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There are no neutrinos in an atom. Neutrinos are the product of various kinds of nuclear reactions. Stable atoms have a nucleus of protons and neutrons, surrounded by electrons. That's all.

As others have said, the electron-volt is a unit of energy. If you have a particle's mass in kilograms, its energy equivalent is

$E=mc^2$

where c is the speed of light. The above equation gives the energy in Joule. To convert Joule to electron-volts, divide by $1.6 \times 10^{-19}$.

No. The rest-mass energy of a particle has nothing to do with it moving at any speed. The rest-mass energy is a straight conversion according to $E=mc^2$.

12. ### James RJust this guy, you know?Staff Member

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If photons had rest mass, then quantum electrodynamics would be wrong. Instead, it is one of the most accurate scientific theories we have.

13. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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Mod Note:

How is this hard?

14. ### zephirBannedBanned

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390
Not necessarily, if the photon rest mass will remain bellow 10E-52 kg, i.e some 3E-17 eV. By Popper's methodology all theories should be considered invalid - not matter, how exactly they're verified by now.
OK, but why was my answer deleted, after then?

Last edited: Mar 23, 2008

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16. ### QuarkHeadRemedial Math StudentValued Senior Member

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1,533
OK. First this:

I have no way of knowing whether or not I being fed soup; I trust the mods to make sure that only mainstream answers to my questions are being offered (of course I can make no judgment about whether the non-mainstream has any validity).

So I am now OK with using units of electron-volts to describe the mass of uncharged particles. Thanks.

James pointed out that the neutrinos are not part of the structure of the atomic nucleus, rather that they arise as a result of "various kinds of nuclear reaction". I don't fully understand this, but I assume it means, as I read a biog of Pauli, that he postulated the existence of the neutrino, such that as an an unstable atom decays, then not only are electrons ejected, but so are neutrinos, these in some sense arising as a result of the decay process. Sort of makes sense I guess, since it seems the sums don't add up otherwise.

OK now bosons. I understand they are "mediators" of force. I further understand that only three bosons are fully characterized; those that mediate the strong nuclear force, those that mediate the weak nuclear force and the photon. Is this correct? Alpha called the two former as W and Z bosons.

I also read that, since everything in sight can be quantized, it seems thought likely there are bosons (the graviton) that mediate the gravitational "force" and another (the Higgs) that assigns mass to any massive particle.

So here is today's dumb question. If a particle is known a priori to have zero mass, and another particle is known to have non-zero mass, how do you guys know that the latter "couples" (is this the right term) to the quantized scalar field that is the hypothetical Higgs field, whereas the former doesn't?

I'm sorry to appear impertinent, but this seems ever so slightly circular reasoning to me.

17. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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I will work to make this the case.

Maybe the best way to think of it is in terms of interactions. Neutrinos don't live in the neucleus, but a neutron can decay'' into a proton, and the interaction produces an electron and a neutrino.

This is correct. But remember photons are also bosons.

But this is a tricky thing---another abuse of terminology. There is an internal quantum number called spin. Bosons are things with integer spin. So, for example, a helium-4 nucleus is also a boson.

I don't want to obfuscate this with details, and this is probably why mathematicians hate physicists

I would call W, Z and photons gauge bosons.

So a bit of history. It was Weinberg, Glashow, and Salam who predicted the W and Z bosons before they had observed them, even though we knew there was a massless photon. So we predicted how the higgs would couple.

The way to see it is that the electroweak sector of the standard model has four gauge bosons, called the W and the B. There are three W's and one B (this corresponds to the adjoint rep of SU(2)xU(1), but we can come to that later...I always end up screwing up the math, as you well know).

At high energies, this is a long range force, like gravity or electromagnetism. However, when gauge bosons become massive, the force is broken, and becomes a short range force. The range of the force is given by it's Compton wavelength. In God's units (which you asked about!), the range is just 1/mass of gauge boson.

When the higgs gets a vev (like AN said), it couples to two of the W's (W1 and W2), plus a linear combination of one of the W's and the B (W3 + B). This is called the Z boson. The orthogonal linear combination is the photon, and does not couple to the higgs.

I could probably pull some equations out of my ass, but when I get back to my office on Tuesday I will make this a bit more explicit if you want.

18. ### James RJust this guy, you know?Staff Member

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What happens is that neutrinos are created in a decay process. The most common way that happens is in beta decay of a radioactive nucleus. In a beta decay, for example, a neutron can split into a proton, an electron and a neutrino. Before the decay, there was only a neutron. After it, their are three particles instead of the one that was there before.

Note: this does NOT mean that a neutron is "made of" a proton, an electron and a neutrino. Things are a little more complicated than that.

You're thinking of what are called "gauge bosons". The term "boson" itself is much wider than that. A boson is any particle that has integral spin (i.e. its spin is 0, 1, 2, 3, etc.). The "opposite" of a boson is a fermion, which has half-integral spin (1/2, 3/2, 5/2, 7/2, etc.).

Most familiar particles are fermions - e.g. neutrons, protons, electrons.

The "mediators" of fundamental forces, in quantum field theories, all turn out to be bosons, but they're just another type of particle.

19. ### QuarkHeadRemedial Math StudentValued Senior Member

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Once again, thanks James and Ben, I appreciate your patience!

So let's leave aside spin for now as I find it really confusing, but we can come to it later. OK. In addition to the electron, proton and neutron in an atom, we also have the possibility of neutron decay to a proton, electron and a neutrino. I'm cool with that.

But there are a bewildering number of other animals in the particle zoo, pi, tau, mu and so on. May I, as a first approximation, say something like this: the results of experiments with atom smashers show the existence of such particles. Moreover, since Earth is constantly bombarded by ultra-high energy "cosmic rays" (whatever these might be) resulting form the violent nuclear reactions in our Sun, one may assume these guys occur naturally when these cosmic rays interact with atoms in our upper atmosphere.

Anyway, I was premature with my question about the Higgs, I see that. But, can you please explain what "gauge boson" is. What does the term gauge mean?

Ben, you said this
What did you mean by "at high energies"? Are we talking about experiments here or what? What does it mean for a gauge boson to "become massive"? Surely a boson has mass or it doesn't?

You guys are too clever for me.

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Ok Quark. I'm sitting in an airport in Houston, Texas after a weekend tryst with my girlfriend in San Antonio. This is a $9.95 answer because that's the cost of internet access. And there's nothing better than eating barbeque, drinking Shiner Bock beer, and talking about physics. This is correct. As always, there are subtleties, but in general this is correct. This may be a long forray Please Register or Log in to view the hidden image! The long answer is that the term gauge boson'' comes from the fact that the physics is gauge invariant''. Essentially, the gauge boson is a consequence of the fact that the physics is invariant under a local symmetry. The first thing to know is that all of the physics can be derived from the action, which is a functional of the fields in the problem. I think mathematicians call it a functor'', but I could be mistaken---either way, I don't want to confuse you by trying to speak your language and failing miserably! The action is written in terms of fields (which are functions of space-time coordinates), and describes all of the interactions that a set of fields can have, as well as how the fields get from point A to point B. So the action is a function(al) of fields, and the fields are functions of space-time coordinates x. Typically we write the action as $S[\phi(x)]$. There is nothing particularly physical about the field---one can write a local expression to describe a particle, in terms of fields. The field itself should be thought of as an operator---it has no meaning by itself, and needs a good Hilbert space to act on. (I don't know if the definition of Hilbert space is the same as the definition you'd use---I would say that a Hilbert space is a vector space with a positive definite inner product.) Also, to keep things simple, we will work with scalar fields, which have spin 0. Ok. So we have an action which tells us about our fields. It turns out that sometimes the fields have some symmetry associated with them. What does this mean? Well, for example, let me take a stupid example. Take the function $f(x) = x^2$. There is a symmetry of this function that we all probably learned after we learned to read---that is, $x\rightarrow -x$ leaves f alone. To use the parlance of a physicist (pretending to know something about math), we would say that there is a discrete symmetry $\mathbb{Z}_2$ of this function. Further, this symmetry is global in the sense that we have to apply this symmetry at every point at the same time, in order for the symmetry to make sense. Now we can wave our hands and generalize this. Suppose the action is invariant under a symmetry which acts on fields. Let's take another stupid example. Suppose we have a boson, spin 0 particle with an easy action: $S[\phi] = \partial_{\mu}\phi\partial^{\mu}\phi^* -m^2\phi^*\phi + \lambda (\phi^*\phi)^2$. The $\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}$, and the indices, when they are repeated are automatically summed over. The first term is called the kinetic term, and the second two terms are interaction terms. (This is why I said earlier that it is always useful to think of things in terms of interactions. The action describing nuclear decay can be written in a similar way, with a terms connecting a neutron, a proton, an electon, and a neutrino.) Also, this action describes the interaction of TWO fields (again, a subtlety!), $\phi$ and $\phi^*$. Let's not worry right now that the two fields are related by complex conjugation, and treat them as independant fields. Now let's ignore some interesting physics, and just look at the symmetries of this action. Clearly $\phi \rightarrow -\phi$ is a symmetry of this action. We can try and be a bit more sophisticated about this, by writing $\phi \rightarrow e^{-i \alpha}\phi$, and $\phi^* \rightarrow e^{i \alpha} \phi^*$. It is now clear that the phase is a GLOBAL phase, which means that the symmetry I talked about is a global symmetry. In other words, the symmetry does not depend on the space-time coordinate x. Further, the symmetry is good for ANY phase $\alpha$. What we have discovered is a GLOBAL U(1) smmetry, because $e^{i\alpha}$ is the generator of U(1) transformations. Here, I mean the group U(1), not the algebra (I think). But we can do better. What about LOCAL symmetries? What if we want our symmetry to depend on local functions of the space-time coordinate x? We might be worried about this because, for example, we know that the Lorentz transformations are local---we might expect that there are other transformations that we can do that are also local. So let's see what happens. Suppose we want to look for symmetries of the form $\phi \rightarrow e^{i\alpha (x)}$. We see that there are no real problems except in the derivatives in the kinetic term of the action. The transformation is now a local function of x, so the derivative acts on it in a non-trivial way. You might think that we are screwed, but as good theoretical physicists, we know that if at first we don't succeed, we should add shit until it works. The way to add shit is the following: just work it out. We should see that the transformation $\phi \rightarrow e^{i\alpha (x)}\phi$ gives: $S = \left[\partial_{\mu}e^{-i\alpha(x)}\phi\right] \cdot \left[\partial^{\mu}e^{i\alpha(x)}\phi^*\right]$ $\Rightarrow S = \partial_{\mu}\phi\partial^{\mu}\phi^* + \phi^{*}\phi \partial_{\mu}\alpha (x) \partial^{\mu}\alpha (x) -i \partial_{\mu}\phi\partial^{\mu}\alpha(x) + i\partial_{\mu}\phi^*\partial^{\mu}\alpha(x) + \cdots$. Now look what we have found! Not only do we have a scalar field, but the fact that we promoted our U(1) to a LOCAL symmetry gave us another interaction. In fact, it gave us several new interactions! The new field we have found is $\partial_{\mu} \alpha(x)$. We can just rename this to $A_{\mu}$. Because $\alpha$ is a local transformation (or, a GAUGE transformaiton), the fied $A_{\mu}$ is a GAUGE field. Specifically, we have discovered a theory with a charged scalar field (spin 0 boson, called a scalar because it transforms trivially (i.e. spin 0) under the Lorentz group) interacts with another boson, except this one with a (space-time) vector index. The field $A_{\mu}$ is called a vector boson. You may ask, why are we justified in adding things to our action? Remember---we want to use math to describe physics. We only want to add the minimum amount of shit---this is called (by some) Occam's razor...the simplest solution is typically the correct one. Either way, you may be a bit more comfortable if I redefine what I mean by derivative. Remember---the real trouble came in when we looked at the trnasformation properties of the derivative terms in the action. We can define a covariant derivative'' that transforms trivially under the symmetry: $\mathcal{D_{\mu}} \equiv \partial_{\mu} - i A_{\mu}$ So I have probably gone a bit far in answering your question. BUT now (hopefuly) you know what gauge invariance is: it is a local symmetry of the action. Gauge invariance (which is my second favorite invariance) of the action IMPLIES (or, predicts, if you like) the existence of a gauge boson. A gauge boson always mediates a force. So, putting it all together, a force'' in nature is caused by the existence of a symmetry in the action, which IMPLIES the existence of a gauge boson. Perhaps I got a bit ahead of myself. But let's consider our new, gauge invariant action, written in terms of the field $A_{\mu}$: $S = \partial_{\mu}\phi\partial^{\mu}\phi^* + \phi^{*}\phi A_{\mu}A^{\mu} -i \partial_{\mu}\phi A^{\mu} + i\partial_{\mu}\phi^*A^{\mu}-m^2\phi^*\phi + \lambda (\phi^*\phi)^2$. A few things should pop out at you. First, there is no kinetic term for $A$. We can add that by hand (it would have fallen out more naturally if I had done this in another way). But there is also no MASS term for A. That is, A is a massless field, as opposed to $\phi$, which has a mass term. This means that this boson, $A$ is massless. This is good, because we WANT to describe massless bosons, like the photon. Now, as zephir pointed out, we COULD add a mass term for the photon, but there is ABSOLUTELY no reason to do so, and this is utterly something that we would do by hand. This is opposed to the kinetic term that we added (seemingly) by hand for the vector field $A$. But one can motivate that by doing all of this derivation in another manner. Last edited: Mar 24, 2008 21. ### QuarkHeadRemedial Math StudentValued Senior Member Messages: 1,533 Bliss! Once again I thank you. I must say you do a better job of explaining stuff that many other physicists I know (I predict a successful teaching career for you). But there is still some stuff I'm not clear on. Let's see. It took me a while to figure out what you are doing here. I think I may paraphrase as follows; see if you can agree. Multiplication of a vector by an element of the underlying (scalar) field is another vector in the same space. Since the basis for my space is itself a set of vectors, this implies that any arbitrary vector is found by scalar multiplication of the basis vectors. We know this is true: it's part of the usual definition of a vector! And, provided only that we agree to keep the basis fixed, the action of an operator only messes with the scalar coefficients on the basis. From which it seems to follow, from the usual field axioms, that a field can be thought of as a set (vector space, actually) of operators on my original space. Funny way to put it, but not wrong, I believe. Incidentally, in the definition of an Hilbert space, I think I might include square integrability, but this is not a subject I know a lot about. Lemme stop you right here. If I am allowed to think of the $x^{\mu}$ as local coordinates on some manifold, then I will interpret $\partial_{\mu}\equiv \frac{\partial}{\partial x^{\mu}}$ as a co-tangent vector. This is standard notation - is that what you mean here? Or, without getting fancy, is this beast a co-vector? Likewise, is $\partial^{\mu}\equiv \frac{\partial}{\partial x_{\mu}}$ a vector? So, if my interpretation is correct, which I strongly doubt, we have the right-action of an operator on a vector. This is OK, these sorts of constructions make sense, as we both know, but it seems a weird way to do it. Have lost the plot? Eek! Again, if the above is correct, again I doubt that it is, then your first term is of necessity scalar (since the action of a co-vector on a vector is an element of the field). So, is this merely a definition of "kinetic" (I note you didn't add "energy" there) Clear? Damme, it's not. $\phi \rightarrow e^{-i \alpha}\phi$, and $\phi^* \rightarrow e^{i \alpha} \phi^*$ imply rotations to me. What's that got to do with it? I'd better stop and await clarification. PS, by edit: belatedly I discovered this which looks like what I need. I'll look at it when I can Sorry about the$9.95, I'd refund you if I could. Sorry also I'm not being very bright today - I think I murdered my last few brain cells over the holiday weekend.

Last edited: Mar 25, 2008
22. ### temurman of no wordsRegistered Senior Member

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I would say $\partial_\mu$ is a tangent vector, $\phi$ is a scalar on some manifold, and $\partial_\mu\phi$ is thought of as the tangent vector $\partial_\mu$ as a derivation acting on $\phi$.

23. ### BenTheManDr. of Physics, Prof. of LoveValued Senior Member

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I've been thinking about this for a while, and can't find any problem with it. So tentatively I think this is correct.

Yes, it's just called the kinetic term, and it is an artifact of classical mechanics.

Hmm. They certainly are rotations. They are global in the sense that the phase $\alpha$ doesn't depend on a space-time coordinate.

Also, I am hesitant to comment on this:

Because I don't know what most of those words mean