The QH QM QA thread.

OK, so I have now figured out the mathematics of whole gauge thing, at least as far as I wanted to go. I shan't post my reasoning, as it would look like another of my boring monologues; I promised myself I wouldn't do that here again.

Suffice it to say, I had a hard time deriving the connection in the present context (if anyone is interested, it has to do with something called the "horizontal lift" of a tangent vector space). And, suffice it again; my derivation makes explicit use of the unitarity of the relevant Lie groups, which is nice. Oh, and finally, the derivation also makes use of the Lie algebra, as well as the unitary group transformations.

With that boastful statement out of the way, here's another question. What is meant by a "coupling constant". Obviously I know what a constant is, but what is coupling to what, exactly?

Please.
 
Despite my failings so far in this thread, I will give this a shot :)

So you've worked out in detail what happens in the gauge sector of a theory. That is, untill now you haven't done anything that REALLY resembles nature (unless it's photon scattering experiments at energies below the electron mass!).

What you need to do now is add some fermions. Let's take the simple case first of QED and add some electrons. You know what the derivative looks like for a U(1) gauge theory:

$$D^{\mu}=\partial^{\mu} + ieA^{\mu}$$.

I will state that the Dirac lagrangian is given by

$$\mathcal{L} = \bar{\psi} D\cdot \gamma \psi - m\bar{\psi}\psi$$.

where $$m$$ is the fermion's mass.

WTF is $$\gamma_{\mu}$$, you may ask. Well, it is a (4x4) Dirac Matrix, and the dot stands for contracting the space-time index $$\mu$$. Just take my word for it (for now!) that there are four of them, one for each space-time direction. This (I'm sure) will interest you more in a bit, but let's get to the coupling constant part first. We could write the equation above the long way:

$$D\cdot \gamma = D^0\gamma_0 - D^1\gamma_1 - D^2\gamma_2 - D^3\gamma_3$$

If you write out the Dirac lagrangian you find

$$\mathcal{L} = \bar{\psi} \partial\cdot \gamma \psi + ie\bar{\psi} A \cdot \gamma \psi - m\bar{\psi}\psi$$.

Notice the term that goes $$ie\bar{\psi} A \cdot \gamma \psi$$...this term describes an interaction between a photon, an electron, and an anti-electron (positron). The interaction strength, or ``coupling constant'' is $$e$$.

So a coupling constant just tells you the way that a particle interacts with a gauge field.

Now we'll need Alpha to tell us some about the Dirac matrices, because anything I say about them will probably burn your mathematical ears :)
 
Ben, I thank you for that attempt. As usual, you are streets ahead of me - remember I am just a simpleton in this area. In particular, in fact particularly in particular, I am am bit hazy on Lagrangians (they are part of mechanics/dynamics, as far as I understand?). Do I need to understand them to progress here?

Maybe some kind person might give me a gentle guide?

Anyway, frankly, I didn't understand your post well enough to even frame a sensible question. Sorry. Leave it with me a while........
 
Dammit. Apologies of course.

Hopefully Alpha can phrase this is a manner that speaks to you:)
 
Would a simple description be that a coupling constant is like a tension or elasticity in the connection?
 
I dug this up from somewhere:
"
Early work in nuclear beta decay seemed to show that the relevant interaction was a mixture of S, T, and P.
Only after the discovery of parity violation, and the undoing of several wrong experiments, did it become clear that the weak interactions were in reality V-A. The synthesis of Feynman and Gell-Mann and of Marshak and Sudarshan was a necessary precursor to the notion of a gauge theory of weak interactions.
Bludman formulated the first SU(2) gauge theory of weak interactions in 1958. No attempt was made to include electromagnetism. The model included the conventional charged-current interactions, and in addition, a set of neutral current couplings.
These are of the same strength and form as those of today’s theory in the limit in which the weak mixing angle vanishes.

Of course, a gauge theory of weak interactions alone cannot be made renormalizable. For this, the weak and electromagnetic interactions must be unified.
...
We know now that neutral currents do exist and that the electroweak gauge group must be larger than SU(2).

Another electroweak synthesis without neutral currents was put forward by Salam and Ward in 1959.
Again, they failed to see how to incorporate the experimental fact of parity violation. Incidentally, in a continuation of their work in 196 1, they suggested a gauge theory of strong, weak, and electromagnetic interactions based on the local symmetry group SU(2) x SU(2). This was a remarkable portent of the SU(3) x SU(2) x U(1) model which is accepted today.
...
We finally saw that a gauge group larger than SU(2) was necessary to describe the electroweak interactions.

Salam and Ward were motivated by the compelling beauty of gauge theory. I thought I saw a way to a renormalizable scheme. I was led to the group SU(2) x U(1) by analogy with the approximate isospin-hypercharge group which characterizes strong interactions. In this model there were two electrically neutral intermediaries: the massless photon and a massive neutral vector meson which I called B but which is now known as Z.
The weak mixing angle determined to what linear combination of SU(2) x U(1) generators B would correspond.
The precise form of the predicted neutral-current interaction has been verified by recent experimental data."

--Sheldon Lee Glashow
TOWARDS A UNIFIED THEORY - THREADS IN
A TAPESTRY
Nobel Lecture, 8 December, 1979
 
Ben: I was too hasty in my last response (I was in the middle of something else, and didn't give your considered post the attention it deserved. I will do so this evening - for as long as I am relatively sober!)

However I did find a foot-note in a Lie group text that mentions Dirac matrices in the context of supergroups. It also mentions these guys are important in super-symmetry theory, which I gather is your bag.

So, I learned from this foot-note, that the Dirac $$\gamma$$ matrices take the form $$\begin{pmatrix} 0&\sigma^{\mu}\\
\\
\tilde{\sigma}^{\mu}&0
\end{pmatrix},\quad \tilde{\sigma}^{\mu} = (\sigma^0, -\sigma^i),\quad \sigma^i = \text{Pauli}$$.

But they don't say what the $$\sigma^0$$ is, neither do they say what is has to do with coupling.

However, you did rightly guess roughly what I was asking; I know that classically the coupling constant is nothing more exotic than a number associated to an interaction. Newton's G-constant springs to mind. So I should have asked; what is the explanation for this in QFT (and yes, I am aware that there is, as yet, no QFT for gravity). Does the concept transfer simply like, say:

the coupling constant is that number that describes the "coupling" of a (?matter?) particle to the (bosonic - is this a word?) gauge field that describes the interactions between these (?matter?) particles? Or something? Edit: This is somewhat tautological; lemme try again.

The coupling constant describes the coupling of a particle to the gauge field, this coupling being mediated by a gauge boson with a characteristic coupling constant. Any better?

We can do some arithmetic later, if you like
 
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Ben, I am sooooo sorry (and so stupid); you said
BenTheMan said:
So a coupling constant just tells you the way that a particle interacts with a gauge field.
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Then I said, feeling really clever and original
QuarkHead said:
The coupling constant describes the coupling of a particle to the gauge field, this coupling being mediated by a gauge boson with a characteristic coupling constant. Any better?,
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Doh. It looks like I didn't read your post properly. That was wicked of me - in fact it was unforgivable.

I obviously have some work to do here (but bear in mind I just do this stuff for fun!)
 
OK, Alpha, Ben, temur, and others, I realize I am totally out of my depth here.

In a sycophantic attempt to talk to you physics guys on your own terms, I persuaded an ex-girlfriend at ICL to lend me a couple of texts that explain the Lagrangians in classical and quantum field theories, neither of which I am familiar with.

I am reading furiously, but I cannot promise to come back on this. As I said, I think I am in over my head
 
I am reading furiously, but I cannot promise to come back on this. As I said, I think I am in over my head
Click to expand...

I think I take for granted the learning curve on these things. It took me a long time to be comfortable with these things. It doesn't mean they're hard or anything, it just takes some time to be comfortable with this language, at least. So it is partly my fault because many times I didn't take time to explain things properly---I just assumed that you would catch things as I threw them to you, regardless of how fast I was pitching.

So don't quit! Please! I love consolidating things here, and at the very least you are teaching me how to be a bit more rigorous about things!
 
OK, Ben, at the risk of humiliating myself even further, I will continue with a couple of questions that arise from my reading.

But first, thanks to Alpha for his almost legible notes. They were a great adjunct to my texts. So here are my questions:

It seems that, when applying the classical Lagrangian formalism to fields (classical or otherwise) one introduces the quantity $$\mathcal{L}$$ by

$$ L \equiv T - V = \int d^3 x \;\mathcal{L}$$, which is called the Lagrangian density, and is also!! called the Lagrangian. (I swear I shall end up serving time for physicists' notational abuses). One text leaves this as a definition. The other is slightly (but only slightly) more forgiving, and says it is a density since its space integral is the true Lagrangian. Leaving aside the pathetic fact that I don't know what a "space integral" is - I only know of line, surface and volume integrals - my question is:

How do I interpret this integral? Usually one sees $$\int \text{{something}}\quad dx$$ where the dx just gives the derivative that' being integrated. And why to 3rd order? Surely I haven't carried out successive differentiations for each spatial dimension? Wouldn't I usually use partials? The above looks bonkers to me.

However, I am willing to swallow it whole if I must, as the rest seems to follow rather nicely.

My second question is slightly trivial, in that it refers to an example given. For what one text calls the Klein-Gordon Lagrangian I am given

$$\mathcal{L} = \partial_{\mu}\varphi^{\dag} \partial^{\mu} \varphi - m^2 \varphi^{\dag}\varphi$$. (This is reminiscent of an equation Ben once showed us). So what is the adjoint of a field? Here's my totally ad hoc answer.

The Klein-Gordon equation is found in EM field theory. Here, as elsewhere, from the quantum perspective, elements of the field are vectors - state vectors - and are elements in a normed $$L^2$$ space, known to its mates as Hilbert.

The inner product on such a space is defined as $$\int^b_a \overline{f}(x)g(x)\;dx = \langle f|g \rangle$$. This of course is real, positive-definite. The norm is defined in the obvious equivalent way.

So I have the conjugate covered. But I may, by taking the analogy with bra = row vector, ket = column vector, switch from one to the other by matrix transposition of their respective representations. Hence the adjoint.

And in order to preserve symmetry, if $$\partial_{\mu} \equiv \frac{\partial}{\partial x^{\mu}}$$ is a differential operator acting on the bra, then I must have $$\partial^{\mu} \equiv \frac{\partial}{\partial x_{\mu}}$$ acting on my ket.

I guess there is a sense in which these operators can also be thought of as transposes of each other?
 
QuarkHead said:
$$ L \equiv T - V = \int d^3 x \;\mathcal{L}$$, which is called the Lagrangian density, and is also!! called the Lagrangian. (I swear I shall end up serving time for physicists' notational abuses). O
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The equations of motion etc come out to be the same. Basically you're working on derivatives etc of the density.
QuarkHead said:
And why to 3rd order? Surely I haven't carried out successive differentiations for each spatial dimension? Wouldn't I usually use partials? The above looks bonkers to me.
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$$d^{D-1}x = dV$$. If you're working with dx = $$dx_{i}$$ then you're just integrating all the $$dx_{i}$$ one at a time. Physicsts prefer that notation rather than dV.

A space integral is just integrating over all the spacial dimensions. If you know how to do volume integrals, you know how to do higher order ones.

The scary stuff comes when you do renormalisation and you work in non-integer dimensions. See the pdf I linked to 2 posts ago.
QuarkHead said:
So what is the adjoint of a field? Here's my totally ad hoc answer.

The Klein-Gordon equation is found in EM field theory. Here, as elsewhere, from the quantum perspective, elements of the field are vectors - state vectors - and are elements in a normed $$L^2$$ space, known to its mates as Hilbert.

The inner product on such a space is defined as $$\int^b_a \overline{f}(x)g(x)\;dx = \langle f|g \rangle$$. This of course is real, positive-definite. The norm is defined in the obvious equivalent way.

So I have the conjugate covered. But I may, by taking the analogy with bra = row vector, ket = column vector, switch from one to the other by matrix transposition of their respective representations. Hence the adjoint.
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Again, see the pdf I linked to for a worked example.

If you were working with a single complex field, it would be $$\mathcal{L} = \partial_{\mu}\varphi^{\ast} \partial^{\mu} \varphi - m^2 \varphi^{\ast}\varphi$$. But if you have $$\varphi = \left( \begin{array}{c} \varphi_{1} \\ \vdots \\ \varphi_{n} \end{array} \right) $$ then to get a valid notion of the inner product and it being positive def you need to take the dagger, not the asterix.
QuarkHead said:
I guess there is a sense in which these operators can also be thought of as transposes of each other?
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If $$\varphi$$ is a real field, then it is just the transpose. If it's complex then you need to take the complex conjugate too so it's pos def.
 
Thanks Alpha. As you can see, my distaste for calculus is exceeded only by my incompetence at it!

I had (wrongly I now believe) interpreted $$d^3x$$ as an indicator that I was to integrate over a single variable that had been subject to 3 successive differentiations like, for example, $$\frac{d^3x}{dy^3}$$. It seems from what you said it should be read as, say, $$dx dy dz$$.

This makes sense, if true, since we may assume that $$V$$ depends on the spatial $$x_i$$ and (usually?) not at all on the $$\frac{dx_i}{dt}$$.
 
OK, here is todays dumb question. My author is deriving the Maxwell equations (in tensor form) from a variational principle. The action being varied is that of the vector potential $$A_{\mu}$$.

All goes swimmingly, until we get to this step: he finds (convincingly) that

$$-\frac{1}{4}(\delta F_{\mu \nu})F^{\mu \nu} = -\frac{1}{4}(\partial_{\mu}F^{\mu \nu} \delta A_{\mu}- \partial_{\nu}F^{\mu \nu} \delta A_{\nu}$$

($$F^{\mu \nu}$$ is the antisymmetric field strength tensor, btw)

At this point he decides he doesn't like the minus sign in the RHS, so points out, quite correctly, that the indices on $$F^{\mu \nu}$$ in the 2nd term above have no real meaning - they are labels, dummies - and feels fee to switch them; $$F^{\mu \nu} \to F^{\nu \mu}$$.

He then says Oooooh look, $$ F^{\nu \mu} = -F^{\mu \nu}$$, by antisymmetry, so he transposes, and plugs his "new" $$F^{\mu \nu}$$ in the above, thereby changing sign. He then blithely proceeds as if the $$F^{\mu
\nu}$$ in the 1st term is the same as his "new" $$F^{\mu \nu}$$ in the second term.

I doubt the legality of this trick, since , merely by re-labeling indices, he is not actually changing the matrix. Effectively all he is done is to declare that $$F^{\mu \nu} = -F^{\mu \nu}$$, which we know is not true.

I'm sorry not to give more details, but I am assuming any kind soul willing to comment on this will be familiar with the exercise.
 
QuarkHead said:
I doubt the legality of this trick, since , merely by re-labeling indices, he is not actually changing the matrix. Effectively all he is done is to declare that $$F^{\mu \nu} = -F^{\mu \nu}$$, which we know is not true.
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Check the other indices---I think you have a typo. There should always be as many up $$\mu$$s as there are down $$\mu$$s.

Either way, when I work it out I get (up to a constant)

$$\partial_{\mu} F^{\mu\nu}\delta A_{\nu} - \partial_{\nu} F^{\mu\nu}\delta A_{\mu}$$.

After that, try FIRST permuting the indeces:

$$\partial_{\mu} F^{\mu\nu}\delta A_{\nu} + \partial_{\nu} F^{\nu\mu}\delta A_{\mu}$$,

and THEN renaming them:

$$\partial_{\mu} F^{\mu\nu}\delta A_{\nu} + \partial_{\mu} F^{\mu\nu}\delta A_{\nu}$$.

Then you claim that $$\delta A$$ is arbitrary so that $$\partial_{\mu} F^{\mu\nu} = 0$$, and then go have a beer.

For extra credit, you can do it all in terms of differential forms :)
 
Thanks Ben you were right about the typo. For
QH said:
$$-\frac{1}{4}(\partial_{\mu}F^{\mu \nu} \delta A_{\mu}- \partial_{\nu}F^{\mu \nu} \delta A_{\nu})$$
Click to expand...
read $$-\frac{1}{4}(\partial_{\mu}F^{\mu \nu} \delta A_{\nu}- \partial_{\nu}F^{\mu \nu} \delta A_{\mu})$$.

As to permuting, let's see that in the second term;

$$\partial_{\nu}F^{\mu \nu} \delta A_{\mu} \to \partial_{\mu} F^{\nu \mu} \delta A_{\nu}$$.

Which is precisely as my author wrote! (Gosh, I hope he's not reading this shite)

My bad, as they so inelegantly say. Anyway, it all falls nicely into place - thanks, man
 
BenTheMan said:
Then you claim that $$\delta A$$ is arbitrary so that $$\partial_{\mu} F^{\mu\nu} = 0$$, and then go have a beer.
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Consider the beer bit done.

In fact my text says that I need to show that $$\partial_{\mu} F^{\mu\nu} - J^{\mu} = 0$$, where the second term on the LHS is "current density".

For extra credit, you can do it all in terms of differential forms :)
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Oh, clever, clever. Let's make a start.

The electric field tensor (let's recall it is an antisymmetric type (0,2) tensor) can also be defined as follows.

An arbitrary 2-form is given by the tensor product of 2 1-forms as $$\tilde{X}=\omega \otimes \pi - \pi \otimes \omega \equiv \omega \wedge \pi$$. Self evidently this implies that $$\tilde{X}$$ is antisymmetric.

But we know that the field strength tensor satisfies this condition, so let's define the 2-form $$\tilde{F} \equiv F_{\mu \nu}$$ and, for purposes of distinction, call it the "Faraday 2- form".

Umm, now I need the Hodge operator, and to be blunt I can't be arsed. Mary-Bernadette is away for the weekend, so I can indulge myself with no fear of sniping! But yes, I will claim those extra credits, just see if I don't!
 
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