The QH QM QA thread.

OK I'll stop with the questions, and let the original seeker after quantum truth take over, as should be the case.

BTW I'd say I'm somewhere beyond high-school with QM, but admit, apart from undergrad chemistry and physics the only post-grad stuff I have under my belt, are a couple of CS courses, one parallel processing, and one other which is kind of algorithmical, but there it is, that's my lot.
As for Berry phase, I have a copy of his article in SA from way back in the 80s, and I've done a bit of snooping to see what's up with it lately. It appears to be a useful kind of mathematical insight, and an interesting one, like the Aharanov-Bohm effect is interesting, and the Hall effect is.

AlphaNumerix said:
...how to go from a representation of a vector bundle to it's equivalent principle bundle or how to use a connection to seperate the tangent space of a bundle into it's fibre's tangent space and the uplifted path section from the base manifold...
Click to expand...
Not bad, could have been a bit pithier.
"Base manifold" is a projection, like say a circle is the base space, or projection of a mobius loop or a sphere? This is an important bit, the other two bits are the projected set of lines, (tangent bundle) and the space itself - the double loop or the sphere, or any abstract space?. The ideal representation would be a line segment? Look it up in Wikipedia?
 
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OK, folks, rather sooner than expected, I think I may be making progress with gauge theory.

First fibre bundles. Let $$M$$ be a manifold; we will assume it is connected and Hausdorff. Hey, what the heck, assume also it is compact and $$C^{\infty}$$, though I don't think we need these here.

Then to each point $$p \in M$$ I may associate a space of vectors $$T_p(M)$$ called the tangent space at p. Notice that there is no possibility whatever of vectors in $$T_p(M)$$ and $$T_q(M)$$ "talking" to each other. However, I can form the set-theoretic disjoint union of all such spaces $$\bigsqcup \nolimit_i T_p_{(i)}(M) \equiv TM$$ which is called the tangent
bundle over M.

The co-tangent bundle is found equivalently.

So the fibre bundle is merely a generalization of this concept, that is the fibres need not be vector spaces. But we need to exercise some caution here; there may be, and generally is, more than one one way to form such a bundle (c.f. the bundle on $$S^1$$ vs. the bundle on the Moebious). So, the difference between all such possible bundles is given by what's called its "structure group", which I do know how to explain, if anyone wants

Now to the matter at hand. Suppose that $$\psi$$ is a scalar-valued function on (possibly "non-flat") spacetime $$M$$. Let $$U(1)$$ be an element in the fibre bundle over Minkowski spacetime. And since $$U(1)$$ is the complex unit circle group, a section of this bundle defines a global gauge transformation on Minkowski space $$\psi \to \psi^{i \phi}$$ in the obvious way. I cannot convince myself this is true for non-flat space, however. Help?

But I now require a local gauge transformation $$\psi_p \to \psi^{i \phi(p)}$$. I offer this for starters:

Define a curve $$\mathcal{C} \in M$$ parametrized by elements $$p \in M$$. Let's say this is an integral curve. Then "travelling along" this curve I am merely following the "direction pointed to" by some vector in $$T_p(M)$$.

But psi defines a scalar field, so I will also encounter a sequence of co-vectors (one-forms), which, allowing for "non-flatness", I will assume to be the gradient at that point, $$ (\nabla \psi)_p$$

For each point on the corresponding curve $$\mathcal{C}^*$$ on the $$U(1)$$ bundle, there will be a choice of elements of $$ U(1)$$ such
that $$\mathcal{C} \to \mathcal{C}^*,\;\; \psi_p \to \psi e^{i \phi(p)}$$.

That is, since I am not assuming "flatness", for each point p, there is a choice of gauge transformations, dependent on the value of phi at that point. I am going to require that $$\mathcal{C}^*$$ is in some sense "parallel" to $$\mathcal{C}$$, that is the transformation $$\psi_p \to \psi e^{i \phi (p)}$$ is constant for any point p.

So it looks like I require some transformation acting on phi, and dependent on the various gradients, that somehow "corrects" for the variations that otherwise prevent my U(1) curve from being parallel to my M curve.

Do I know how to do this? No, not yet. Don't tell me! (but please correct the obvious errors). I'm having (masochistic) fun trying to figure it out
 
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Well the description I gave of fibre bundles was so vague as to be useless. I want now to into more detail, because there is information in there which I think may be relevant.

A fibre bundle is a space $$B$$ equipped with a base manifold $$M$$, a typical fibre $$F$$, a projection $$\pi: B \to M$$ and a set of homeomorphisms $$h_i:F \to F$$.

The bundle $$B$$ is said to be globally trivial if $$B = M \times F$$. As part of the definition, I will require the bundle to be locally trivial. That is, for any neighbourhood $$U$$ of $$M$$, the bundle $$B_U = U \times F_U$$.

A number of things should leap out at you. First is that, since any homeomorphism has a true inverse (by definition), the set of all homeomorphisms $$F \to F$$ has the structure of a group.

Second is that the pre-image of some point $$p \in M$$ under the projection $$\pi$$ is in fact the fibre at $$p$$. That is $$F_p = \pi^{-1}(p)$$.

Let's see a simple example, which will be useful. Let the base manifold $$M = S^1$$, the unit circle, and let a typical fibre $$ F = R^1$$. Let's call $$B = TS^1$$. Then local triviality ensures that, for $$U_i \subset S^1$$ that $$TU_i = F_{U_{(i)}} \times U_i$$. The projection will be $$\pi: TS^1 \to S^1$$, and locally $$\pi:TU_i = F_{U_{(i)}} \times U_i \to S^1,\; \pi(x,p) = p$$ where $$x \in R^1,\; p \in S^1$$.

Stare at this a while......

Notice that the domain of the projection is the set of ordered pairs of real numbers (x,p), i.e. $$R^2$$. Thus, the local triviality condition implies that $$TS^1$$ is locally homeomorphic to $$R^2$$, and therefore is, by definition, a manifold!

Now notice this; the homeomorphism $$h: R^1 \to R^1$$ sends real number to real numbers, self-evidently. Suppose there is some $$p \in U_i \cap U_j \in S^1$$. Then the fibre at p has 2 homeomorphisms $$h_i(p) \to R^1,\; h_j(p) \to R^1$$.

But homeomorphisms are invertible, so $$h_j \cdot h_i^{-1}$$ is also a homeomorphism $$R^1 \to R^1$$. From this it follows that $$h_j(p) \;\cdot \;h_i^{-1}(p) = \gamma_{ij} = \alpha_j/\alpha_i$$, all real, so none of these guys is allowed to be zero. Hence the structure group is given as $$R^1\setminus \{0\}$$, the non-zero reals.

I'd better get back to work, so quickly this: the frame bundle is defined to be the bundle of all possible bases for each fibre, which in the present case is again the structure group $$R^1\setminus\{0\}$$.

This is part of the definition of the principal bundle; for this latter case, I need to impose some more restrictions, but duty calls.
 
Do you not need to define precisely what your h is doing? For instance, all you need to know is that it's invertible and it maps R to R. You seem to be restricting it to multiplication, hence why h(x) = kx isn't allowed to have k=0. But that's too restrictive. What's wrong with all strictly increasing (or descreasing, and thus invertible) functions on the reals?

For instance, h(x) = sinh(x) is invertible. You just need an invertible function which has the domain and range of the entire reals.

For instance, h(x) = x+k for some real k is fine, you're just doing a translation in your 1d space, rather than a dilation as you were considering with h(x) = kx, |k|>0.

/edit

And Reiku, this is the kind of fundamentals expected to be within the grasp of anyone doing research level material in relativity or string theory. It's the foundation of rigorous differential geometry. So not only is it 'nice', it's something you should have come across by now, if you aren't lying about your progress through education ;)
 
To nitpick, I think you are talking about vector bundle, not general fibre bundle since you introduce frame bundle. And I think one should require local trivialization on some open cover of the manifold, not on any neighborhood U.
 
temur: Yes and yes, particularly I should have said that the set U, as I introduced it, was in an open cover of the base manifold.

Alpha: Well it's true my argument for the structure group of the fibre bundle over the unit circle was weak (worse, it was circular!). But even if I am wrong, I don't feel compelled to "do the decent thing", since I merely wanted to show what role the structure group plays and that the bundle has the structure of a manifold, as I think this is key - we'll see.

So a principal bundle is simply a fibre bundle in which a typical fibre is the structure group (well, up to homeomorphism). Thus, suppose $$M$$ is a connected Hausdorff $$C^{\infty}$$ manifold.

Then the principal bundle $$P(G,M)$$ over $$M$$ consists of a manifold $$P$$ (the total space), together with a structure group $$G$$, a projection as before and local triviality as before, with the following constraints:

I will require there to be an action of $$G$$ on $$P$$, that is $$P \times G \to P$$ called (merely by convention) the right action such that $$R_g(p) = pg$$;

I require this action to be "free" in the sense that $$R_g(p) = p \Rightarrow g = e$$, the identity on $$G$$. (I note in passing, this means there is no fixed point - tough tits, Herr Brouwer!).

I desire the group $$G$$ to have a topology. This is easily arranged for arbitrary groups by invoking the powerset (i.e. discrete) topology. In the present case, I am going to be greedy, and say I want this group to have the additional structure of a manifold, i.e. to be a Lie group.

So suppose that, in the above construction, $$M$$ is Minkowski, and that $$G = U(1)$$, i.e. we are doing EM. Define a curve $$\mathfrak{C} \in M$$. For my gauge transformation to be kosher, I want the transformed curve $$\mathfrak{C}' \in P$$ to be in some sense "parallel", or more strictly, I want an arbitrary gauge transformation to respect the gradients we talked about t'other day.

In the case that $$P$$ were Riemann, I would use the affine connection together with the covariant derivative to establish parallelism. But it's not, as far as I can tell, and in any case, parallelism is not really what I am after; stated strictly, I require that the transformation $$\psi \to \psi'$$ to respect the gradient, but I think I will need something similar.

Umm, first bit is easy (I think) Define a vector potential $$A$$ such that, for some vector $$V = \nabla \times A$$. Now suppose that $$A' = A + \nabla \phi$$. Then
$$V' = \nabla \times A' $$
$$= \nabla \times (A + \nabla \phi)$$
$$ = \nabla \times A +\nabla \times \nabla \phi= \nabla \times A$$

since my last equality contains the term curl(grad) = 0 on the LHS. Therefore V' = V under this transformation.

So this takes care of phi, which, unless I am told not to be so silly, I will think of as the "phase" (or do I mean "angle"?) at each point $$p' \in \mathfrak{C}'$$ I will think of it as my "connection", that is, as $$\psi \to \psi e^{i \phi(p)},\; A \to A + \nabla \phi$$

But now I need to find a derivative $$D_{\mu}$$, depending on both $$A$$ and $$\psi$$ such that $$D_{\mu}(\psi e^{i \phi (p)}) = e^{i \phi (p)}D_{\mu}\psi$$. That is,I need to be able to actually specify the curve $$\mathfrak{C}' \in P$$ such that at each point $$p'$$ the phase is the same as at the corresponding point $$p \in \mathfrak{C} \in M$$

Short of an ugly trial and error session, I can't see where to start.

I am skiving again - this must stop!
 
Ha! On a quick review of this entire thread, I see my last post merely re-iterates some of what Ben and Alpha told me yonks ago! Sorry guys; my excuse is that I don't really understand stuff until I have worked it out for myself from first principles
 
QuarkHead said:
Short of an ugly trial and error session, I can't see where to start.
Click to expand...
Well start with a connection, so that $$D_{\mu} = \partial_{\mu} + gA_{\mu}$$ and see what transformations you'd have to have on $$A_{\mu}(x)$$ to give $$D_{\mu}(e^{i\alpha(x)}\phi) = e^{i\alpha(x)} D_{\mu}\phi $$ (or whatever your 'kinetic' term is) gauge invariant.
Reiku said:
Well, i've not seen this work... Forgive me if my course is doing something else.
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This stuff is end of degree level material, if not post grad. The book I just emailed to Quarkhead, Nakahara, is a textbook on geometry and topology for physicists, aimed at postgrads, and connections and bundles aren't covered till the last 1/3 of the book.

So not only is your course 'doing something else', it's about 6 years behind this!
 
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This stuff is end of degree level material, if not post grad.
Click to expand...
Not bad considering this started 7 pages ago with protons and neutrons, electrons and neutrinos with "mass" (and "charge"). Particles whizzing around somewhere.

To a well-understood symmetry group, and how it maps to the EM field, and you can see that it's really a "fibre", or a connection in an abstract space, a discrete mathematical object, which is pretty cool.
I've studied the EM symmetry group in a Physics course to 3rd year, but they didn't call it a fibre bundle, or mention the word tensor once (we looked at Laplace and Lagrange transforms though, and how to use matrix algebra and hyperbolic geometry in circuit analysis) - I didn't happen across fibre bundles until I read that article that I quoted, by Bernstein and Phillips. The EM model extends to nuclear interactions, if you extend the model a little.

But isn't it kind of meaningful that the whole thing is mathematically describable, in a kind of exact way, or that quantum reality is somehow fundamentally mathematical?
 
So, when all else fails...... cheat. Looking into the covariant derivative in Lovelock & Rund (Section 3.3, pp70 - 71, Dover edn), I remind myself that, for a type (1,0) tensor field $$X^j$$I will have

$$DX^j = dX^j + \Gamma_h^j_k X^h dx^k$$

And for a type (0,1) field $$Y_h$$, I will have

$$DY_h = dY_h - \Gamma_h^l_k Y_l dx^k$$

where the Gammas in the last term in each of these identities is the (Christoffel) affine connection.

So, remembering that the gradient of a scalar function is a type (0,1) field, let's see if I can simply "import" this last equality into the present case.

Let $$A$$ be our connection, $$\psi$$ our scalar field, and try $$D\psi = \nabla_{\mu}\psi - A_{\mu} \psi$$. But now recall that we are working on the manifold $$P$$, where a right action of the group $$U(1)$$ is required. Let's assume the group action is addition, and that we may as well work with this group's generator. Let's assume this is $$1+1i$$.

Then the above becomes $$D \psi = \nabla_{\mu}\psi- iA_{\mu} \psi$$.

And finally recalling that as $$\psi \to \psi e^{i \phi (p)},\; A \to A + \nabla
\phi$$, and that $$A = A + \nabla \phi$$, let's see if this works. Deep breath, fingers, legs, toes and eyes crossed........

$$D(\psi e^{\phi(p)}) =\nabla_{\mu}\psi e^{i \phi(p)} - i(A_{\mu} + \nabla _{\mu} \phi)\psi e^{i \phi(p)}\\
= e^{i \phi(p)}(\nabla_{\mu} \psi - iA_{\mu} \psi)\\
=e^{i \phi(p)}D \psi $$.

Yay!! Good old me!

Now a serious physics question: how come, when it's so hard to put a fresh quilt cover on your bed, when you put it in the washing machine everything always ends up inside it? Must be some serious research money in this conundrum
 
QH said:
...how come, when it's so hard to put a fresh quilt cover on your bed, when you put it in the washing machine everything always ends up inside it? Must be some serious research money in this conundrum
Click to expand...
I think it's related to the same principle that applies to tangled power cords or garden hoses - if you try some shaking to untangle them, it makes a bigger tangle instead.
Er, the "random fibre knotting" theory, perhaps...?
 
Quarkhead, you need to be careful about how you're ordering your gauge change.

As I'm certain you know, right action and left action are different for anything other than U(1) so you should be doing the following :

If $$\psi \to e^{i\alpha(x)}\psi$$ then you need to show that $$\nabla_{\mu}\psi \to e^{i\alpha(x)}\nabla_{\mu}\psi$$

If $$\psi \to \psi e^{i\alpha(x)}$$ then you need to show that $$\nabla_{\mu}\psi \to \left(\nabla_{\mu}\psi\right) e^{i\alpha(x)}$$

Otherwise gauge invariant terms like $$\psi^{\dag}\psi $$ would end up going to $$e^{i\alpha(x)}\psi^{\dag}\psi e^{-i\alpha(x)}$$.

It's particularly apparent when you start considering vectors of things and spinors because $$\psi$$ and $$\psi^{\dag}$$ are fundamentally different.
 
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Yes, you are right; in my search for an aesthetic presentation, I was rather blindly assuming commutativity. Thanks for that pointer.

Now suppose I were crazy enough to try and find the gauge transformations for what Ben called the electroweak sector. First I note there are 3 generators for SU(2); I assume this will mean I want 3 transformations that are, in some sense, related to that for U(1)?

Second, the neonatal physicist in me says I will need to include a mass term in there somewhere?

Obviously I will need to specify a different field - or not?

As to whether I can use the same connection will require some thought on my part - let that hang a while.

But lookee here; you physics jocks seem to be miserly bunch. Considering that when I started this thread I didn't understand what an electronvolt is, and that I thought the neutrino was structural in the atomic nucleus, I think it might be fair to say "the boy done good".

Ah well, self-satisfaction is a sin, after all, so forget it
 
Now suppose I were crazy enough to try and find the gauge transformations for what Ben called the electroweak sector. First I note there are 3 generators for SU(2); I assume this will mean I want 3 transformations that are, in some sense, related to that for U(1)?
Click to expand...

Well, I'll give you a hint. Use an exponential function as before, except this time you will have three $$\alpha(x)$$'s, and they will carry an SU(2) index a = 1,2,3.

But lookee here; you physics jocks seem to be miserly bunch. Considering that when I started this thread I didn't understand what an electronvolt is, and that I thought the neutrino was structural in the atomic nucleus, I think it might be fair to say "the boy done good".
Click to expand...

Indeed:)
 
Yes Ben, thanks. But this presents me with a problem; as Alpha rightly (if somewhat tartly) pointed out, matrix operations are not in general commutative.

So. I seem to have (at least) two equally ugly choices. I could think of the operations on the SU(2) generators as elements of the symmetric group $$S_3$$, which is, of course, of order 3! = 6. I cannot say if this makes sense in physics terms, but it would a nasty construction, mathematically.

Or, I could index the generators by some ordered field, but this again would involve an ugly choice of assignment of the generators to this field.

Or, I suppose I might make use of a gadget I once encountered, but have now forgotten, that looks a bit like

$$\epsilon_{ijk} =\begin{cases}+1\quad \text{if}\; (i,j,k ) \quad \text{is an even permutation of (1,2,3)}\\
-1 \quad \text{if}\; (i,j,k) \quad \text{is an odd permutation of (1,2,3)}\\
0 \quad \text{if any indices are equal}\end{cases}$$

I will look it up tomorrow, but I have no intention of doing anything serious tonight; it's Friday, and I am about to meet my mates, get hammered and talk about sport and tits (as one does!)
 
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Hmm, nobody want to help me out here? Anywyay, it looks like my introduction of the $$\epsilon_{ijk}$$ as defined above may have been a lucky guess. Ok, I did this; it follows relatively straightforwardly that

$$\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} = -\epsilon_{kji}=-\epsilon_{jik} =-\epsilon_{ikj}$$.

Then knowing that to get at the $$SU(2)$$ gauge transformation I would need the right action of this Lie group on my principal bundle, I set the generators of $$SU(2)$$ as

$$G_1 = \begin{pmatrix}0&1\\
\\1&0 \end{pmatrix} \; G_2 = \begin{pmatrix}0&-i\\
\\i&0\end{pmatrix}\;G_3 = \begin{pmatrix}1&0\\
\\ 0&-1 \end{pmatrix} $$.

Then, following Ben's hint (as far as I understood it at all), I worked out the group commutators. Without boring you with details they came up as

$$[G_i,Gj] = \pm i2G_k$$, i,j,k = 1,2,3. Ignoring the annoying factor of 2, this suggests the following;

$$[G_i,G_j] = i \epsilon_{ijk} G_k$$.

Edit: On the off-chance that any one is still trying following this gibberish, let me explain the preceding, which was poorly worded. Here any of the i, j or k can be either 1 OR 2 OR 3, hence the permutation rule in my last post. Any better?

So, if I am allowed to model on the $$U(1)$$ case, but without, for now, assuming my fields are the same, I will offer the following;

as $$\varphi \to \varphi^{i \alpha(p)},\; B \to B + \nabla \alpha$$, then the relevant gauge transformation will be (or maybe include the terms?)

$$D \varphi =\nabla_{\nu}\varphi -B_{\nu}\varphi \;\cdot \; i \epsilon_{ijk} G_k$$

where the centre dot merely indicates that I am not too sure what the operation should be, nothing more.

C'mon, guys, help me out, as I have a sinking feeling that I am being far too simplistic.
 
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As always I end up on this site talking to myself. This is OK for a period, but gets pointless after a while.

I don't know why the experts will no longer engage; if they are too polite to tell me I am an idiot, let me tell them they need have no such manners. As I said earlier - on this subject, I have no pride.

So, in spite of the fact that this thread seems to be moderately popular, in order that I waste no more time looking for nonexistent replies, let alone the work I put into trying to figure out the basics of this subject, may I respectfully request this thread be closed
 
QuarkHead said:
Ok, I did this; it follows relatively straightforwardly that

$$\epsilon_{ijk} = \epsilon_{jki} = \epsilon_{kij} = -\epsilon_{kji}=-\epsilon_{jik} =-\epsilon_{ikj}$$.
Click to expand...
A more short hand way of writing that is $$\epsilon_{ijk} = \epsilon_{[ijk]}$$.
QuarkHead said:
$$[G_i,Gj]=\pm i2G_k[\tex]$$
Click to expand...
$$Why the $$\pm$$?
QuarkHead said:
where the centre dot merely indicates that I am not too sure what the operation should be, nothing more.
Click to expand...
It's not written as anything. Since everything other than the gauge group generators are abelian, they are just multiplicative factors.

You've generally get something like $$D_{\mu} = \partial_{\mu} + A^{a}_{\mu}T^{a}$$. Sure, the form of $$A^{a}_{\mu}$$ might change but it'll be nothing but scalars.

I've vaguely forgotten what you're trying to do? Are you trying to find a transformation rule which makes $$D_{\mu}(e^{i\alpha(x)}\psi) = e^{i\alpha(x)}D_{\mu}(\psi)$$? Expand out $$D_{\mu}$$ and do the various differentiations, collect the terms which give the required result and remember that $$A_{\mu}$$ changes too. In the non-abelian case I think it goes to 3 terms.$$
 
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