Meaning of a phrase from "The algebra of grand unified theories". - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:54:43Z http://mathoverflow.net/feeds/question/118362 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118362/meaning-of-a-phrase-from-the-algebra-of-grand-unified-theories Meaning of a phrase from "The algebra of grand unified theories". Pablo Lessa 2013-01-08T14:37:22Z 2013-01-08T21:18:12Z <p>Motivated by an answer to <a href="http://mathoverflow.net/questions/57656/standard-model-of-particle-physics-for-mathematicians" rel="nofollow">this mathoverflow question</a> I've been making an effort to understand Baez and Huerta's article <a href="http://www.ams.org/journals/bull/2010-47-03/S0273-0979-10-01294-2/home.html" rel="nofollow">"The algebra of grand unified theories"</a>.</p> <p>As far as I can tell, mathematically, the first 30 pages or so describe one concrete finite dimensional representation of the group $U(1)\times SU(2)\times SU(3)$. The vectors of this space are to be interpreted as the elementary particles which do not mediate forces (i.e. fermions) and their anti-particles, while composed particles (made up of several fermions and anti-fermions) are represented by vectors of tensor powers of this space which must satisfy some additional conditions (such as being fixed by the action of elements in $1 \times 1 \times SU(3)$). Finally, the elementary force mediating particles (elementary Bosons) are represented by elements of the Lie algebra of the group.</p> <p>Assuming I've got that right there are still some phrases which make absolutely no sense to me, or worse, I can make sense of them only as completely trivial statements. The most extreme example I can find is the following (page 490 third paragraph):</p> <blockquote> <p>"... in 1936 a paper by Cassen and Condon appeared suggesting that the nucleon’s Hilbert space $\mathbb{C}^2$ is acted on by the symmetry group $SU(2)$."</p> </blockquote> <p>The literal interpretation is that Cassen and Condon suggested that it is possible to multiply a $2\times 2$ matrix with a $2\times 1$ vector. Obviously this can't be what that authors intended to communicate. </p> <p><strong>So my question is: What's the meaning of this phrase?</strong></p> <p>The paper refered to is "On nuclear forces" from Phys. Rev. 50 (1936) which I don't have access to. From <a href="http://books.google.fr/books?id=3U2HSMHsouMC&amp;lpg=PA410&amp;ots=WXCS_1nBC8&amp;dq=on%2520nuclear%2520forces%2520cassen%2520condon&amp;pg=PA410#v=onepage&amp;q=on%2520nuclear%2520forces%2520cassen%2520condon&amp;f=false" rel="nofollow">this page</a> it seems they proposed some model accounting for the experimental fact that the strong nuclear force doesn't depend on the electrical charge of the particles involved. I still don't get it. Any help?</p> http://mathoverflow.net/questions/118362/meaning-of-a-phrase-from-the-algebra-of-grand-unified-theories/118364#118364 Answer by Jeff Harvey for Meaning of a phrase from "The algebra of grand unified theories". Jeff Harvey 2013-01-08T15:00:42Z 2013-01-08T16:28:09Z <p>I suspect part of the confusion is due to the fact that the $SU(2)$ appearing in the Standard Model gauge group $U(1)\times SU(2) \times SU(3)$ is different from the $SU(2)$ of the Cassen-Condon paper. The latter is usually called isospin and is an approximate global symmetry of nuclear interactions. It is only an exact symmetry in the limit that one ignores electromagnetic interactions and the mass difference between the up and down quarks. The $SU(2)$ of the standard model gauge group on the other hand is a (local) gauge symmetry. I'm assuming here that you understand the difference between global and local symmetries as there phrases are used in the physics literature. If not, please consult any book on quantum field theory.</p> <p>The particular phrase you are asking about is simply the statement that the isospin part of the nucleon (i.e. (neutron, proton)) Hilbert space is an $SU(2)$-module with $SU(2)$ the approximate isospin symmetry of the nuclear interactions. </p> http://mathoverflow.net/questions/118362/meaning-of-a-phrase-from-the-algebra-of-grand-unified-theories/118388#118388 Answer by Igor Khavkine for Meaning of a phrase from "The algebra of grand unified theories". Igor Khavkine 2013-01-08T19:48:00Z 2013-01-08T19:48:00Z <p>Some physical background is needed to interpret phrases like these. In quantum mechanics, interactions between two particles (either composite or elementary) are described by an operator, called a Hamiltonian $V$ that acts as a linear operator on the tensor product $H_1\otimes H_2$, where $H_1$ and $H_2$ are Hilbert spaces corresponding to individual particles (or rather, their possible states). After much simplification, these Hilbert spaces can be reduced to the finite dimensional ones discussed in the paper by Baez and Huerta.</p> <p>Starting already in the very early days of quantum mechanics, and progressing even to this day, physicists have spent a lot of time trying to work out the right expression for $V$ that would be consistent with all the experimental data available to them. In many hypotheses were of the form, $V=V_a+V_b+\cdots$, where $a,b,\ldots$ indicates the kind of force is responsible for the interaction (in the early days of Cassen and Condon, mostly electromagnetic or nuclear forces were known). Such hypotheses worked reasonably well, and there were several candidate interaction Hamiltonian $V_n$ hypothesized to express the nuclear forces.</p> <p>Now, any group that acts linearly on a Hilbert space, induces an action on linear operators on that space (by conjugation). In particular, if for two nuclear particles we have $H_1\cong H_2\cong \mathbb{C}^2$, there is an action of $SU(2)$ on either Hilbert space, as well as the standard product action on the tensor product $H_1\otimes H_2$. Question, what is then the induced action on $V_n$? If $V_n$ is fixed under this action, then $SU(2)$, with the given action on $H_1$ and $H_2$ is called a symmetry of the interaction.</p> <p>As far as I understand, Cassen and Condon proposed that a $V_n$ that is <strong>symmetric</strong> under the defining action of $SU(2)$ would fit well with the available experimental data on nuclear interactions, provided the effects of electromagnetic interactions were ignored.</p> http://mathoverflow.net/questions/118362/meaning-of-a-phrase-from-the-algebra-of-grand-unified-theories/118394#118394 Answer by John Baez for Meaning of a phrase from "The algebra of grand unified theories". John Baez 2013-01-08T21:01:30Z 2013-01-08T21:18:12Z <p>Our sentence indeed loses its point taken out of context. Condon and Cassen were not merely claiming that we can multiply a vector in $\mathbb{C}^2$ by an element of $\mathrm{SU}(2)$. They were claiming that when $\mathbb{C}^2$ is used as the Hilbert space for nucleons, this transformation is a <em>symmetry of the laws of physics</em>. In other words: if we have a solution of the equations of motion (whatever they are), we can apply this transformation and get a new solution. </p> <p>More physically: since $\mathbb{C}^2$ is the Hilbert space for a nucleon, with one basis vector standing for a proton and one for a neutron, it means that we can do things like replace all the protons in the universe by neutrons and all the neutrons by protons, and this will have as little effect as, say, rotating the whole universe, or moving everything in the universe two feet to the left. </p> <p>Of course this is <em>false</em>, taken literally. Protons and neutrons behave differently with respect to the electromagnetic force, since protons are electrically charged and neutrons are neutral! But of course Cassen and Condon knew this. They (and other people, like Heisenberg) were trying to focus on the nuclear force and temporarily ignore the electromagnetic force. They hoped that in a universe like ours, but with the electromagnetic force turned off, and only the nuclear force remaining, the laws of physics would have this $\mathrm{SU}(2)$ symmetry.</p> <p>If you read on, you'll see this hope turned out to be <em>false</em>. Nonetheless it was incredibly fruitful, because it got physicists thinking about Lie groups, and it got Yang and Mills to invent the Yang-Mills equations, which are very important in our current thinking about particle physics.</p> <p>Summary: the difference is between a mere action of a group on a Hilbert space, and an action as symmetries of the laws of physics. That's why we said "symmetry group", not just "group".</p>