Suppose we're looking at an $SU(N)$ symmetry in the adjoint representation. The kinetic part of the Lagrangian yields a term   $$ \Delta \mathcal{L} = -g^2 Tr\left[[T^a, \Phi][T^b, \Phi]\right]A^a_{\mu}A^{b\mu} $$  When $\Phi$ acquires a VEV (call it $\Phi_0$) the broken generators are those whose associated gauge bosons acquire mass, via the above term of the Lagrangian. This amounts to saying that the unbroken generators are those whose commutator with the VEV vanishes:  $$ [T^a, \Phi_0] = 0 $$  Let's say generically the VEV has the form  $$ \Phi_0 = diag(a, a, ..., a, b, b, ... b) $$  where there are $n_a$ of $a$, $n_b$ of b, and $Tr[\Phi_0] = 0$. My understanding is that there are effectively three cases for generators that will commute with the VEV:  1) $$ T^a =  \begin {bmatrix} A &amp; 0 \\ 0 &amp; 0 \\ \end {bmatrix} $$  where A is an $n_a \times n_a$ block  2) $$ T^a =  \begin {bmatrix} 0 &amp; 0 \\ 0 &amp; B \\ \end {bmatrix} $$  where B is an $n_b \times n_b$ block  3) $T^a$ is a diagonal matrix.  I understand why generators from case 1) would produce generators of SU($n_a$), and why generators from case 2) would produce generators of SU($n_b$). What I can't figure out is why diagonal generators (often proportional to the VEV) are generators of U(1). My understanding was that the only generator of U(1) was the identity. Why is it that any traceless diagonal matrix can produce U(1)?
122297
In GUTs one starts with some larger group, like $SU(5)$, which is then broken into smaller groups, for example  $$SU(5) ~\longrightarrow~ SU(3) \times SU(2) \times U(1)$$  This can be seen, for example, by looking at the Dynkin diagram for $SU(5)$: Removing one node leaves us with the Dynkin diagrams for $SU(3)$ and $SU(2)$.  My problem is understanding where $U(1)$ comes from. I've read several statements about that, but couldn't fit the puzzle pieces together.  Dynkin diagrams are one the level of Lie algebras. Removing one node means we remove one generator. For example, in  from the the course Symmetries in Physics by Michael Flohr:     By removing a node, the rank of the subalgebra is reduced by one, and the simple roots are a subset of the original simple roots.  On the level of the groups, we thus find, $$G=G_1\times G_2 \times U(1),$$ where the additional $U(1)$ factor comes from the left out Cartan generator. For example,   $SU(n+m)$ can be reduced in this way into   $$SU(n)\times SU(m)\times U(1).$$ This is the classical ansatz for a GUT:   $SU(5)$ gets broken into $$SU(3)\times SU(2)\times U(1).$$    In  by John Baez it is claimed that $SU(3) \times SU(2) \times U(1)$ is no subgroup of $SU(5)$ and some homomorphism is used the justify why $U(1)$ appears.     EDIT:   In Lie Algebras In Particle Physics: from Isospin To Unified Theories Georgi writes:     "The Cartan generator that have been left out generate $U(1)$ factors"   and my problem is understanding why this is the case.  Any ideas to clarify this would be awesome!
50221
Suppose we have a particle of mass $m$ that is in an eigenstate $|\psi\rangle$ of the Hamiltonian $\hat{H}=\hat{T}+\hat{V}$, where $\hat{T}$ is the kinetic energy operator and $\hat{V}=V(\hat{r})$ is the potential energy operator. If the potential has lower bound $V_0$, then is it necessary for the energy eigenvalue $E$ of $|\psi\rangle$ to be greater than $V_0$? Classically this is true, since we regard negative kinetic energy as physically unrealizable/meaningless. However, I don't know if I can necessarily say the same in the quantum case. For example, what I'm tempted to do is write  $$\langle\psi | \hat{T}|\psi \rangle = \langle\psi | E-\hat{V}|\psi\rangle$$  and say "If $E&lt;V_0$, then the RHS is necessarily negative, implying the LHS is as well, which we will regard as physically meaningless. If $E=V_0$, then $|\psi\rangle$ is trivial.", but I'm not sure if that's right.   Confused by that, I then wanted to show that, if $E&lt; V_0$, then a non-trivial $|\psi\rangle$ is non-normalizable. However, I'm not entirely sure how to do this.
109903
I'm having problems understanding the excersice: E) Define $D_n$ as the group of symmetries of a regular n-gon. Name the vertices $V=\{V_0,V_1,...,V_{n-1}\}$ so that $$V_{k}=\exp({i\cdot\dfrac{2\pi k}{n})}$$ Let $S$ be the symmetry along the x-axis, and $R$ the rotation by $\dfrac{2\pi}{n}$. Show that $\{R,S\}$ generate $D_n$.  I tried to define something like $$R:\exp(i\cdot\dfrac{2\pi k}{n})\mapsto \exp(i\cdot\dfrac{2\pi (k+1)}{n}) $$   $$S:\exp(i\cdot\dfrac{2\pi k}{n})\mapsto \exp(i\cdot\dfrac{2\pi (-k)}{n}))$$ but these are not binary operation so $(V,R) ; (V,S)$ are not groups. To me it's clear that there are n rotations and n reflection so there are 2n symmetries on the n-gon; also fixing one reflection we can get the others by some combination of $r,s$.  I tried to solve the problem by defining $R,S$ as permutation of vertices and the vertices as coordinates in a lattice and it solves nicely, but I don't seem to understand this way of approaching the definitions.
94795
The casimir operator $\textbf{L}^2$ commutates with the elements $L_i$ of the angular momentum operator $\textbf{L}$:  $$ [\textbf{L}^2, L_i] = 0. $$  However, the $L_i$ do not commute among themselves:  $$ [L_i, L_j] = i\hbar\epsilon_{ijk}L_k. $$  This makes sense so far, but it leaves me wondering how their eigenspaces relate to each other. I remember some theorem that diagonalizable, commuting matrices share their eigenspaces. If those operators could be expressed as complex matrices (in the finite-dimensional case), they surely are diagonalizable. So it follows that $\textbf{L}^2$ has the same eigenspaces as the three $L_i$, but that would imply that they commute among themselves, which is not the case.  What am I missing? What is the relation between the eigenspaces of these operators?
117696
This post is an exact copy of one that I posted in Math's site. I do this copy because people there suggested me to do it since, apparentely, in Mathematics and Physics we use different conventions for this kind of topics and none there could understand mine. (You can see first post here: )  Imagine you have the Young tableu and the Dynkin numbers, $(q_1, q_2, ..., q_r)$, of the Lie algebra of $SU(n)$ which has $r$ simple roots. The way I assign Dynkin numbers is increasing its value from left to right so the $k$-th Dynkin number is the number of columns with $k$ boxes: $q_k$ columns made of $k$ boxes.  The Young tableau is the 'usual' one with columns that decreases in boxes from left to right. The calculation of the dimension gives you some number $d$ that is given by  $$d = \frac{N}{H}$$  Where $N$ is the product of the following numbers: in the highest left box for $SU(n)$ write an $n$ and going to the right, increase this number in one unit box per box. Going down, decrease the number in the same amount box per box. $N$ is the product of all those numbers. $H$ is the product of the hook numbers: in each box write the number of boxes that you cut going from right (out of tableaux) to left till you reach that box and then keep cutting boxes going down from that box. Do this for each box and the product of these numbers (hook numbers) is $H$.  Now, my question is: how can I know if this Young tableau corresponds to the representation $d$ or to the complex conjugated $\bar{d}$ since both of them have the same dimension?    My source is:  sections 6.6.1 and 6.11
2666
In QED, according to Schwinger-Dyson equation $^{[1]}$, $$\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)\langle 0|\mathcal{T}A_{\nu}(x)...|0\rangle = e\,\langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle + \text{contact terms}.$$ And the term $\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)$ is just the inverse bare photon propagator$^{[5]}$, so if we put the photon on shell, then the l.h.s will yield the complete $n$-point Green function with the complete photon propagator removed and also multiplied by a factor $Z_3$, the vector field renormalization constant. But the r.h.s gives, according to the Ward Identity associated with global symmetry of the Lagrangian, $$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)...|0\rangle = \text{contact terms}^{[2]}$$ which is the sum of complete $(n-1)$-point complete Green functions each multiplied by a certain $\delta$ function.  So if we truncate all the $n-1$ external complete propagators, then we are left with the proper vertex Ward identity.    The problem is, now the constant $Z_3$ appeared.    For, example, if we apply this to $j^\mu=\bar\psi\gamma^\mu\psi$, $$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle = -ie[\delta(x-x_1)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle-\delta(x-x_2)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle]^{[3]},$$ we will get an identity containing $Z_3$ RATHER THAN the well known proper vertex Ward identity (which is derived from the gauge or local symmetry of the Lagrangian), e.g.   $$q_\mu\Gamma^\mu_P(p,q,p+q)=S^{-1}(p+q)-S^{-1}(p)^{[4]}$$ which doesn't contain $Z_3$.    Where went wrong? Please help.  [1]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page308, eq.(9.88) [2]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page310, eq.(9.97) [3]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page311, eq.(9.103) [4]. Lewis H.Ryder, Quantum Field Theory, page 263-266, eq.(7.112) [5]. Peskin&amp;Schroeder, An Introduction to Quantum Field Theory, page297, eq.(9.58)
35431
I've been studying different scattering processes (from Mandl &amp; Shaw QFT's book, chapter 8) and there's always a purely-mathematical common step I do not understand: the showing-up of the trace. Let me give two specific examples.  $$A_{(l) \alpha \beta}=\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}=Tr\Big[\frac{\not{\!p_2'}+m_l}{2m_l} \gamma_{\alpha} \frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big] \ \ \ \ (1)$$  $$X= \frac 1 2 \Lambda_{\delta \alpha}^+ (\vec p') \Gamma _{\alpha \beta} \Lambda_{\beta \gamma}^+ (\vec p) \tilde \Gamma _{\gamma \delta}=\frac 1 2 Tr \Big[\Lambda^+ (\vec p') \Gamma \Lambda^+ (\vec p) \tilde \Gamma \Big] \ \ \ \ (2)$$  Where:   $\not{\!A} := \gamma^{\alpha} A_{\alpha}$ $\Gamma$ is a $4 \times 4$ matrix $\tilde \Gamma := \gamma^0 \Gamma^{\dagger} \gamma^0$ $u, v$ are the Dirac Spinors (which are $4 \times 1$ matrices) The Dirac-$\gamma$-matrices are $4 \times 4$ matrices and the adjoint is defined as $\bar w := w^{\dagger}\gamma^{0}$ $\Lambda^+$ is the positive energy projection operator defined as  $$\Lambda_{\alpha \beta}^+ (\vec p) := \Big( \frac{ \not{\!p}+m}{2m} \Big)_{\alpha \beta}$$ $\Lambda^+$ has the following property   $$\Lambda_{\alpha \beta}^+ (\vec p) = \sum_{r=1}^2 u_{r \alpha} (\vec p) \bar u_{r \beta} (\vec p)$$  But what I do not understand is why does the Trace show up in $(1)$, $(2)$   Any help is appreciated.  EDIT  Let's write out the spinor indices explicitly for $(1)$  $$A_l = \Big(\sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')\Big) \gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Big(\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')\Big)\gamma_{\color{gray}{\beta}\color{red}{\delta}} \ \ \ \ (3)$$  We know the following properties  $$\Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')= \sum_{s_1} u_{s_1 \color{red}{\delta}}(\vec p_2') \bar u_{s_1 \color{blue}{\alpha}}(\vec p_2')=\Big(\frac{\not{\!p_2'}+m_l}{2m_l}\Big)_{\color{red}{\delta}\color{blue}{\alpha}} $$  $$\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')= -\sum_{s_2}v_{s_2 \color{green}{\nu}}(\vec p_1') \bar v_{s_2 \color{gray}{\beta}}(\vec p_1')=-\Big(\frac{\not{\!p_1'}-m_l}{2m_l}\Big)_{\color{green}{\nu}\color{gray}{\beta}}$$  Thus we get  $$A_l = \Lambda^+_{\color{red}{\delta}\color{blue}{\alpha}}(\vec p_2')\gamma_{\color{blue}{\alpha}\color{green}{\nu}}\Lambda^-_{\color{green}{\nu}\color{gray}{\beta}}(\vec p_1')\gamma_{\color{gray}{\beta}\color{red}{\delta}}=-\operatorname{Tr}\Big[\frac{\not{\!p_2'}+m_l}{2m_l}\gamma_{\alpha}\frac{\not{\!p_1'}-m_l}{2m_l}\gamma_{\beta}\Big]$$  Now I have two questions:  1) Why are we allowed to manipulate $\sum_{s_1} \sum_{s_2} \Big[ \bar u_{s_2} (\vec p_2') \gamma_{\alpha} v_{s_1} (\vec p_1'))(\bar v_{s_1}(\vec p_1')\gamma_{\beta} u_{s_2} (\vec p_2'))\Big]_{(l)}$ in such a way to get $(3)$? What I mean is that I do not see what mathematical properties allow us to do so.  2) I get a negative sign. I guess it gets cancelled out due to an antysymmetric swap of certein indices, but what pair specifically?  Thank you :)
26262
I know that derivative couplings in a Lagrangian interaction, such as   $$\mathcal{L}_{int} = \lambda \phi (\partial_{\mu}\phi)(\partial^{\mu}\phi)$$  bring down two momentum factors into the matrix element $\mathcal{M}$.   Why doesn't this extend to a typical kinetic term  $$\mathcal{L}_{kin} = (\partial_{\mu}\phi)(\partial^{\mu}\phi)\quad ?$$  It seems odd that derivatives in $\mathcal{L}_{int}$ contribute a factor of momentum$^2$ to $\mathcal{M}$, but the derivatives in $\mathcal{L}_{kin}$ contribute a factor of the propagator $$\frac{1}{\mbox{momentum}^2}$$  to $\mathcal{M}$.
75677
This might be a stupid question, but I just don't get it. In Hamiltonian mechanics when examining conditions for a $(\boldsymbol{q},\boldsymbol{p})\rightarrow(\boldsymbol{Q},\boldsymbol{P})$ transformation to be canonical one starts with $$ \dot{q}_ip^i-H(\boldsymbol{q},\boldsymbol{p},t)= \dot{Q}_iP^i-\bar{H}(\boldsymbol{Q},\boldsymbol{P},t)+\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)$$ where $\bar{H}$ is the transformed Hamiltonian, and $W$ is the generating function (now a function of $\boldsymbol{q}$ and $\boldsymbol{Q}$). This term shouldn't break Hamilton's principle, since $$ \delta\int_{t_1}^{t_2} dt\frac{d}{dt}W(\boldsymbol{q},\boldsymbol{Q},t)=\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_2}-\delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=0-0=0 .$$ But I don't see why the variation of $W$ should disappear at the endpoints (say at $t_1$). Expanding leads to: $$ \delta W(\boldsymbol{q},\boldsymbol{Q},t)|_{t_1}=\left(\frac{\partial W}{\partial q_i}\right)_{t_1}\underbrace{\delta q_i(t_1)}_{=0}+ \left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1)=\left(\frac{\partial W}{\partial Q_i}\right)_{t_1}\delta Q_i(t_1).$$ $\boldsymbol{Q}$ is itself a function of $\boldsymbol{q}$ and $\boldsymbol{p}$, so $$ \delta Q_i(t_1)=\left(\frac{\partial Q_i}{\partial q_k}\right)_{t_1}\underbrace{\delta q_k(t_1)}_{=0}+\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1)=\left(\frac{\partial Q_i}{\partial p_k}\right)_{t_1}\delta p_k(t_1). $$ It seems as if we also needed the variation of $\boldsymbol{p}$ to vanish at the endpoints, and I don't get this because (at least in cartesian coordinates) $\boldsymbol{p}=m\dot{\boldsymbol{q}}$ and the velocity can be different along the original and the varied orbitals even at the endpoints (they can point in totally different directions), so in general $\delta \dot{\boldsymbol{q}}(t_1)\neq 0$. What am I doing wrong? Can someone help me with this, please?
39829
I feel like this question has been asked before but I can't find it. would the Euler Lagrange equation for, say, the standard model Lagrangian be $$\frac{\partial L}{\partial \phi}=\partial_\mu \frac{\partial L}{\partial (\partial_\mu \phi)}$$ Where $\phi$ is whatever field is an question and $\mu$ is (I believe) being summed from 0 to 3. Or, ist the correct equation $$\frac{\partial L}{\partial \phi}=D_\mu \frac{\partial L}{\partial (D_\mu \phi)}$$ Where $D$ is the covariant derivative of the theory. My intuition tells me its the second eqn but i just wanted to be sure, and I think I once saw someone say that the two were equivalent.
85625
In the context of group theory (in my case, applications to physics), I frequently come across the phrase "the ${\bf N}$ of a group", for example "a ${\bf 24}$ of $\mathrm{SU}(5)$" or "the ${\bf 1}$ of $\mathrm{SU}(5)$" (the integer is usually typeset in bold).  My knowledge of group theory is pretty limited. I know the basics, like what properties constitute a group, and I'm familiar with simple cases that occur in physics (e.g. rotation groups $\mathrm{SO}(2)$, $\mathrm{SO}(3)$, the Lorentz group, $\mathrm{SU}(2)$ with the Pauli matrices as a representation), but not much more. I've got a couple of related questions:   What is meant by "${\bf N}$ of a group"? Is is just shorthand for an ${\bf N}$ representation? If so, what exactly is an ${\bf N}$ representation of a given group? How can I work out / write down such a representation concretely, like the Pauli matrices for $\mathrm{SU}(2)$? I'd be grateful for a simple example. What does it mean when something "transforms like the ${\bf N}$"?
14073
My major is physics. I need to use some tools in group theory, but I am really  confused by the trace in compact infinite groups. The following is my question:  In discrete group theory, the irreducible representation of identity group element $e$ is always an identity matrix. So the trace of $e$ under  regular representation is the order of the group : $\chi(e) = |G|$. I hope to get  similar result for Lie group. For example, SO(2) has infinite number of irreducible representations, $D^{(m)}$, where $m=0,\pm 1, \pm 2,\cdots $. All of them are 1-dimensional $D^{(m)} = e^{im\phi}$. Here $\phi$ is the rotation angle. For element $R(\phi)$ in SO(2), the trace in $m$th irreducible representation should be  $\chi(R(\phi))=e^{im\phi}$, if we sum all these traces up, we get the trace in regular representation. So what is the trace of $e$ in the regular representation ? It seems to be infinite.
41321
The $\mathrm{U(1)}$ QED case has two physical degrees of freedom, which is easy to understand because the free electromagnetic field must be transverse to the direction of propagation. But what are the physical degrees of freedom the $\mathrm{SU(N)}$ Yang-Mills theory?   I believe that the $\mathrm{SU(N)}$ Y.M. theory has four degrees of freedom, and it is easy to see that by gauge fixing (such as the Lorentz condition or the Coulomb condition) that we can always remove one redundant degree of freedom. But then we still haven't completely fixed the redundant degrees of freedom. Thus my question is:     How many physical degrees and how many are redundant degrees of freedom does the Y.M. theory have?    I would be interested to understand how we can determine this mathematically, and also to understand what the physical intuition behind this is?
37300
Show that the Lie algebras of $SO(4)$ and $SU(2) \times SU(2)$ are isomorphic.  I managed to show that $so(4) = \{A \in M_4(\mathbb{R}) \mid A^T = -A\}$ and $su(2) = \{A \in GL_2(\mathbb{C}) \mid A^\dagger = -A, \text{tr}(A) = 0\}$ with $su(2) \times su(2)$ being the Lie algebra of $SU(2)\times SU(2)$. However I am not able to find any mapping that preserves the Lie bracket structure. I am aware of the Pauli matrices being the generators of $su(2)$ but I can't find similar generators of $so(4)$ that preserve the bracket.
94808
If we have a complex scalar $\phi$ we know that the gauge-invariant interaction with $A$ is given by $A^\mu J_\mu$, where $J$ is the Noether current of the $U(1)$ symmetry of the Lagrangian $$ J_\mu\sim \phi^\dagger\partial_\mu\phi-\phi\partial_\mu \phi^\dagger \tag{1} $$  If we have a real scalar instead, $J=0$ and the field doesn't couple to $A$: real particles aren't charged.  My question is: what if we took $$ J_\mu\sim \phi\partial_\mu\phi \tag{2} $$ to couple $\phi$ to $A$? The interaction is Lorentz invariant and renormalisable, but it is not gauge-invariant, which is probably a bad thing. At what point would this theory break down?  This is probably very naïve, but I think that the Feynman rules for this theory are straightforward. So I guess the theory makes sense at least perturbatively. The theory is probably flawed at a more fundamental level, but I cant't seem to find where (some kind of gauge anomaly perhaps?)
115742
The Hamiltonian   $$H =  \left[ \begin{array}{cccc} a &amp; 0 &amp; 0 &amp; -b \\ 0 &amp; 0 &amp; -b &amp; 0\\ 0 &amp; -b &amp; 0 &amp; 0\\ -b &amp; 0 &amp; 0 &amp; -a \end{array} \right] $$  commutes with the qubit exchange operator  $$ P = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{array} \right] $$  So I would expect the two to have the same eigenvectors. The eigenvectors of $P$ are easily seen to be $(1,0,0,0)^T; (0,0,0,1)^T; (0,1,1,0)^T ; (0,1,-1,0)^T$. The latter two are also eigenvectors of $H$, but the first two are not. Why? I thought commuting operators shared the same eigenbasis?
122488
Suppose a torus $T=(\mathbb{C}^\ast)^n$ acts on a finite dimensional vector space $W$, and define for $m \in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by  $$W_m = \{w \in W \mid t\cdot w = \chi^m(t)w \text{ for all }t \in T \}$$ i.e. for $w \in W_m$ is a simultaneous eigenvector for all $t \in T$, with eigenvalue $\chi^m(t)$ depending on $t \in T$. Then it is a famous fact $$W=\underset{m \in M} \bigoplus W_m$$ Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.