I come aross an OPE between stress tensor components in CFT which is  \begin{equation} T(z)\bar{T}(\bar{w})\sim -\frac{\pi c}{12}\partial_{z}\partial_{\bar{w}}\delta^{(2)}(z-w)+... \end{equation} I am confused about this OPE. Because in general in CFT chial part does not correlated with anti-chiral part, for example in free fermion and boson. So how to derive this OPE?
123873
In his lectures on String Theory (), Tong gives a proof of the Weyl anomaly, using equation $(4.36)$. It seems wrong to me.  Here he uses the OPE between the stress-energy tensors $T_{zz}T_{ww}$  obtained when trace vanishes, i.e. $T_{z \bar{z}} = 0$: this implies that they are holomorphic functions $T_{zz} = T_{zz}(z)$. But in this proof he starts from the fact that $T_{z \bar{z}} \neq 0$ (we want to proof this thing after all) and so $T_{zz}$ is not a holomorphic function anymore! In the OPE $(4.36)$ I should have also terms with $(\bar{z}- \bar{w})$.  I can't also understand why he uses in the rest of the proof only the singular term $(z-w)^{-4}$, neglecting the other ones $(z-w)^{-2}$, $(z-w)^{-1}$.  (The same proof is given in these lectures  on conformal field theory, equation $(6.9)$).  I'll be really thankful if someone could explain me this proof :)
118336
I come up with this conclusion after reading some books and review articles on conformal field theory (CFT).     CFT is a subset of FT such that the action is invariant under conformal transformation of the fields and coordinate but leave the metric unchanged.   Is this correct?   Let me explain further and take the $\phi^4$ theory in $4$-dim as an example (just discuss classical invariance, I know that loops break the invariance). In $4$-dim consider a scalar field with conformal weight $\Delta=1$ such that  \begin{align} x \to x' = \lambda x,\\ \phi'(x')=\phi'(\lambda x) = \lambda^{-1}\phi(x). \end{align} Then the action is unchanged \begin{align} S'&amp; = \int d^4 x' \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\partial'_{\mu}\phi'(x')\partial'_{\mu}\phi'(x')-\phi'^4(x')\right\}\\ &amp;=  \int d^4 x \lambda^4 \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\lambda^{-4}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\lambda^{-4}\phi^4(x)\right\} \\ &amp;=  \int d^4 x \sqrt{g} \left\{\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\phi^4(x)\right\}. \end{align} Note that I did not use $g'^{\mu\nu} = \lambda^2g^{\mu\nu}$, all metrics are unprimed. In this example we see that conformal invariance is realized without changing the metric. I was confused at the beginning since all textbooks and articles derive the conformal group and representations by considering the change of the metric.  If we use $g'_{\mu\nu}x'^{\mu}x'^{\nu} = g_{\mu\nu}x^{\mu}x^{\nu}$, the physical distance does not change at all and we are just choosing a new coordinate chart. My interpretation is, as what we meant by a physical scaling or transformation, we really change the distance between two points. Another reasoning is, the metric in CFTs are just background (not being integrated in the path integral) thus we do not change them. If we consider a theory including the metric as a dynamical field (we path integrate it and perhaps quantize it), the actions has to be invariant including the transformation of the metric.   Is the above correct? Please give me some comments and point out the wrong concepts if there is any. Thank you very much.  If you have time, could you please take a look at .
117372
Given sectional curvature as a constant, i.e. $\dfrac{R_m(X,Y,Y,X)}{|X|^2|Y|^2-&lt;X,Y&gt;^2} = C$, I want to compute the curvature tensors $R(X,Y)Z$ and $R_m(X,Y,Z,W)$.  I believe I need to use the identity $$-6R_m(X,Y,Z,W) = \partial_t\partial_s\{R_m(X+sZ, Y+tW, Y+tW, X+sZ) - R_m(X+sW, Y+tZ, Y+tZ, X+sW)\}$$  I can almost see what to do for $R_m(X,Y,Z,W)$ and how this identity and the defintion of sectional curvature are related but can't really see how to finish it off. I don't really even know where to begin with $R(X,Y)Z$.
9321
$PHP^{-1}=-H$ (particle-hole symmetry) and   $\Gamma H \Gamma^{-1}=-H$ (chiral symmetry)   I understand why we get the negative signs but im just a bit confused as to why such equalities mean $H$ is particle hole and chiral symmetric.   When we say $H$ is symmetric under some operation, like time reversal,  don't we usually mean that $H$ is invariant under the said transformation?  $THT^{-1}=H$ if $H$  is time reversal symmetric.  On the contrary, we say $H$ is symmetric under particle hole and chiral operations when $H$ picks up a negative sign. This seems to contradict our usual convention of what we mean by $H$ being symmetric(invariant).
50009
Considering  $$0\to \Bbb Z \to \Bbb Z\to \Bbb Z/(2)\to 0$$ We want to show that the tensor product is not left exact by tensoring with $\Bbb Z/(2)$, which apparently gives us what we want.  But I did this, and I get  $$0\to \Bbb Z/(2)\overset{\sim}{\to}\Bbb Z/(2)\overset{0}{\to} \Bbb Z/(2)$$   Where I can't see why this isn't exact? Is there a reason why I can't have the zero map on the right?
14908
This is essentially a follow-up on this question . Specifically, I'm studying the construction of the tensor product of two vector spaces as indicated by the OP's 2nd definition. Namely, that the tensor product of two vector spaces $V$ and $W$ is the quotient space $V \times W / E$ where $E$ is the subspace of $V \times W$ generated by linear combinations of the form  $$ (v_1 + v_2, w) - (v_1, w) - (v_2, w) $$ $$ (v, w_1 + w_2) - (v, w_1) - (v, w_2) $$ $$ (av, w) - a(v,w) $$ $$ (v,aw) - a(v,w) $$  We also denote an element of this quotient space by $v \otimes w$. Now, I'm trying to understand precisely why it follows from this construction that, for example, $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$. I believe it follows from the following argument:  First, we note that   $$ (v_1 + v_2, w) - (v_1, w) - (v_2, w) = (v_1 + v_2, w) - ((v_1, w) + (v_2, w)) $$  By definition of what it means to be a quotient space, this last equation means that  $$ [ (v_1 + v_2, w)] = [(v_1, w) + (v_2, w)] $$  where $[ \cdot ]$ denotes the induced equivalence class. But, according to the definition of the $\otimes$ operator, we can express this equality as  $$ (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w $$  Is this argument valid and is this the appropriate way to think about these things?
7834
$$Q=\int_{\text{all space}} \frac{\hbar \nu_g \mathbf{k}\mathbf{k}}{\exp[(\hbar \nu_g |\mathbf{k}|-\mathbf{k}\cdot\mathbf{u})/k_B T]-1}d\mathbf{k} $$   Please help me to integrate the above tensor expression in the infinite domain of the vector $k$.  I have tried to let $u$ in the direction of $k_z$ and then transform the current integral into a spherical coordinate with the following relation:  $$ k_x = k \sin \varphi \cos \theta $$ $$ k_y = k \sin \varphi \sin \theta $$ $$ k_z = k \cos \varphi $$  And I find $Q$ is a symmetric tensor, and $Q_{xy}=0,Q_{xz}=0,Q_{yz}=0$. However,$Q_{xx}=Q_{yy}\neq 0,Q_{zz}\neq 0$,and $Q_{xx}\neq Q_{zz}$. I guess $Q$ has relevance to the tensor $uu'$, the assumption that $u$ is in the direction of $k_z$ is just a simple case.
19824
I am new to QFT, so I may have some of the terminology incorrect.  Many QFT books provide an example of deriving equations of motion for various free theories. One example is for a complex scalar field: $$\mathcal{L}_\text{compl scaclar}=(\partial_\mu\phi^*)(\partial^\mu\phi)-m^2\phi^*\phi.$$ The usual "trick" to obtaining the equations of motion is to treat $\phi$ and $\phi^*$ as separate fields. Even after this trick, authors choose to treat them as separate fields in their terminology. This is done sometimes before imposing second quantization on the commutation relations, so that $\phi$ is not (yet) a field of operators. (In particular, I am following the formulation of QFT )  What is the motivation for this method of treating the two fields as separate? I intuitively want to treat $\phi^*$ as simply the complex conjugate of $\phi,$ not as a separate field, and work exclusively with $\phi$.  Is it simply a shortcut to obtaining the equations of motion $$(\square +m^2)\phi=0\\ (\square + m^2)\phi^*=0~?$$  I also understand that one could write $\phi=\phi_1+i\phi_2$ where the two subscripted fields are real, ; perhaps this addresses my question in a way that I don't understand.
65274
I'm trying to understand the energy-momentum tensor $T^{\mu\nu}$ but I'm confused about the units. My textbook says the components of $T^{\mu\nu}$ are $\mathrm{Jm^{-3}}$. Four-momentum is is given by$$P^{\mu}=\left(E/c,\mathbf{p}\right)=\left(E/c,p_{x},p_{y},p_{z}\right)$$  The $E/c$ component of $P^{\mu}$ has units $\mathrm{Jsm^{-1}}$. The definition of $T^{\mu\nu}$ is “the rate of flow of the $\mu$ component of four-momentum across a surface of constant $\nu$.” Using this definition, how do you get the rate of flow of the $E/c$ component of four-momentum across a surface of constant time (ie the $T^{00}$ component) to have the correct units of $\mathrm{Jm^{-3}}$? Surely you need to multiply $E/c$ by something with $\mathrm{s^{-1}m^{-2}}$ units, but what exactly?  Thank you
6496
I am stuck with an exercise in Sean Carroll's Spacetime and Geometry (Chapter 4, Exercise 3). The goal is to show that the continuity of the energy-momentum tensor, i.e. \begin{equation} \nabla_\mu T^{\mu\nu}=0\tag{1} \end{equation} is equivalent to the geodesic equation in the case of a free particle. The energy-momentum tensor of a free particle with mass $m$ moving along its worldline $x^\mu (\tau )$ is \begin{equation} T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2} \end{equation} Taking the covariant derivative of this tensor gives $$\begin{align} \nabla_\mu T^{\mu\nu}=&amp;m\int d \tau \nabla_\mu\left[ \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\right]\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\cr &amp;+m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\nabla_\mu\left[\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}\right].\tag{3} \end{align}$$ The first covariant derivative of the right-hand side of the above equation reduces to an ordinary partial derivative, as the argument is a scalar. This allows us to apply partial integration to this term. The second covariant derivative has an argument that is not explicitly dependent on $y^\sigma$, so the covariant derivative can be written as a multiplication of this tensor with the appropriate Christoffel symbols. This finally leads us to $$\begin{align} &amp;-m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{d^2x^\nu}{d\tau^2} \cr &amp;+  m\int d \tau \frac{\delta^{(4)}(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\left[ \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau} \right].\tag{4} \end{align}$$ The continuity equation requires \begin{equation} -\frac{d^2x^\nu}{d\tau^2} + \Gamma^\mu_{\mu\sigma}\frac{dx^\sigma}{d\tau}\frac{dx^\nu}{d\tau} + \Gamma^\nu_{\mu\sigma}\frac{dx^\mu}{d\tau}\frac{dx^\sigma}{d\tau}=0.\tag{5} \end{equation} This is the geodesic equation with an extra term, i.e. the term in the middle and with an incorrect sign for the first term. Can I get rid of this term in the middle by changing the parameter $\tau$ of the worldline? What about the incorrect sign? What did I do wrong?
23257
I am trying to derive Eq. (7.25) (p. 117) of Polyakov's book:  $$ \delta \Psi (C) ~=~ \int_{0}^{2\pi} {\rm P} \left(F_{\mu\nu}(x(s)) \exp \oint_C A_\mu dx^\mu \right)\dot{x}_\nu \delta x_\mu(x) \, {\rm d} s, \tag{7.25} $$  where the non-abelian phase factor around a closed loop $C$ is defined as  $$ \Psi(C) ~=~ {\rm P}\exp \left(\oint A_\mu dx^\mu \right) = {\rm P}\exp \left(\int_{0}^{2\pi} A_\mu \dot{x}_\mu\, {\rm d}s \right). \tag{7.1} $$  It seems that he is using the relation given on p. 116:  $$ \delta \, {\rm P} \exp \int_{0}^{2\pi} M(\tau) {\rm d}\tau ~=~ \int_{0}^{2\pi}{\rm d}t\,{\rm P} \left(\delta M(t) \exp \int_{0}^{2\pi}M(\tau){\rm d}\tau\right). \tag{7.24b} $$  Matching with (7.25) I find $\delta A_\nu = F_{\mu\nu} \delta x_\mu$. This relation seems to be saying that if I change the position of the loop at the parameter $s$ by $\delta x_\mu(s)$ then the vector potential changes by $\delta A_\nu(x(s)) = F_{\mu\nu}(x(s)) \delta x_\mu(s)$.    I don't know how to derive this relation. Is it legitimate?
81851
If $\Omega=\{a,b,c\}$ give an example that $E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)$   If I choose the $\sigma$-algebras disjoint (or better said: one is not fully contained in the other) is the inequality always satisfied ?
4014
You can think this of the following as a 3d QFT where we try to calculate the self-energy of two fields. $I$ is a this external self-energy and let us assume it does not depend on the loop momenta which I denote a $\vec{x}=(x,y)$. Consider the following propagator  $$ P(\vec{r}) = \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} \,\, \gamma_i. $$ Then we want to calculate $$ I(x) = \int \frac{\,dx \,dy\, dz}{(2\pi)^3} \text{Tr} \left( \frac{x-ia}{2i^2[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,}\gamma_x \gamma_x  \right), $$ where Tr$(\gamma_i \gamma_j)=2\delta_{ij}$. Therefore  I am trying to resolve the following integral where $x,y,z \in \mathbb{R}$ and $a,b,c$ are constants (internal self-energies if you wish) in $\mathbb{R}$.  Then, after simplifying the propagator $P(x,y)$ we end up with the following expression for the $x$ component:  $$ I = \frac{1}{2i^2} \int \frac{dxdy}{(2\pi)^2} (x-ia) \int \frac{dz}{2\pi} \frac{1}{[\,(x-ia)^2+(y-ib)^2+(z-ic)^2]\,} $$ but it can be shown that  $$ \int \frac{dq}{(q+iv)^2+w^2} = \frac{\pi}{|w|}\theta(|w|-|v|).  $$ (e.g. see here ) Thus we get that for $|(y-ib)^2+(x-ia)^2|&gt;|a|$, $\theta=1$ (for this is what we are interested about) and then we are left with $$ I = \frac{1}{4i^2} \int \frac{dxdy}{(2\pi)^2} \frac{(x-ia)}{|(x-ia)^2+(y-ib)^2|}.   $$ The question is wether I can make the transformation $$ (x-ia) = q \cos \theta  $$ $$ (y-ib) = q \sin \theta $$ in polar coordinates. Then $dx dy = qdqd\theta$ and the previous integral becomes, assuming $|q|&gt;0$, $$ I = \int_0^{\infty} dq \int_0^{2\pi} d\theta \frac{q \cos \theta}{|q|} =0. $$ Is this correct? I am confused since one can find complex poles in the $I$. If not, what would the correct way to resolve this be? I have a suspicion that this integral should indeed give zero but how can we find this?
87592
If one were to derive the stress-energy tensor for a metric with $(+,-,-,-)$ signature would it be different from the stress-energy tensor derived from the same metric but with $(-,+,+,+)$ signature?
109578
How do I find the values for the structure constants ($f^{abc}$) of a gauge group, as a mean to find the tree-level amplitude for a process in which one of the tree-level diagrams is:    the process being: $$Z_L(p_1) +Z_L(p_2) \to Z_L(p_3) + Z_L(p_4) $$
42564
The title says it all: what is the diagram that defines a tensor product?  (I'm using the term diagram here in the  it has in category theory.)  Edit: This question was motivated by the following statement:  "...all limits and colimits are initial objects somewhere, and we can even sort of take the view that all universal properties are initial objects somewhere", from  (@~9:15)  Thanks!
105550
Recently I came across the following result in a paper, which I can't seem to find right.  Anyway the authors had cited  by Johnson &amp; Kotz, page 20 as the source. Since I do not have access to this book I am unable to check on the derivation of this result. I was hoping someone could shed some light on how to prove this.  Let $X$ and $Y$ denote a pair of bivariate log normal random variables with their means and variances given by the ordered pairs $(\zeta_{X}, \eta_{X})$ and $(\zeta_{Y}, \eta_{Y})$ and with correlation coefficient $\rho_{(X, Y)}$, derived from the bivariate normal with marginal distributions having parameters $(\xi_{X}, \nu_{X}^2)$ and $(\xi_{Y}, \nu_{Y}^2)$ and correlation coefficient $\rho_{N}$.  The expression for $\rho_{(X, Y)}$ is given as \begin{equation} \rho_{(X, Y)} = \dfrac{\mathrm e^{(\rho_{N} \nu_{X} \nu_{Y})} - 1}{\sqrt {(\mathrm e^{\nu_{X}^2} - 1)(\mathrm e^{\nu_{Y}^2} - 1)}} \end{equation}  The below is a manipulation I came up with, any error in that would be completely mine.  Upon rearranging we get \begin{equation}\begin{aligned} \rho_N &amp;= \dfrac{\ln (1 + \rho_{(X, Y)} \sqrt {(\mathrm e^{\nu_{X}^2} - 1)(\mathrm e^{\nu_{Y}^2} - 1)})}{\nu_{X}\nu_{Y}}\\ &amp;= \dfrac{\ln (1 + \rho_{(X, Y)} \frac{\eta_{X}\eta_{Y}}{\zeta{X}\zeta{Y}})}{\sqrt {\ln ((\frac{\eta_{X}}{\zeta_{X}})^2 + 1)\ln ((\frac{\eta_{Y}}{\zeta_{Y}})^2 + 1)}}. \end{aligned} \end{equation}  I am very interested in learning if this result would hold true between $X$ and $Y$'s correlations if we had a quad-variate situation as in there were 2 more random variables which are also correlated with $X$ and $Y$ would this result still be true ? I suspect it should be, but I am newbie and due to lack of a credible source I was wondering if some one would be kind to confirm it.