As asked in the title, how do the Weyl spinors $(\frac{1}{2},0)$ and $(0,\frac{1}{2})$ differ from dotted and undotted spinors?
121956
I have had an introduction to QFT following the book of Mandl and Shaw. However, I have been asked to write a report on the CPT theorem. For this, the main reference I'm using is PCT, spin and statistics and all that, by Streater and Wightman.  I find that the introduction of dotted and undotted indices is really unclear in this book. I also found a reference on dotted indices in Local Quantum Physics by Haag, but I do not understand what he is talking about.  Can someone explain, what the indices are, how they are defined and most importantly, how, for a given spinor, do you know what their indices are supposed to be?  For example, what is the difference between $\psi$, $\psi_\alpha$, $\psi_\dot{\alpha}$, $\psi^\alpha$, $\psi^\dot{\alpha}$ and $\psi^{\alpha \beta \dot{\gamma}}_{\delta \dot{\epsilon} \dot{\zeta}}$?  The problem I think, lies with the fact I have no experience with representation theory, nor has the course I had ever mentioned such things. So if you could give or reference an explanation of these things, without the need for representation theory, that would be most helpful. Thanks!
108904
I have done problem 4.50 from Griffiths' Quantum Mechanics (2nd ed), quoted below, and got the correct answer. What would be a physical situation where this would be useful?  I have already had some people tell me to look at Bell's Theorem in Ch 12, which I did. But that is seven chapters ahead of where I am at in the book, and contains things that are over my head at this point. Can anyone tell me, in plain undergrad terms, a physical situation where this would be useful?     Problem 4.55 Suppose two spin-$1/2$ particles are known to be in the singlet configuration (Equation 4.178). Let $S_a^{(1)}$ be the component of the spin angular momentum of particle number 1 in the direction defined by the unit vector $\hat{a}$. Similarly, let $S_b^{(2)}$ be the component of 2's angular momentum in the direction $\hat{b}$. Show that   $$\langle S_a^{(1)} S_b^{(2)} \rangle = -\frac{\hbar^2}{4}\cos\theta,\tag{4.198}$$   where $\theta$ is the angle between $\hat{a}$ and $\hat{b}$.
23047
Let's say I have the ability to crunch a given quantity of particles into a black hole.  I now use a sizeable quantity of electrons to do this, with the restriction being that all of them have $+\frac{1}{2}$ spin.   Apart from the charge, does this intrinsic angular momentum exhibit itself in the black hole?  Would the resulting black hole resemble a Reissner–Nordström black hole or a Kerr-Newman black hole?   Would it be different from a black hole that I make out of equal amounts of $+\frac{1}{2}$ and $-\frac{1}{2}$ electrons? It should be, shouldn't it? (because the information about those spins should be conserved)
43754
Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined? For example, given the  for a  $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian? As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a  equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?  [1]: From Matthew D. Schwartz's , p.114:  A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.
86362
Any rank 2 tensor would do the same trick, right?  Then, which is the motivation for choosing the metric one?    Also, if you help me to prove that $g^{kp}g_{ip}=\delta^k_i$, I would be thankful too ^^
109577
Basically what I'm trying to show is $\forall h_1, \ h_2 \in \mathscr{H}$ and $U: \mathscr{H} \rightarrow \mathscr{K}$ then $\langle Uh_1, \ Uh_2\rangle_\mathscr{K} = \langle h_1, \ h_2 \rangle_\mathscr{H}$  (So your standard definition). Now I figure the easiest way to do so is by polarization.  Since we have an isometry then   $\begin{eqnarray*} \langle h_1, \ h_2 \rangle_\mathscr{H} &amp;=&amp; \frac{1}{4} \bigg( \|h_1 + h_2\|_\mathscr{H}- \|h_1-h_2\|_\mathscr{H} +i\|h_1+ih_2\|_\mathscr{H} - i\|h_1 - ih_2\|_\mathscr{H} \bigg) \\&amp;=&amp;\frac{1}{4} \bigg( \|Uh_1 + Uh_2\|_\mathscr{K}- \|Uh_1-Uh_2\|_\mathscr{K} +i\|Uh_1+iUh_2\|_\mathscr{K} - i\|Uh_1 - Uih_2\|_\mathscr{K} \bigg) \\&amp;=&amp; \langle Uh_1, \ Uh_2\rangle_\mathscr{K} \end{eqnarray*} $  But what I'm wondering is where surjectivity comes into play?
77326
Let $ w \in \Bbb C $, and let $ \gamma, \delta : [0,1] \rightarrow \Bbb C $ be closed curves such that for all $ t \in [0,1], |\gamma(t) - \delta(t)| &lt; |\gamma(t) - w| $. By computing the winding number $n_\sigma(0)$ about the origin for the closed curve $\sigma(t) = (\delta(t) - w)/(\gamma(t) - w) $, show that $n_\gamma(w) = n_\delta(w)$.  This seems intuitively clear to me. (Informally, the inequality tells us that $\delta$ and $\gamma$ can never be on "opposite sides" of $w$, and hence their winding numbers must be equal. Making this rigorous isn't enough, however, since I need to use the winding number of $\sigma$ about 0.  I can also show that $n_\sigma(0) = n_\gamma(w) - n_\delta(w)$ (fairly elementarily), so it remains to show that $n_\sigma(0) = 0$. This is the bit that I'm stuck on.  Can someone give me a (small) hint (not a major hint)? If I still can't get anywhere, then I'll probably ask for a larger hint!  There is also a second part to this question - dependent on whether I'm able to do the first part with a hint, I may well update this question to include the second part.  (This is an example sheet question - completely non-examinable.)  Thanks! :)
61693
Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.
56133
Every now and then a question comes up about the status of the momentum operator in the infinite square well, and while we have two good answers on the topic  and , I'm generally not satisfied by their level of detail and by how easy (not very) it is to find them when you need them. I therefore posit that it is high time we settle things and build a canonical Q&amp;A thread for this, so, with that in mind:  What's the deal with momentum in the infinite square well?   Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$  when restricted to the compact interval of the well? Are there any subtleties in its definition, via its domain or similar, that are not present for the real-line version? Is the momentum operator $\hat p=-i\hbar \frac{\mathrm d}{\mathrm dx}$  in these conditions? If not, why not, and what are the consequences in terms of the things we normally care about when doing one-dimensional QM? If it is not self-adjoint, does it admit a ? If it does, is that extension unique? If the extension is not unique, what are the different possible choices, and what are their differences? Do those differences carry physical meaning / associations / consequences? And what is a self-adjoint extension anyways, and where can I read up about them? What is the spectrum and eigenvectors of the momentum operator and its extensions? How do they differ from each other? Is there such a thing as a momentum representation in this setting? If not, why not? What is the relationship between the momentum operator (and its possible extensions) and the hamiltonian? Do they commute? Do they share a basis? If not, why not? Do these problems have counterparts or explanations in classical mechanics? ()   And, more importantly: what are good, complete and readable references where one can go and get more information about this?
101271
I was going through my notes on the unitary irreducible representations of the Poincare group and the subsequent construction of one particle states and I stumbled across the following steps in the method of induced representations: ($\mathscr{H}$ is the Hilbert space and $\mathscr{H}_p$ is the subspace of one-particle states with momenta p. $\mathcal{S}_k$ is just the little group of the standard momentum $k$. $L(p)$ is the standard Lorentz boost; $L(p)k=p$)   (1) Let $|k,\sigma\rangle$ be a basis for $\mathscr{H}_k$ and let $\mathcal{D}$ be the (finite) unitary irrep of $\mathcal{S}_k$ induced by the unitary irrep $\mathcal{U}$. Then $\mathcal{D}$ acts on $\mathscr{H}_k$ by mixing the spin polarization states,  \begin{align} \mathcal{U}(h)|k,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}(h)|k,\sigma'\rangle,~~~ \forall h\in\mathcal{S}_k,~~~\mathcal{U}(h)\equiv\mathcal{U}(h,0).  \end{align} (2) Define the basis for $\mathscr{H}_p$ by \begin{align} |p,\sigma\rangle=\mathcal{U}(L(p))|k,\sigma\rangle. \end{align} (3) The action of Lorentz transformations on these states is then \begin{align} \mathcal{U}(\Lambda)|p,\sigma\rangle=\sum_{\sigma'}\mathcal{D}_{\sigma'\sigma}\big(h(\Lambda,p)\big)|\Lambda p,\sigma'\rangle  \end{align} where $h(\Lambda,p)=L^{-1}(\Lambda p)\Lambda L(p)\in \mathcal{S}_k$ is known as the Wigner rotation.  (4) Finally, we extend this action to an arbitrary state in the Hilbert space $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$ by, \begin{align} \mathcal{U}(\Lambda)|\Psi\rangle&amp;=\sum_{p,\sigma}\big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big( p\big)| p,\sigma\rangle,\notag\\ \text{ where } \big[\mathcal{U}(\Lambda)\cdot\Psi\big]_{\sigma}\big(p\big)&amp;=\sum_{\sigma'}\mathcal{D}_{\sigma\sigma'}\big(h(\Lambda,\Lambda^{-1}p)\big)\Psi_{\sigma'}(\Lambda^{-1}p). \end{align}  These are my own notes which are largely influenced by Weinbergs construction, so they may be incorrect. I understand all of the steps except for one part of step 4; The part where I say $|\Psi\rangle=\sum_{p,\sigma}\Psi_{\sigma}(p)|p,\sigma\rangle\in\mathscr{H}$ for some coefficients $\Psi_{\sigma}(p)$. What are these coefficients? Are they the wavefunctions? Do they depend continuously on $p$, if so should the sum over $p$ be replaced by an integral? Should these coefficients be square-integrable; $\Psi_{\sigma}(p)\in L^2(\Sigma_m,\mu,\mathbb{C})$ where $\Sigma_m$ is the mass shell and $\mu$ a Lorentz invariant measure? Is this expansion of an arbitrary state even correct?
101593
How can the  be calculated in Yang-Mills theory?
15573
For a free particle of mass $m$, with Hamiltonian  $$\hat{H} = \frac {\hat{P}^2} {2m},$$  where $$\hat{P} = -i \hbar \frac{\partial} {\partial x}.$$  The commutative relation is given by  $$[\hat{X}, \hat{H}] = \frac {i\hbar} {m} \hat{P}\tag{1}$$  In the common eigenstate of $\hat{H}$ and $\hat{P}$, $|e, p\rangle$, can we do the following?  $$\langle e, p| [\hat{X}, \,\hat{H}] |e, p\rangle = \langle e, p|\hat{X} (\hat{H}|e, p\rangle) -  (\langle e, p|\hat{H}) \hat{X}|e, p\rangle  \\ = \langle e, p|\hat{X} (e|e, p\rangle) -  (\langle e, p|e) \hat{X}|e, p\rangle \\ = e( \langle e, p|\hat{X}|e, p\rangle -  \langle e, p|\hat{X}|e, p\rangle ) \\ = 0 $$  Since the $\hat{H}$ is Hermitian, the above derivation doesn't seem to show any flaw. Given the commutative relation, Eq (1), we know the result is wrong. What's wrong with the above derivation?  [EDIT]  Following the comment by Luboš Motl, I have worked out the solution and would like to share it here. The link provided by Qmechanic had the solution closely related to this question.  $$ \langle e', p'| [\hat{X}, \,\hat{H}] |e, p\rangle \\ = \langle e', p'|\hat{X} (\hat{H}|e, p\rangle) -  (\langle e', p'|\hat{H}) \hat{X}|e, p\rangle  \\ = (e - e') \langle e', p'|\hat{X}|e, p\rangle $$  Note that:   $$ e - e' = \frac{p^2}{2m} - \frac{p'^2}{2m} = \frac{(p+p')(p-p')}{2m} $$  $$ \langle e', p'|\hat{X}|e, p\rangle = -i\hbar \delta'(p - p') $$  where $\delta'(\cdot)$ is the derivative of the Dirac function, with respect to $p$.  Then we get  $$ (e - e') \langle e', p'|\hat{X}|e, p\rangle \\ = -i\hbar \frac{(p+p')}{2m} \cdot (p - p')\delta'(p - p') \\ = - \frac{i\hbar (p+p')}{2m} \cdot (-\delta(p - p')) \\ = \frac{i\hbar (p+p')}{2m} \delta(p - p') $$  As we take the limit $p \rightarrow p'$:  $$ lim_{p \rightarrow p'}    \frac{i\hbar(p+p')}{2m} \delta(p - p') \\ \rightarrow \frac{i\hbar}{m} p \delta(p - p') $$
87938
TL;DR  When we take the transpose of two Grassmann-valued spinors (fermions), should we add a minus sign because we end up anticommutating the two spinors?    More details.  I'm studying the behavior of 1+1-dimensional Majorana fermion. I use the same representation for the gamma matrices as in Lectures on string theory by Lüst and Theisen, chapter 7 appendix B. That is  $$\gamma^0 = i \sigma^2 \qquad \gamma^1 = \sigma^1 \qquad C= \gamma^0 \qquad \eta^{\mu\nu}=diag(-1,+1)$$  Equation B.3 of these lectures says that Majorana fermions obey the following indentity:  $$ \bar{\psi}_1 \gamma^{\alpha_1} \gamma^{\alpha_2} ... \gamma^{\alpha_n} \psi_2 = (-1)^n \bar{\psi}_2 \gamma^{\alpha_n} ... \gamma^{\alpha_2} \gamma^{\alpha_1} \psi_1.$$  However I cannot prove this identity, I always get the wrong sign on the RHS, even for simple cases such as $n=0,1$ or $2$. For instance for $n=1$, knowing that Majorana fermions are real and therefore $\bar{\psi} = \psi^\dagger \gamma^0 = \psi^T \gamma^0$, we can write the LHS as:  $$\psi_1^T \gamma^0 \gamma^\alpha \psi_2.$$  This is just a number, so it should be equal to its transpose. When we take the transpose, $\gamma^0$ takes a minus sign. If $\gamma^\alpha=\gamma^0$ then this one also takes a minus sign which cancels the first one and we get no overall sign. If $\gamma^\alpha=\gamma^1$ then the 2nd minus sign comes when we anticommute the two matrices. In every case I get:  $$ \bar{\psi}_1 \gamma^{\alpha} \psi_2 =  \psi_1^T \gamma^0 \gamma^\alpha \psi_2 = \psi_2^T \gamma^0 \gamma^\alpha \psi_1 = \bar{\psi}_2 \gamma^{\alpha} \psi_1,$$  even though the identity above says there should be a minus sign. My guess is that when I took the transpose of the LHS, I should have added a minus sign since I inverted two Grassmann-valued spinors in the process, otherwise I don't know what I did wrong.  On a related note, if the sign changes when we swap to spinors, does that imply that the trace of a fermion bilinear is no longer cyclic?  $$\text{Tr}\ \bar{\psi}\psi \neq \text{Tr}\ \psi \bar{\psi} ?$$     but the answer is unclear.  is actually pretty much the same question, but then why this "transpose antisymmetry" only rarely mentioned? Does it only hold for Majorana fermions?
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 :)
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?
54599
I'm a beginner of QFT. Ref. 1 states that      [...] The Lorentz group $SO(1,3)$ is then essentially $SU(2)\times SU(2)$.    But how is it possible, because $SU(2)\times SU(2)$ is a compact Lie group while $SO(1,3)$ is non-compact?   And after some operation, he says that the Lorentz transformation on spinor is complex $2\times2$ matrices with unit determinant, so Lorentz group becomes  $SL(2,\mathbb{C})$. I'm confused about these, and I think there must be something missing.   References:   L.H. Ryder, QFT, chapter 2, p. 38.
100132
I have seen in the in the Dirac equation $$\gamma_0,\gamma_1,\gamma_2,\gamma_3.$$ Then I have seen the definition of a new matrix $$\gamma_5=i\gamma_0\gamma_1\gamma_2\gamma_3.$$ Now my question is why the name of the new matrix has not been given as $$\gamma_4.$$ Is there any historical reason behind this or it is simply taken as $$\gamma_5$$ without any special reason skipping $$\gamma_4~?$$