On the page 28 of the book "An Introduction to Quantum Field Theory" by Michael E. Peskin and Daniel V. Schroeder in the last paragraph it says:     "When $(x-y)^2&lt;0$  we can perform a Lorentz transformation taking $(x-y)\rightarrow -(x-y)$. Note that if  $(x-y)^2&gt;0$  there is no continuous Lorentz transformation that takes  $(x-y)\rightarrow -(x-y)$"   Why is this happening?
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Related post   In Peskin and Schroeder's QFT p28, the authors tried to show causality is preserved in scalar field theory.  Consider commutator $$ [ \phi(x), \phi(y) ] = D(x-y) - D(y-x) \tag{2.53} $$ where $D(x-y)$ is the two-point correlation function,  $$D(x-y):= \langle 0 | \phi(x) \phi(y) | 0 \rangle  = \int \frac{d^3 p}{ (2\pi)^3} \frac{1}{ 2E_{\mathbf{p}}} e^{-ip(x-y)}\tag{2.50}$$  P&amp;S argued that each term in the right-hand-side of (2.53) is Lorentz invariant, since $$\int \frac{d^3p }{ (2\pi)^3} \frac{1}{2E_{\mathbf{p}}} = \int \frac{ d^4 p }{ (2\pi)^4} (2\pi) \delta(p^2-m^2)|_{p^0&gt;0} \tag{2.40}$$ is Lorentz invariant.  Since there exists a continuous Lorentz transformation in the spacelike interval $(x-y)^2&lt;0 $ such that $(x-y) \rightarrow - (x-y) $ and $D(y-x)=D(x-y)$, (2.53) equals zero in the spacelike interval. In timelike interval, since such continuous Lorentz transformation does not exist, (2.53) is non-zero in general.  My question is, consider a non-continuous Lorentz transmation in the timelike interval, $PT$, namely time reversal times parity transformation. I can also let  $(x-y) \rightarrow - (x-y) $. Why (2.53) in the timelike interval is non-zero?  I guess $PT$ will let (2.40) go to $p^0&lt;0$ branch. But I am not sure if it breaks the Lorentz invariant of (2.40) and (2.50).
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In Weinberg's book The Quantum Theory of Fields, Volume 1 on p.388 (Chapter 9), the following identity is used (with $f$ being any "reasonable" function):  $$f(+\infty) + f(-\infty) = \lim_{\epsilon \rightarrow 0^+} \epsilon \int_{-\infty}^{+\infty} d\tau f(\tau) e^{-\epsilon |\tau|}.\tag{9.2.15} $$  I don't understand the identity in a qualitative / heuristic way.
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The action of a free relativistic particles can be given by $$S=\frac{1}{2}\int d\tau \left(e^{-1}(\tau)g_{\mu\nu}(X)X^\mu(\tau)X^\nu(\tau)-e(\tau)m^2\right).$$ If we then make an infinitesimal transformation of the parametrization parameter $\tau$ this would be $$\tau\to\tau^\prime=\tau-\eta(\tau),$$ for an infinitesimal parameter $\eta(\tau)$.  Of course we can describe the system as it pleases us so we know that $$X^{\mu^\prime}(\tau^\prime)=X^\mu(\tau).$$ From this relation we see that $X^{\mu^\prime}(\tau)$ must be  $$X^{\mu^\prime}(\tau) \approxeq X^{\mu^\prime}(\tau^\prime+\eta(\tau)) \approxeq X^{\mu^\prime}(\tau^\prime)+\eta(\tau)\frac{d X^{\mu^\prime}(\tau^\prime)}{d\tau^\prime} \approxeq X^{\mu}(\tau)+\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}$$  Which all can be summarized as $$\delta X^{\mu}(\tau)=X^{\mu^\prime}(\tau)-X^{\mu}(\tau)=\eta(\tau)\frac{d X^{\mu}(\tau)}{d\tau}.$$ Now this is all well I hope. But if one does the same argument for $e(\tau)$ one gets the wrong transformation. It is a scalar function so it has to obey $$e^\prime(\tau^\prime)=e(\tau).$$ Which would give the same transformation.  The right transformation are written in  on page 13, eq. 1.10. The transformation is $$\delta e(\tau)=\frac{d}{d\tau}\left(\eta(\tau) e(\tau)\right).$$ Could someone show me how this is done and elaborate a little on how one knows how different object transforms?
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1.  Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an explanation that I need some help with: that the de Broglie wavelength and the wavelength of an elastic wave do not show similar properties under a Galilean transformation. He basically says that both are equivalent under a gauge transform and also, separately by Lorentz transforms. This, accompanied with the observation that $\psi$ is not observable, so there is no "reason for it being real". Can someone give me an intuitive prelude by what is a gauge transform and why does it give the same result as a Lorentz tranformation in a non-relativistic setting? And eventually how in this "grand scheme" the complex nature of the wave function becomes evident.. in a way that a dummy like me can understand.  2.   A wavefunction can be thought of as a scalar field (has a scalar value in every point ($r,t$) given by $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{C}$ and also as a ray in Hilbert space (a vector). How are these two perspectives the same (this is possibly something elementary that I am missing out, or getting confused by definitions and terminology, if that is the case I am desperate for help ;)   3.   One way I have thought about the above question is that the wave function can be equivalently written in $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{R}^2 $ i.e, Since a wave function is complex, the Schroedinger equation could in principle be written equivalently as coupled differential equations in two real functions which staisfy the Cauchy-Riemann conditions. ie, if $$\psi(x,t) = u(x,t) + i v(x,t)$$ and $u_x=v_t$ ; $u_t = -v_x$ and we get $$\hbar \partial_t u  = -\frac{\hbar^2}{2m} \partial_x^2v + V v$$ $$\hbar \partial_t v  = \frac{\hbar^2}{2m} \partial_x^2u - V u$$ (..in 1-D) If this is correct what are the interpretations of the $u,v$.. and why isn't it useful. (I am assuming that physical problems always have an analytic $\psi(r,t)$).
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I've started today the book of Landau and Lifshitz Vol.2: The Classical Theory of Fields $\S 2$. They start from the invariance of the speed of light, express it as the fact that $$c^2(\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2=0$$ is preserved when we change inertial frame, so they consider $$ds^2=c^2dt^2-dx^2-dy^2-dz^2,$$ and say     We have observed that if $ds=0$ in one frame then $ds'=0$ in another frame. But $ds$ and $ds'$ are infinitesimal of the same order. So it follows that $ds^2$ and $ds'^2$ have to be proportional that is $ds^2=ads'^2$...    and he goes on to prove that $a=1$.  How to translate this argument in a rigorous one? I'm really interested in this, both to understand this deduction and also to be able in future to make similar ones.
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Assume one has a continuous transformation of fields, and also of coordinates - in case if we consider coordinate transformations as well. Global internal symmetries, rotations, translations, dilatations, whatever, or any other. How do we generally check whether the action is invariant under such a transformation?  I feel like my question is way more basic than Noether theorem, charges, currents, etc.  To be more precise. Let's change the fields (and probably the coordinates) as: \begin{align}     x &amp;\to x + \xi(x)\\     \phi(x) &amp;\to \phi(x) + \delta\phi(x) \end{align} where $\xi(x)$ and $\delta \phi(x) = F[\phi(x),\partial\phi(x)]$ have certain functional form. I would like to know, whether the action \begin{equation}     S = \int \operatorname{d}^D x\, \mathcal{L}[g^{\mu\nu}, \phi(x), \partial\phi(x)] \end{equation} is form-invariant under such a transformation. In other words, whether it will go into something like \begin{equation}     \tilde{S} = \int \operatorname{d}^D \tilde{x}\, \mathcal{L}[\tilde{g}^{\mu\nu}, \tilde{\phi}(\tilde{x}), \partial\tilde{\phi}(\tilde{x})] \end{equation} (notice no tilde above $\mathcal{L}$! I guess, that is form-invariance...)  So far, it seems to me that the obvious option - to calculate $\delta S$ does not give anything (that's what we do when deriving Noether currents). On classical trajectories, we end up with smth like $\delta S = \partial(\ldots)$ - which if of course correct, but does not help  to answer my question at all.  I feel like my question should be related to Killing vectors... In various textbooks I found how the Lie derivatives are applied to the metric in order to determine which transformations preserve it. Seems like I'm interested in a similar procedure for the action.  As usually, any references greatly appreciated.  UPDATE  Since the original question can be answered in a brute force way - "just plug your transformations into the Lagrangian and see how it goes", let me ask it in a slightly more general way:  Given the action, what is the procedure to find all continuous symmetries which leave it form-invariant?
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To quantize a scalar field theory with the action:  $$S=\int \mathrm d^Dx\mathscr{L}(\phi,\partial_\mu\phi)=\int \mathrm dx^0L(\phi,\partial_0\phi)$$  we promote $\phi(\vec{x})$ and $\pi=\frac{\delta L}{\delta(\partial_0\phi)}=\pi(\vec{x})$ to be field operator at fixed time $x^0$ (with the canonical commutation relation $[\phi(\vec{x}),\pi(\vec{y})]=\delta^{D-1}(\vec{x}-\vec{y})$), then we use the classical equation of motion to find $\phi$ and $\pi$ at all time.  What principles lie behind this? Why can we assume that the field operators satisfy the classical equation of motion?    By example if we have a classical theory with $S=\int \,\mathrm dtL(\phi,\partial_t\phi)$, momentum $\pi=\frac{\delta L}{\delta(\partial_t\phi)}$ hamiltonian $H=\pi\phi-L$, the classical equation of motion is:  $$\frac{\mathrm d\phi}{\mathrm dt}=\frac{\delta L}{\delta(\partial_t\phi)}$$  When we do the canonical quantization, we promote $[\phi,\pi]=\delta(\vec{x}-\vec{y})$ and the equation of motion for $\phi$ is $$\frac{\mathrm d\phi}{\mathrm dt}=i[H,\phi]$$  Why should $i[H,\phi]$ has the same form as $\frac{\delta L}{\delta(\partial_t\phi)}$? This fact comes from where?
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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 :)
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I have almost no background in physics and I had a question related to Schrodinger's Equation. I think, it is not really research level so feel free to close it, but I would request you to kindly suggest some existing literature which can help me develop a better understanding for the same.   While reading up about it from an introductory text on Quantum Physics, I wondered for a little while how does one derive this equation. Very soon, the author () added the detail that he just took it on faith when he was himself a student. He goes on to say Schrodinger equation rightly predicts the behavior of atomic transitions etc and people believe that it is a fundamental law of nature for quantum systems.  Then he raises the question himself which I wanted to know the answer to. Namely     What made Schrodinger write such an equation which became a fundamental equation, not to be derived from more fundamental equations?   He then adds that it will be an interesting topic for students of history of science which does not answer my question. Could you please try answering this question (or is this question really useless) ?  Thanks for your time
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Unlike rotations, the boost transformations are non-unitary. Therefore, the boost generators are not Hermitian. When boosts induce transformations in the Hilbert space, will those transformation be unitary? I think no. If that is the case, what is the physical significance of such non-unitary transformations corresponding to boosts in the Hilbert space?
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It is usually said that existence of discrete spacetime violates Lorentz symmetry. What quantity is used to quantify such violation? I mean could someone points a reference for a derivation that shows such analysis.  My other question is: if Lorentz symmetry is violated, does that imply space-time is discrete? or not necessarily?)
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In almost all proofs I've seen of the Lorentz transformations one starts on the assumption that the required transformations are linear. I'm wondering if there is a way to prove the linearity:  Prove that any spacetime transformation $\left(y^0,y^1,y^2,y^3\right)\leftrightarrow \left(x^0,x^1,x^2,x^3\right)$ that preserves intervals, that is, such that  $$\left(dy^0\right)^2-\left(dy^1\right)^2-\left(dy^2\right)^2-\left(dy^3\right)^2=\left(dx^0\right)^2-\left(dx^1\right)^2-\left(dx^2\right)^2-\left(dx^3\right)^2$$  is linear (assuming that the origins of both coordinates coincide). That is, show that $\frac{\partial y^i}{\partial x^j}=L_j^i$ is constant throughout spacetime (that is, show that $\frac{\partial L_j^i}{\partial x^k}=0$).  Thus far all I've been able to prove is that $g_{ij}L_p^iL_q^j=g_{pq}$ (where $g_{ij}$ is the metric tensor of special relativity) and that $\frac{\partial L_j^i}{\partial x^k}=\frac{\partial L_k^i}{\partial x^j}$. Any further ideas?
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There seems to be something very strange about the relationship between quantum field theory and quantum mechanics.  It is bothering me; perhaps somebody can help.  I'll consider a free Klein-Gordon field.  In standard treatments (e.g. Peskin &amp; Schroeder and Schwartz) the one-particle momentum eigenstates $| \vec{k} \rangle$ are normalized so that  $$ \langle \vec{p} | \vec{k} \rangle = 2 \omega_{\vec{p}} (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k}), \qquad 1 = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 \omega_{\vec{p}}}| \vec{p} \rangle \langle \vec{p} |. $$ Now, assuming $\langle \vec{x}' | \vec{x} \rangle = \delta^{(3)}(x-x')$ as usual, it follows that $$ \langle \vec{x} | \vec{p} \rangle = \sqrt{2 \omega_{\vec{p}}} e^{i \vec{p} \cdot \vec{x}}. $$ Now, one can compute (here in the Schrodinger picture; see Schwartz 2.76 or P&amp;S 2.42) that $$ \langle 0 | \phi(\vec{x}) | \vec{p} \rangle = e^{i \vec{p} \cdot \vec{x}}. $$ This is supposed to mean that $\phi$ creates a particle localized at position $\vec{x}$.  P&amp;S are a little cautious about the details, but Schwartz claims that the calculation implies  $$ \phi(\vec{x}) |0 \rangle = | \vec{x} \rangle. $$ But this is false because $\langle \vec{x} | \vec{p} \rangle \neq e^{i \vec{p} \cdot \vec{x}}$ with the normalization conventions used.  I suppose it could be true with some weird normalization of $| \vec{x} \rangle$, but I can't see what that might be (and at the very least this is not spelled out on the text).    Even if this works out, it seems extremely strange for there to be a relative normalization between the one-particle states of field theory and the states of one-particle relativistic quantum mechanics.  One ought to be able to redo the correspondence to make the normalization work out, but I don't see how.  (Note that the normalizations can be easily made to agree in the non-relativistic limit $\omega \approx m$, but that's besides the point.  Even if fully relativistic quantum mechanics is inconsistent [as some texts claim without reference], at the very least the perturbative corrections for $v \ll 1$ should be recoverable from field theory.)  [Edit: This seems to go beyond normalization.  We can get a feel for what kind of state $\phi(\vec{x})|0\rangle$ is by computing its wavefunction as a function of $\vec{x}'$, $$ \langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}. $$ This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do).  I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another.  This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states.  Words of wisdom would be appreciated.]
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We consider in Lorentz spacetime, $(x^0,x^1,x^2,x^3)=(t,x,y,z)$, choose the unit of time such that $c=1$.   Given a four vector $A_\mu$, and let the Lagrangian  $$L(x^i,\dot x^i,t)=-m\sqrt{1-\dot x_i\dot x^i}+qA_0+qA_i\dot x^i,\tag{1}$$ where we use Einstein's convention for summation. (See  with electron charge $q=-e$.)    By the Euler-Lagrange equation $$\frac{d}{dt}\frac{\partial L}{\partial\dot x^i}-\frac{\partial L}{\partial x^i}=0\tag{2}$$ we can get three equations, and if we use the fact that  $$\frac{d}{d\tau}=\frac{1}{\sqrt{1-\dot x_i\dot x^i}}\frac{d}{dt}\tag{3}$$  then these equations are, for $i=1,2,3$ $$m\frac{d^2x^i}{d\tau^2}=q(\frac{\partial A_\mu}{\partial x^i}-\frac{\partial A_i}{\partial x^\mu})\frac{dx^\mu}{d\tau}\tag{4}$$  My question follows, can we obtain a fourth equation which is just letting $i=0$?   The  says that this fourth equation is from the property that the action is invariant under Lorentz transformation. But I think this is not very convincing. I tried to view the first three equations as two four vectors having three equation space components, but I do not think this fact can lead to the equality for the last time component.  References:   L. Susskind, .
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This question is based on  problem II.3.1 in Anthony Zee's book Quantum Field Theory in a Nutshell      Show, by explicit calculation, that $(1/2,1/2)$ is the Lorentz Vector.    I see that the generators of SU(2) are the Pauli Matrices and the generators of SO(3,1)is a matrix composed of two Pauli Matrices along the diagonal. Is it always the case that the Direct Product of two groups is formed from the generators like this?   I ask this because I'm trying to write a Lorentz boost as two simultaneous quatertion rotations [unit quaternions rotations are isomorphic to SU(2)] and tranform between the two methods. Is this possible?      In other words, How do I construct the SU(2) representation of the Lorentz Group using the fact that $SU(2)\times SU(2) \sim SO(3,1)$?  Here is some background information:  Zee has shown that the algebra of the Lorentz group is formed from two separate $SU(2)$ algebras [$SO(3,1)$ is isomorphic to $SU(2)\times SU(2)$] because the Lorentz algebra satisfies:   $$\begin{align}[J_{+i},J_{+j}] &amp;= ie_{ijk}J_{k+} &amp; [J_{-i},J_{-j}] &amp;= ie_{ijk} J_{k-} &amp; [J_{+i},J_{-j}] &amp;= 0\end{align}$$  The representations of $SU(2)$ are labeled by $j=0,\frac{1}{2},1,\ldots$ so the $SU(2)\times SU(2)$ rep is labelled by $(j_+,j_-)$ with the $(1/2,1/2)$ being the Lorentz 4-vector because and each representation contains $(2j+1)$ elements so $(1/2,1/2)$ contains 4 elements.
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Noether's theorem usually considers coordinate/field transformations which leave the Lagrangian invariant up to a divergence term, i.e.  $$\mathcal{L} \rightarrow \mathcal{L} + \partial_{\mu}f^{\mu}$$  However there is a more general class of transformations which leave the equations of motion invariant, and that is a divergence term along with an overall scaling:  $$\mathcal{L} \rightarrow \alpha\mathcal{L} + \partial_{\mu}f^{\mu}$$  but Noether's theorem does not seem to deal with these types of symmetries, which are exhibited by, for example, the Klein-Gordon Lagrangian:  $$\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi - \frac{1}{2}m^2\phi^2$$  under the transformation $\phi \rightarrow \alpha\phi$ (giving $\mathcal{L} \rightarrow \alpha^2\mathcal{L}$).    The action similarly transforms as $S \rightarrow \alpha^2 S$.  I feel this is important to emphasize as there are cases in which the Lagrangian scales by a factor but the action remains strictly invariant.  This is not the case here, as the action scales by $\alpha^2$ but the equations of motion and extrema of the action are invariant, which is ultimately the only physical thing that matters.    Is there a generalization to Noether's theorem for these "scaling-type" symmetries?  I.e. is there a conserved quantity/current arising from scale invariance?
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The  is always used to add an EM field to the Schrödinger/Pauli/Dirac equations. Why does one not use the ? As far as I can see, the difference is a factor i in front of the vector potential. I know I'm combining two seemingly unrelated things, but they seem very similar, an the covariant form seems much "better" with respect to the inherent gauge freedom in the EM field. I can also see that with the canonical momentum form, the equations remain unchanged after an EM and a QM (phase) gauge transformation. Suffice to say my field theory knowledge is not that impressive.