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2

I didn't look over your work (check-my-work questions are off topic), but the conceptual error is still obvious. The gravitational field isn't uniform, so the force acts on the center of gravity, not the center of mass. For uniform fields these are the same thing, but in general they are not.


2

There is no anomaly problem with this system --- except that as written it does not have a continuous $U_A(1)$ symmetry. You need to include a term $i\bar\psi \gamma^5 \psi$ term in addition to the $\bar\psi\psi$ term. With that included it is a simple model that can be be used for illustrating chiral symmetry breaking.


3

I'm not sure I completely understood your question, but perhaps it will become clear if I explain what the last equation means. The left hand side of the equation means: Take a state $|\mathbf{x}'\rangle$ Translate it by some infinitesimal amount $\text{d}\mathbf{x}'$ using the translation operator Act on it with the position operator, $\mathbf{\hat{x}}$. ...


0

It turns out to be easier to compare both versions if one uses the functional changes $\bar{\delta}\phi(x):=\phi'(x)-\phi(x)=\phi(x-\delta x)+\delta\phi(x-\delta x)-\phi(x)=-\delta x^\mu\partial_\mu\phi(x)+\delta\phi(x)$. It is precisely in terms of this functional changes that the first version of Noether's theorem is written in. The variation in the second ...


1

Maybe the wikipedia article could help you : https://en.wikipedia.org/wiki/Symmetry_in_quantum_mechanics You could think of it as $|x'>=T|x>$ with some translation operator $T$ that maps $|x>$ onto $|x'>$ and $T^{-1}$ the mapping back $T^{-1}|x'>=|x>$. We could then take $Q'=TQT^{-1}$ and evaluate the action of $Q'$ on a state $|x'>$ as ...


0

What you have done is correct. In my answer I will show why the terms with $l>0$ vanish. As you have shown, $$\Phi = \frac{1}{4\pi\epsilon_0r}\sum^{\infty}_{l=0}\int \rho(r') P_{l}(\cos{\alpha})\bigg(\frac{r'}{r}\bigg)^{l} dV'$$ Since the distribution is spherically symmetric I have considered the origin to be the centre of the sphere. The cosine of the ...


2

The transformation $\phi\rightarrow\phi+\delta \phi$ is assumed to be small (infinitessimal). You can write $\delta\phi$ as \begin{align}\delta\phi&=\phi'-\phi\\ &=e^{i\delta \alpha}\phi-\phi\\ &\approx(1+i\,\delta\alpha)\phi-\phi\\ &=i\,\delta\alpha\,\phi \end{align} So this means $\frac{\delta\phi}{\delta\alpha}=i\phi$. Why does this ...


1

No, it definitely would not look like the identity matrix! Are you sure this is not an extended homework question? M=1 means m=-1 a combination of the spin 2 and the three spin 1 states. It appears your Hamiltonian is proportional to the total $\hat j_z$, which commutes with $\hat j \cdot \hat j=(j)(j+1) 1\!\! 1$, so it does not mix the 4 multiplets ...


3

In order to calculate the electric field using only the Gauss law (rather than by solving the Poisson equation) one usually exploits the symmetry of the problem. In textbook cases this usually means that one can guess such a Gaussian surface, that the electric field has the same magnitude and direction in respect to the surface everywhere on the surface. ...


2

The first formula in OP's post is Noether's theorem for vertical variations only, i.e. when the field values are varied at a fixed point. Moreover, it is only for strict symmetries, when the Lagrangian is invariant under the symmetry. There is also a formula for when the independent variables (the $x^\mu$) are also varied, which has an added term, and there ...


0

For what it's worth, a special conformal transformation $K=I\circ T\circ I$ can e.g. be understood as a composition of the inversion $I: \vec{x}\mapsto \vec{x}/|\vec{x}|^2$, a translation $T$, and the inversion $I$.


1

There is a lot of truth behind OP's observations, which are backed up by the following facts: An infinitesimal symmetry $\delta$ with symmetry parameter $\epsilon$ is generated by a Noether charge $\hat{Q}$ in the sense that $\delta=\epsilon [\hat{Q},\cdot]$, cf. e.g. this Phys.SE post. The symmetry parameter $\epsilon$ can often be associated with a ...


0

Suppose our system (which could be any kind of system) depends on some parameters in a topological space $P$ and there is a group $G$ which acts on $P$, such that for each $p \in P$, the theory described by the parameter value $p$ has symmetry whatever subgroup of $G$ fixes $p$. Let's assume that $G$ acts continuously on $P$. Then if $S \subset P$ is a ...


0

In principle, I would say yes it is possible. But that would be totally theoretical. Say you have some anisotropic potential to start with such as, $V(x,y,z) = ax^2 + (a+\epsilon)y^2 + az^2$, where, $\epsilon$ is a small parameter. Now if you add perturbation such that, the perturbation potential is $V_{pert} = - \epsilon\, y^2$, then after adding this ...


1

When we diagonalise a matrix we are in a process of determining it's Eigenvalues (and Eignevectors). It is similar story for block-diagonal matrices. The difference between diagonal and block-diagonal matrix forms lies in the notion of degenerate Eigenstates. That is, different states with similar Eigenvalues (in this case, energies). If we have an operator $...


0

One strong reason for choosing a unit cell is that shares crystal symmetries is that it gives us more intuition about the crystal structure, which might be very helpful in doing practical calculations. Good examples are diamond-like lattices and graphene (which is easier to visualize, since it is two-dimensional). Moreover, the choice of the primitive unit ...


3

Closed strings are ubiquitous to string theories. Open strings are the objects that encode the weakly coupled description of D-branes, and D-branes are optional, some backgrounds may contain them, some others don't. Type II theories and M-theory are perfectly consistent without open strings in flat ten and eleven dimensional spacetimes respectively. The same ...


1

Frankly, If you only sat down and wrote down all 16 states involved, and confirmed the multiplets, you'd see them all together, including their symmetries, and dispel your magical misconception on symmetry and antisymmetry. (You'd have 6 states of total M=0, of which two combinations would amount to your singlets!) In any case, distributing $$ (1\oplus 0)\...


0

Maybe not the answer you are looking for but, remember that (Wilsonian) QFTs are defined at a certain scale $\mu$.For example we can take Yang-Mills theory with various matter fields added with a certain set of coupling constants/masses $a_i$. Classically this theory can be made to have conformal symmetry by choosing the couplings in such a way that all the ...


0

There is a general formula for the Kronecker multiplication of an arbitrary number of doublets. Applied to your case, it yields $$ \frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}\otimes\frac{1}{2}= 2\oplus 1\oplus 1\oplus 1\oplus 0\oplus 0, $$ a total of 16 states. These multiplicities are specified by the Catalan triangle. The general formula for n doublets ...


0

So the related question is not so much help to you because you have a much more fundamental misunderstanding. Once we clear that up, any further questions should be answered by the above. There is no such $-1$ or $-1/2$ system. For this we have to be clear on the difference between two different angular momentum quantum numbers, $\ell$ and $m$. They come ...


1

This equation is an equation for representations of the Lie algebra of rotations. It is a fact proven in every quantum mechanics book that the irreducible representations of the angular momentum algebra $[J_i,J_j]=\epsilon_{ijk}J_k$ are spin $j$ systems where $j\in \mathbb{N}/2=\{0,1/2,1,3/2,2,\dots\}$. We thus call this representation by the integer $j$. ...


0

I am not sure which $SU(2)$ symmetry you are refering to. There is the $U(1) \times U(1)$ symmetry under $$ \psi_L \rightarrow e^{i\theta_L} \psi_L, \quad \psi_R \rightarrow e^{i\theta_R} \psi_R $$ where $\theta_L,\theta_R$ at the $U(1) \times U(1)$ parameters. It is easy to show that $\mathcal{L} = \bar{\psi}(i \gamma^{\mu}\partial_{\mu} - m) \psi$ is ...


0

In the second example, you have expressed the Lagrangian in terms of some "shifted coordinates" with origin at the minimum of the potential. You can think of it like expanding the Lagrangian around the (classical) ground state. So the Lagrangian in terms of $(\vec{\pi},\sigma)$ coordinates is manifestly $O(N-1)$-symmetric corresponding to rotations ...


0

Equation 3.33 or its equivalent 3-34 is simply the Euler Lagrange equation for your system The expression 3-34 is the quadridivergence of the moment energy tensor, it is null indicating the energy moment conservation law


0

Something that can help is that you must first formulate the Lagrangian in the most convenient coordinate system, that is, according to its symmetry; Once the Lagrangian is built, look at which coordinate is not explicitly included in the Lagrangian; This coordinate will then be cyclical and its conjugate moment will be an integral of the dynamics of the ...


1

As correctly pointed out in the comments, I was making a mistake in the Taylor expansion of the exponential. The correct coefficient in front of the integral should be $\left(\frac{gh}{4!4}\right)$ The number of contractions is indeed 48.


0

The Lagrangian $L=T-U$ is a function of the generalized coordinates $\vec{q}$ and $\vec{\dot{q}}$. if you don't have applied forces : then the criteria to find a cyclic coordinate is: obtain the vector equation $$\underbrace{\left(\frac{\partial L}{\partial \vec{\dot q}}\right)^T}_{\vec A}= \underbrace{\left(\frac{\partial L}{\partial \vec{q}}\right)^T}_{\...


2

It is simply non-physical. Wigner symmetries deal with rays $[\psi]$, i.e., non-vanishing vectors of the Hilbert space up to multiplicative factors. $$[\psi] := \{a\psi \:|\: a \in \mathbb{C} \setminus \{0\}\}\:, \quad \psi \in H \setminus \{0\}$$ Transition probabilities of couples of rays are defined as $$P([\psi], [\psi']) := \frac{|\langle\psi|\psi'\...


2

The weaker notion will also allow for anti-unitary operators, such as time-reversal and charge conjugation: https://en.wikipedia.org/wiki/T-symmetry#Time_reversal_in_quantum_mechanics These transform the amplitude to its complex conjugate, keeping the norm the same. They can be very interesting, such as in the CPT theorem in relativistic QFT.


4

Consider a generic Hamiltonian $H$ and a unitary operator $U$. This operator is a symmetry of the system if and only if it commutes with the Hamiltonian $[H,U]=0$. I consider only one symmetry operator for simplicity. If a state $|\psi\rangle$ is stationary and non-degenerate, then it is an eigenvector of the Hamiltonian $H$, and since $[H,U]=0$ it is also ...


1

Assume that $O$ is bounded with point spectrum so that it admits a Hilbert basis of eigenvectors $\psi_n$ and, in the strong operator topology, $$O = \sum_n a_n |\psi_n\rangle\langle \psi_n|.$$ If every eigenvector is also eigenvector of each said $P$, then it is easy to see that (notice that the eigenvalues are $\pm 1$) $P$ commute with $O$. That is ...


0

In a sense you are correct: one can always use the BdG trick to write $$ \psi^\dagger H \psi = \frac 12 (\psi^\dagger, \psi)\left(\matrix{ H&0\cr 0 &-H^T}\right)\left(\matrix{\psi \cr \psi^\dagger}\right)+ \frac 12 {\rm Tr}\{ H\}\\ = \frac 12 (\psi^\dagger, \psi)\left(\matrix{ H&0\cr 0 &-H^*}\right)\left(\matrix{\psi \cr \psi^\dagger}\right)+ ...


2

For what it's worth, it is very important at which stage one uses Euler-Lagrange (EL) equations in an application of Noether's (first) theorem. Noether's first theorem has 2 sides: Input: A global off-shell$^1$ (quasi)symmetry. Here one should not use EOM. (An on-shell symmetry is a vacuous notion, because whenever we vary the action $\delta S$ ...


1

I think i got the mistake i was doing . I was considering $C_k$ and $C_1$ to be the expansion coefficient of $\phi$ which isn't. Am I right?


4

The Euler-Lagrange equation is not automatically satisfied by $\mathcal{L}$. It's the other way around. Given $\mathcal{L}$, you can find the classical equation of motion satisfied by the field $\phi$. This is like giving you a formula for the force in Newtonian mechanics. Even if you know $F$, you still need to know Newton's second law $F=ma$ to find the ...


8

You have written your Euler-Lagrange equation correctly. So when you simplify it, you get the equation of motion (just as you mentioned in your question):$$\partial_\mu\partial^\mu\phi=\partial^2\phi=0.$$ Note that there's nothing peculiar about this because if you used the Lagrangian with the potential energy term too, i.e. $$\mathcal{L}=\frac{1}{2}(\...


0

Your first statement is a little ambiguous, let me rephrase it. Any Hermitian Hamiltonian of the form $$ \mathcal{H} = (c_1^\dagger, c_1, c_2^\dagger, c_2,...) \begin{pmatrix} H_{11} & H_{12} & \cdots \\ H_{21} & H_{22} & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix} \begin{pmatrix} c_1 \\ c_1^\dagger\\ c_2 \\ c_2^\dagger \\ \...


2

I don't think there's any specific 'rule' in regards to transformations to cyclic coordinates. Given the Lagrangian in a specific set of coordinates, it may just be by inspection that you see that coordinates in another system become cyclic. It also helps to know what system the Lagrangian is representing. For example, something rotating, cyclic coordinates ...


1

$|\langle \psi_k|\psi_\ell\rangle |^2$ is a real positive number, let us call it $c^2$. Then $|\langle \psi_k|\psi_\ell\rangle |=\pm c$. But also $|\langle \psi_k|\psi_\ell\rangle |$ is a real positive number, so we need to take the positive square root, here $\delta_{k\ell}$.


0

This is an example of induced representation. Consider two groups, $K < G$. Let a representation $D(K)$ act in a vector (usually, Hilbert) space ${\mathbb{V}}$. Based on this, we now wish to construct a representation of $G$. In mathematics, this (so-called "induced") representation is denoted with $\operatorname{Ind}_K^GD$ or simply $D(K)\...


-1

First of all, in your last paragraph, you make references to fermions. BdG hamiltonians may also represent bosonic systems. The particle-hole symmetry is necessary to ensure that the time-evolved creation and annihilation operators are related by the adjoint operation, i. e. $c_j(t)^{\dagger} = c_j^{\dagger}(t)$. Mathematically speaking, time-evolution and ...


2

Every single particle in Earth exerts gravitational attraction to an object in your example, and, all these effects add up/cancel, and the net of these is pointing towards the center. Actually, at the center, all these effects cancel out, and you feel weightless. Correct. If you split the earth up into spherical shells, then the gravity from the shells &...


2

Given a spherically symmetric Earth and a point $P$, the gravitational field vector at $P$ can be determined in two steps: Why does the field point toward the Earth's center $C$? Consider rotational invariance about the axis passing through $C$ and $P$. The mass distribution of the Earth is unchanged by this rotation, and so must be the field -- i.e., the ...


18

All parts of the Earth exert a pull on another body (such as an apple) and all parts of the Earth are pulled by external bodies (such as apples). Let's consider the pull that the Earth exerts on the apple. Imagine the Earth divided up into kilogram portions by imaginary surfaces. The portion nearest the apple will exert the greatest pull, straight downwards ...


4

Because the Earth is spherical (to the same level of approximation, to which we consider that the force of gravity is directed towards its center). In fact, there is a gravity force exerted by every element of the Earth, but when all these forces are added up, the net force is directed towards the center of the Earth.


1

The tricky part is the interpretation of $$ U\left(R_{\theta}, 0\right)=\exp (i \mathbf{J} \cdot \theta) $$ If by $\mathbf{J}$ you mean a single operator $n_xJ_x+n_yJ_y+n_zJ_z$ then your operation generates a one-parameter Abelian subgroup, and you're not gonna get any commutation out of that other than $\mathbf{J}$ as the operator $n_xJ_x+n_yJ_y+n_zJ_z$ ...


1

I looked at page 54 and Weinberg does not say that the $t_{ab}$ are Hermitian, only that the $t_a$ are Hermitian. I have the 7th reprinting of the paperback edition. Maybe it was wrong in earlier editions?


1

$$\mathcal H = \alpha \hat a + \alpha^* \hat a^\dagger\\ =\alpha e^{-i\phi}~\hat a' + (\alpha e^{-i\phi})^* \hat {a'}~^\dagger\\ \equiv \alpha' \hat a' + (\alpha ')^* \hat {a'}~^\dagger.$$ Names, by themselves, cannot affect physical relevance. Arbitrary complex number coefficients present differently in the unprimed and primed representations, which ...


0

I think you might want to have a look at Baez, J. C. (2020). Getting to the Bottom of Noether's Theorem. arXiv preprint arXiv:2006.14741. where I believe he discusses a version of Noether's theorem phrased this way in Section 2. On page 15 specifically, it discusses a version of Noether's theorem which is expressed as the Poisson bracket of two things ...


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