Skip to main content
5 votes

Have all the symmetries of the standard model of particle physics been found?

My answer is also ‘No’ but by direct construction: over the past few years, various research groups have understood new symmetries of the Standard Model. Over the past decade, field theorists have ...
SethK's user avatar
  • 589
3 votes

Designing a thought experiment on Noether's Theorem

Experiments on crystals are translationally variant because the crystal structure is only the same up to translations that reproduce the same structure, in such cases there is "crystal momentum&...
mike1994's user avatar
  • 978
2 votes
Accepted

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

Roughly speaking, in addition to Einstein equations, the (spatial) FLRW metric is constructed by assuming that, at fixed time, in the Riemannian manifold defining space: (a) metric properties are ...
Valter Moretti's user avatar
2 votes
Accepted

Doubts in circuit analysis

Maybe I am loosing some detail here, but assuming that all resistances are equal, and looking at the symmetry of the circuit, all dpps and currents should be the same in module if you flip the circuit ...
ebenezer's user avatar
  • 130
2 votes
Accepted

How does inserting an operator in the path integral change the equation of motion?

The missing conceptual point to appreciate is that when you are calculating a correlation function it can be useful to interpret the same expression in different ways. We wish to calculate the ...
SethK's user avatar
  • 589
2 votes
Accepted

Using particle-hole symmetry of the Hubbard model to study the model at different densities

$\newcommand{\dag}{\dagger}$I realized it's because there are two spins that must be accounted for. Under the particle-hole transformation, we have \begin{align*} N = \sum_{i, \sigma} a^\dag_{i, \...
zeroknowledgeprover's user avatar
2 votes
Accepted

Two contradictory derivations of Killing equation

As an overall comment, I stress that conservation of $Q$ is valid for the Killing vector $\xi$ if the considered curve is a geodesic. Let us come to the issue. First of all, generally speaking, the ...
Valter Moretti's user avatar
2 votes

Two contradictory derivations of Killing equation

Both approaches are fine. In the first approach, the analysis is done at the coordinate/component level of the equations. Simply asking the question how does $Q$ very with $\tau$ if we write ...
TimRias's user avatar
  • 12.5k
1 vote

Which symmetries lead to the ladder operators of the harmonic oscillator?

Promoted from a comment: Ladder operators are a consequence of the root and weight structure of Lie algebras, such as so(3), etc. In the case of the Harmonic oscillator, the relevant algebra is the ...
1 vote

Moment of Inertia of cube about the axis along one of its diagonal

Consider a reference frame centred at the center of the cube with axes perpendicular to the opposite faces. Evidently the three moments of inertia are here identical. This means that the inertia ...
Valter Moretti's user avatar
1 vote

In equation (20) from lecture 10 in Leonard Susskind’s ‘Classical Mechanics’, why is there a summation involved?

All the answers and comments above are brilliant and professional, however I probably get your point here, as I am also very new to these topics and notations. For the summation, it is not about any ...
Marvyn Hsu's user avatar
1 vote

In equation (20) from lecture 10 in Leonard Susskind’s ‘Classical Mechanics’, why is there a summation involved?

In any dimension (not necessarily 3), the angular momentum is a bi-vector $$L^{ij}~=~x^ip^j-x^jp^i,\tag{1}$$ cf. e.g. this Phys.SE post. Together with the canonical Poisson bracket $$\{x^i,x^j\}~=~0,\...
Qmechanic's user avatar
  • 207k
1 vote

Why Consider Only Triplet States for Spin in $2$-Electron Systems?

This is slightly strange. In any case what matters is that the fermionic states should be fully antisymmetric w/r to permutations of the particles, and that means spin and spatial degrees of freedom. ...
ZeroTheHero's user avatar
  • 46.1k
1 vote

Does quasi-symmetry preserve the solution of the equation of motion?

Boundary condition is a part of the very definition of your field theory, classical or quantum. When e.g Peskin & Shroeder leaves out the surface term, "all fields and derivatives vanish at ...
T.P. Ho's user avatar
  • 91
1 vote

Two contradictory derivations of Killing equation

I think the problem lies in the notation. I guess Tong is treating the components $\xi_\mu$ as scalars for which we have $\frac{\mathrm{d}}{\mathrm{d}\tau}=u^\alpha\partial_\alpha$. And the ...
Silas's user avatar
  • 425
1 vote

Is the FRW metric, based on spatial homogeneity and isotropy, rotationally and translationally invariant? If so, how?

If what you're confused about is why the metric seems to depend on the coordinates and thus might not be translationally or rotationally invariant, look at the Minkowski metric (flat spacetime) in ...
controlgroup's user avatar
1 vote

Does all symmetry breaking have corresponding unitary group?

In general you can have broken symmetry groups different from $U(n)$. For example the quantum Ising model has a discrete symmetry (enacted by $P=\prod_j \sigma^z_j$) that breaks spontaneously in the ...
lcv's user avatar
  • 2,474
1 vote

Does all symmetry breaking have corresponding unitary group?

As in this SE post, you can have the Lagrangian with $SO(N)$ symmetry $${\mathcal{L}} = \frac{1}{2}(\partial_\mu \Phi)^T (\partial^\mu \Phi) - \left(\frac{1}{2}\mu^2 \Phi^T \Phi + \frac{1}{4}\lambda (...
Gabriel Ybarra Marcaida's user avatar
1 vote

Probabilistic behavior of quantum mechanics

If you have two identically prepared systems, say two copies of the state $ \alpha \vert \uparrow \rangle + \beta \vert \downarrow\rangle$, with $\vert \alpha\vert^2+\vert\beta\vert^2=1$, you might ...
ZeroTheHero's user avatar
  • 46.1k
1 vote

Variation in the context of symmetries

The Einstein-Hilbert action is $$ S = \frac{1}{2\kappa} \int R \sqrt{-g} d^4x$$ with $\kappa$ the dimensionful constants. If you set $R=0$ at the outset, you have the action $S = 0$, which is a ...
Brick's user avatar
  • 5,097
1 vote

Killing tensor in the Kerr metric

The Killing equation is a overdetermined system of PDEs of finite type, as a result, there is a algorithm to compute all Killing tensors for a given metric. See for example arXiv 1704.02074.
liyiontheway's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible