10 votes

Why are there only three independent rotations and three independent boosts?

When people talk about "number of independent rotations" or "number of independent boosts", what they mean is "the number of real-valued parameters you need to specify to ...
Andrew's user avatar
  • 47.5k
8 votes

What exactly is a quantum field?

Let's address first your title question "What exactly is a quantum field?" The answer is essentially given in the first five chapters of Weinberg's textbook: a quantum field is a convenient ...
Gold's user avatar
  • 35.6k
7 votes

What exactly is a quantum field?

There are formulations of QFT in terms of wavefunctionals but they are often rather useless, both for technical reasons and for more pragmatic ones. For a discussion of the technical issues, see this ...
ACuriousMind's user avatar
  • 124k
5 votes

What exactly is a quantum field?

What exactly is a quantum field? When physicists use the term "field" they usually mean a function of space or a function of space and time. For example, in classical electrodynamics the ...
hft's user avatar
  • 18.9k
4 votes
Accepted

Projector onto Adjoint and Singlet Representations for $SU(N)$

Indeed. Review for su(2), in your notation, where $\vec S$ are the normalized 3-vector generators, so $\vec \sigma /2$ for the fundamental and anti fundamental, so you have $$ \vec{S}_1\cdot \vec{S}_2=...
Cosmas Zachos's user avatar
3 votes

Why are there only three independent rotations and three independent boosts?

Another way to look at this is that rotations are not specified around axes, but within bivectors. In three dimensions, there are three basis bivectors, canonically $xy$, $yz$, and $zx$, and ...
RLH's user avatar
  • 516
3 votes

What exactly is a quantum field?

One certainly makes contact with quantum mechanics via the wave function: $$\langle 0|\Psi|\phi\rangle.$$ Indeed, physical theories must be invariant under the full Poincare group, however, for many ...
Albertus Magnus's user avatar
2 votes

What exactly is a quantum field?

"But then what is the wavefunctional of the system" In simple particle quantum mechanics, the wavefuctional is a function of position $\psi (x)$. Its time derivative is given by Schrodinger ...
Bohan Xu's user avatar
  • 633
2 votes

Degrees of freedom in $(A,B)$ representation of the Lorentz group

Now since this is a direct sum of a $2A+1$ dimensional representation with a $2B+1$ dimensional representation, the matrix representation of $(A,B)$ is $2A+2B+2$ dimensional, which is what I would ...
Cosmas Zachos's user avatar
1 vote

Character Table and Binary Basis

The final column is obtained by representing the abstract group as an isometry subgroup. Each set of quadratic polynomials (separated by a semicolon) span an irreducible representation whose type is ...
LPZ's user avatar
  • 10.9k
1 vote
Accepted

A question from S. Weinberg's book (Sec. 2.7)

Clearly $\phi(T,1)$ is independent of $T$. Let $\phi(T,1)=\phi_0$. You can rescale your unitary operators by a constant phase ${\tilde U}(T) = e^{-i\phi_0} U(T)$. The new set of unitary operators ...
Prahar's user avatar
  • 25.5k
1 vote
Accepted

Lie group symmetry in Weinberg's QFT book

You are correct, (2.B.10) is actually not true. But it's only used through (2.B.11), and (2.B.11) does hold, because the derivative $\frac{\partial^2 f^a(\theta, \theta_1)}{\partial\theta^b \partial\...
Vladimir Lysikov's user avatar
1 vote

Do the solutions to Maxwell's equations form a group?

I will confine my attention to the vacuum Maxwell equations. There are uncountably many solutions $\{\mathbf{E}, \mathbf{B} \}: \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}^3 \times \mathbb{R}^3$ to ...
Michael Seifert's user avatar
1 vote
Accepted

$SU(5)$ Gauge Field Theory, symmetry breaking

There is no "if $\operatorname {tr} G= 0= \operatorname {tr} A +\operatorname {tr} B$": you already assumed it. The traceless matrix diag$(2I_3, -3I_2)$ manifestly commutes with all 12 ...
Cosmas Zachos's user avatar
1 vote

In what sense are fields representations of the Poincare group?

Fields are reducible representations (by which I mean, fields are elements of the vector space on which a reducible representation of the Poincar'e group acts). On the other hand, particles are ...
Prahar's user avatar
  • 25.5k
1 vote

How to obtain the explicit form of Lorentz transformation matrix using Lie algebra?

Addendum: With the six generators in $$l_{ik}\in so(3,1)$$ of the one-parameter subgroups, it remains to show that the total of $SO(3,1)$ is the exponential of any linear combination of 3 boost ...
Roland F's user avatar
1 vote
Accepted

How to obtain the explicit form of Lorentz transformation matrix using Lie algebra?

Lie algebra is the linear approximation of the Lie group at the identity $$L = 1_4 + \epsilon \ l_{ik}$$ where $1_4$ is the 4d identity matrix and $$l_{ik}$$ a linear transformation in the 2d-plane ...
Roland F's user avatar
1 vote

Why the symmetry is not $Pin(1,3)$ or $Pin(3,1)$ in condensed matter physics?

Here is the answer from myself. The group $Pin(1,3)$ has group elements time reversal $T$ and reflection $M_i$ along 3 spatial axes such that $T^2=1$, $M_i^2=-1$ and $TM_i=-M_iT$, $i=x,y,z$. Similarly,...
edittide's user avatar

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