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I saw there are three intrinsic symmetries in physics, $U(1)$, $SU(2)$, and $SU(3)$. What's the $U(1)$ symmetry talking about? I would appreciate it if you can give me some explanation.

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  • $\begingroup$ Are you wondering what U(1) is, which is a mathematical term dealing with symmetry, or are you wondering how it applies to the standard model (which deals with electromagnitism)? $\endgroup$
    – Cort Ammon
    May 7, 2019 at 13:22
  • $\begingroup$ I want to know its formula and some invariances in physics . $\endgroup$
    – karry
    May 7, 2019 at 13:27

2 Answers 2

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Let us first refer to symmetry generically. When we say a theory is symmetric under $G$ ($G$ some group) we mean that the elements of $G$ transform the states, and the operators of a theory, (in the context of the Standard Model (SM)), in such a way that the Lagrangian won't change in form. One could then speak about space-time symmetries, such as Lorentz transformations, or to intrinsic symmetries which transform the fields of your theory. These are the ones you find associated to the SM.

Let us now speak about $U(1)$ in particular. It corresponds to the unitary group of dimension $1$. That is you can think about this group as unitary matrices of dimension 1, which means essentially (complex) numbers of norm 1 or as they are usually parametrized $e^{i\theta}$ with $\theta\in[0,2\pi)$. Observe that $\theta$ is just a parameter that helps us label the elements of the group. You can intuitively see that $U(1)$ corresponds geometrically to a circle, and multiplication among elements is equivalent to adding the parameter $\theta$, that is rotating with some angle around the circle.

Coming back to the physics, one has two cases.

Option 1, it might be the case that your theory doesn't change, is invariant, when you employ a transformation $e^{i\theta}$ where $\theta$ is independent of the point in space-time but otherwise arbitrary. This is called a global symmetry and corresponds to conserved quantities through Noether's theorem.

Option 2, the theory can be invariant through a more complicated rule, $e^{i\theta(x)}$, which is a transformation that depends on the point of space-time. These sort are called gauge symmetries and are associated to gauge freedom and gauge bosons. They essentially tell you that there is some redundancy in the way you are writing your theory and certain states must be identified (they should be considered the same).

There are different interpretations of $U(1)$ depending on what it refers in a given context. Particle number conservation, QED and photons, Hypercharge in SM. I hope I answered your question, and mentioned enough keywords so you can read more on your own.

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    $\begingroup$ One thing I might add, just for completeness: the reason we see the idea of "groups" pop up when we're looking at symmetry is because group theory is mathematics' discipline for modeling the effects of symmetries. Some of the wording can be demanding, but in the end it's nothing more than a very formal way of capturing concepts like "If I spin 360 in place, the world will look exactly like it would have if I didn't spin" or capturing how a kung fu master can spin their staff without twisting themselves up nor having to let go. $\endgroup$
    – Cort Ammon
    May 7, 2019 at 13:53
  • $\begingroup$ @CortAmmon - or how walking down stairs is the same as walking up stairs $\endgroup$
    – Clock
    Mar 1, 2020 at 21:49
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You can intuitively see that U(1) corresponds geometrically to a circle, and multiplication among elements is equivalent to adding the parameter θ, that is rotating with some angle around the circle. Coming back to the physics, one has two cases.

In the first one , The action shown in the question is a functional of ϕ, not of x. A change of the integration variable x is just a relabeling of the index set. It does not transform the dynamic variables ϕ at all, so no: a change of variable does not correspond to a conserved quantity.

In the second case , since the presence of spin is the cause of asymmetry of the canonical energy momentum tensor, it is suggestive that an appropriate use of the spin current should help to construct a new modified momentum tensor $T_{\mu \nu} = \Theta_{\mu \nu } + ∆\Theta_{\nu \mu } $, that is symmetric, while still having the fundamental property of $\Theta_{µν}$ that its spatial integral P µ = R d 3x T µ0 yields the total energy-momentum vector of the system. This is ensured by the fact that $\Delta \Theta_{\mu 0}$ being a three-divergence of a spatial vector. Such a construction was found in 1939 by Belinfante.

As for Non-Relativistic version of U1 Symmetry in the limit that $ N \rightarrow \infty$, the 4-pt correlator of O2, and in fact of all properly normalized Ok, will be given entirely by evaluating pairs of 2-pt functions. Thus in the large N limit, this theory reproduces what we wanted – a generalized free theory with many choices for ∆, the dimension of the operator. It’s that last point that differentiates perturbation theory in 1/N from the more familiar examples of perturbation theory in a small coupling. For further reading, Witten’s ‘Baryons and the Large N Expansion’ [11] is a great intro to the 1/N expansion. An interesting point to note here is that the quantum mechanical expansion of φab is in terms of free field oscillator modes, which are not much like the AdS oscillator modes of a generalized free field. So the relationship between an operator like tr[φ k ] and O isn’t straightforward quantum mechanically. Transformations preserving distances $ dl^2 = dx^2 +dy^2 +dz^2$ ) of this space is the O(1) group (here we do not consider translations), which is also that of orthogonal real $3\times 3$ matrices. Let r~0 = R~r such that r~0 · r~0 = ~r · ~r defines a rotation R for a position vector ~r. Then r 0T r 0 = (Rr) T Rr = r T r so that r T RT Rr = r T r for all vector r. Therefore RT R = 1, which shows that R is an orthogonal matrix. (i) If $R \in O(3)$ and $R_0 \in O(3)$, (RR0 ) T (RR0 ) = R0T RT RR0 = R0T R0 = 1 so that RR0 ∈ O(3): closure. (ii) If R ∈ O(3) then RT ∈ O(3) and RT R = RRT = 1 so that $R^{−1} = RT $ : the inverse of R belongs to O(3). (iii) The unit 3 × 3 matrix I is a neutral element as IR = RI = R (for simplicity we will often write 1 instead of I). It is non-abelian (or non-commutative) as in general RR0 6= R0R, which is a well known property of matrices. As det(RT R) = 1 = (det R) 2 and det R is real so that det R = ±1. SO(3) (special orthogonal group) is a subgroup made of det R = 1 matrices (show it). R ∈ SO(3) a priori depends on 32 = 9 real elements but RT R = 1 means that $R_{ki} R_{kj} = \delta_{ij}$ (summation over repeated indices implied), which gives 6 independent conditions (because the two equations $R_{ki} R_{kj} = 0$ when i 6= j and $R_{ki} R_{kj} = 0$ are the same) so that in the end there are only 3 independent (real) parameters 3 . Derivatives are related to the momentum operator, for example $\frac{p^2} {2m_e} = \frac{\nabla ^2} {2m_e}$ in the Schrödinger equation and $p_\mu = i\partial_\mu$ in the Dirac equation .

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