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You are correct in your interpretation that Weisner's method is geometric in nature: it is a method for finding generating functions for special functions using representation theory of Lie groups and Lie algebras. And as you know, Lie groups play an enormous role in modern geometry, on several different levels. Lie groups are smooth differentiable ...


0

The algebraic form of the quadratic casimir $T^2$ depends only on the structure constants, and are therefore the same in any representation. As in $SU(2)$, of course, it's matrix form is representation dependent. Edit: To see why this is the case, suppose you constructed $T^2$ out of other elements of the algebra, and you did this for say, the fundamental ...


0

Just to elaborate on ACuriousMind's answer in case it it not immediately clear what he means. Think of angular momentum fundamentally being defined as the generator of rotation. If we have any system, in this case an isolated quantum system described by a state, how does this change if we rotate it (or if we rotate the frame from which we describe it). Since ...


0

They do not lie in $\mathfrak{so}(3)$ but they lie in its complexification, which would be $A_1$ in the usual mathematical classification. Much of Lie representation theory is set up this way: you work at the level of the complexification then go back to the real form. For compact groups it's not a big deal; for non-compact groups extra care is needed. So ...


-1

Any angular momentum in QM is quantized. Spin is not an exclusion. A hydrogen atom may also have spin even though "constructed" with spinless particles. And a free atom is not localized in space, just like a free electron. So, the angular momentum is a property of elementary and non elementary "particles" in QM. Spin is the angular momentum in the system ...


0

To say, in non-relativistic QM, that a state has spin $\frac{1}{2}$ means that it transforms in the representation of $\mathrm{SU}(2)$ with highest weight $\frac{1}{2}$, which is a two-dimensional space. In general, to say that a state has spin $s$ means to say that it transforms in the representation with highest weight $s$.


0

This task is more complex than the task to solve a quadratic equation, for example, and one must master a significant portion of a textbook – such as Georgi's textbook – and perhaps something beyond it to have everything he needs. For the 8-dimensional representation of $SU(3)$, things simplify because it's the "adjoint rep" of $SU(3)$ – the vector space ...


2

Let's take the spin, it's the simplest case, $Q = \mathbf{s}$. The operator $\mathbf{s}$ is a vector, \begin{equation}s = \mathbf{i}s_x + \mathbf{j}s_y + \mathbf{k}s_z\end{equation} while the operator s^2 is a scalar \begin{equation}s^2 = s_x^2 + s_y^2 + s_z^2\end{equation}. The operators $s_x$, $s_y$, and $s_z$ don't commute two by two, but $s_x^2$, ...


3

why it is such a prevalent idea. In elementary particle physics and nuclear physics groups and their representations have played a very crucial role in developing the standard models. The elementary particles in the table in the link above have a lot of quantum numbers. These quantum numbers have lead to observed symmetries, that can be described by ...


9

It's an enormous subject, but briefly: Lie groups are smooth groups. Technically, Lie groups are sets that are both a smooth manifold, like a sphere for instance, and also have a group structure (multiplication operator, inverses, and an identity). The group multiplication and inverse must be smooth (differentiable) functions on the manifold. As you ...


0

To find the spin eigenstates corresponding to a multi-particle state, all one needs to do is build the appropriate multi-particle spin operator using the direct product, e.g. $$J_z^{(2)}=J_z \otimes 1 + 1\otimes J_z \\ J_z^{(3)}=J_z \otimes 1 \otimes 1 + 1\otimes J_z \otimes 1 + 1\otimes 1\otimes J_z $$ and then diagonalize the resulting operator to find ...



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