Lorentz algebra elements in an operator representation

[Jason Bennett]

Homework Statement

see below

Relevant Equations

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1) Likely an Einstein summation confusion.
Consider Lorentz transformation's defined in the following matter:

Please see image [2] below.

I aim to consider the product L^0{}_0(\Lambda_1\Lambda_2). Consider the following notation L^\mu{}_\nu(\Lambda_i) = L_i{}^\mu{}_\nu. How then, does L^0{}_0(\Lambda_1\Lambda_2) = L_1{}^0{}_\mu L_2{}^\mu{}_0?

2) To right the J and K generators of the Lorentz group in a compact way, one can write (M^{lm})^j{}_k=i (g^<br /> {lj}g^m{}_k - g^{mj}g^l{}_k) where on the left hand side, it is helpful to think of l and m as indices/labels, and the j and k as rows/columns for the whole matrix.
(Apparently) One can write this in an operator representation as M^{\mu\nu} = i(x^\mu \partial^\nu-x^\nu\partial^\mu).

a) where does this come from?

b) why and how is it used? What is it operating on, x^\mu?

c) how does \partial^\nu x^\sigma= \frac{\partial}{\partial x_\nu} x^\sigma equal g^{\nu\sigma}?

[2]: https://i.stack.imgur.com/uPsLc.png

 

[Antarres]

Regarding your first question, calculating the element of product Lorentz transformation is given by the formula you assumed. It follows from the Lorentz group multiplication rule $L(\Lambda_1)L(\Lambda_2) = L(\Lambda_1\Lambda_2)$, so you have to sum over $\mu$ index in order to get the component you're looking for.

As for the angular momentum generator, generators generate Lorentz tranformations on spaces of functions/fields. We know how Lorentz group acts on coordinates. It's a simple coordinate transformation:
$$x'^\mu = \Lambda^\mu_{\hphantom{\mu}\nu}x^\nu$$
Where $\Lambda$ is defined as matrix that preserves the Minkowski metric, as you've defined it in the picture you've attached. We now ask ourselves how this transformation is going to act on functions of these coordinate. Let's take the example of scalar field, since it is the simplest one(it has zero spin, so there is no additional spin transformation). In order to see how these transformations act on fields, we need to create a mapping that maps our group elements(Lorentz tranformations) into some operators on the space of functions/fields we want to act on. This mapping is called the representation of the group. Let's take scalar field $\phi(x)$. Acting on it with Lorentz tranformation, we get the transformed field $\phi'(x')$, where prime above field indicates the change of form of the function, and prime above x is just the change of coordinates. We wish to investigate how this field differs from the field we started with when the tranformation is infinitesimal. So we define total variation of the field
$$\delta\phi = \phi'(x') - \phi(x) = \phi'(x') - \phi'(x) + \phi'(x) - \phi(x) = \phi'(x') - \phi'(x) + \delta_0\phi(x)$$
We call $\delta_0\phi(x)$ the form variation of the field, that is, the change of function irrespectible of coordinate change. To further clarify the expression, we can use Taylor series to simplify the expression:
$$\delta\phi = \frac{\partial \phi}{\partial x^\mu}(x'^\mu -x^\mu) + \delta_0\phi(x)$$
However, if we look now at the definition of a scalar, scalar is a function that stays invariant under changes of coordinates, that is for scalar fields we have $\phi'(x') = \phi(x)$. So the variation of a scalar field is zero by definition, but that doesn't mean the form variation is zero, too. So, we will now investigate the form variation. We define the difference between old and new coordinates by $\xi^\mu = x'^\mu - x^\mu$. We get:
$$\delta_0\phi(x) = -\xi^\mu\partial_\mu\phi(x)$$
Now if we look what infinitesimal transformation does to coordinates, we get:
$$x'^\mu = x'^\mu(x) = x^\mu + \frac{\partial x'^\mu}{\partial x^\nu}x^\nu = x^\mu + \omega^\mu_{\hphantom{\mu}\nu}x^\nu$$
Where $\omega^\mu_{\hphantom{\mu}\nu}$ is infinitesimal matrix of Lorentz transformation(You get from the law of how coordinates transform that the derivative of new coordinates with respect to old ones equals the Lorentz matrix). Now we rewrite the form variation.
$$\delta_0\phi(x) = -\omega^\mu_{\hphantom{\mu}\nu}x^\nu\partial_\mu\phi(x) = -\frac{1}{2}\omega^{\mu\nu}M_{\mu\nu}\phi(x)$$
We have rewritten it in the form where $\omega^{\mu\nu}$ is an antisymmetric matrix of parameters and $M_{\mu\nu}$ are some operators which now act on $\phi(x)$. These operators are what we call the generators of Lorentz group. Direct comparison gives:
$$M_{\mu\nu} = x_\mu\partial_\nu - x_\nu\partial_\mu$$
The factor of $i$ that appears in your formula is there for Hermiticity reasons and is not the important part, it depends on how you define the representation of the group. So we have here derived the Lorentz generators as operators which change the form of scalar field under the action of Lorentz group. It is simple since scalar field doesn't change under those transformation so total variation is zero. If we do the same with vector and tensor fields, we will see that they transform as some matrix, which is function of these generators, that is, the representation of the group is different. But in general generators generate infinitesimal transformations on spaces of fields.

As for your last question, it is trivial, since you define $\partial^\mu = g^{\mu\nu}\partial_{\nu}$ Acting with that on coordinate ${x^\mu}$ and using $\tfrac{\partial x^\mu}{\partial x^\nu} = \delta^\mu_\nu$ gives the answer to your question.

I hope I didn't expand on this too far, I tried to be comprehensive so that you understand the point of generators. If you have more questions, feel free to ask.