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[princeton118]

What is the rotation transformation generator?
What is the Lorentz group generator?

 

[Fredrik]

I kind of suck at the Lie Algebra stuff, so I'll just mention some of what Steven Weinberg said in The quantum theory of fields, volume 1, chapter 2. (Great chapter, I highly recommend it even if you're not going to read the rest of the book).

A rotation is a linear transformation

x\mapsto Rx : \mathbb{R}^3\rightarrow\mathbb{R}^3

where R is orthogonal. We can let R be a function of three parameters, e.g. Euler angles.

In quantum mechanics, there must be an operator U(R) that acts on state vectors to give us the vector that a rotated observer would use to describe the same physical system. If we take this to be a function of the parameters and Taylor expand it, the result can be expressed as

1-i\vec\theta\cdot\vec{J}+\cdots

where the components of theta are real numbers, the components of J are Hermitean operators, and terms of second and higher order in the parameters have been omitted. The components of J are called spin operators, and they are said to be generators of rotations, because the operators in the higher order terms can be constructed from the operators in the first order terms, so the generators are all that's needed except values of the parameters to construct a particular rotation operator.

It's possible to show that the components of J must satisfy the commutation relations

[J_i,J_j]=i\varepsilon_{ijk}J_k

Most QM books start the discussion about spin with these commutation relations, and don't say much about how they are generators of rotations.

So there are three generators of rotation, called spin. If we do the same with (inhomogeneous) Lorentz transformations instead, we get ten generators instead of three (because there are ten parameters: 3 boost parameters, 3 Euler angles and 4 translation parameters). They are called boost operators, spin operators and components of four-momentum. The zeroth component of the four-momentum is the Hamiltonian. It's the generator of translations in time, and it's eigenvalues are the possible energy levels of the physical system.

The Lie algebra stuff that I suck at, goes something like this: The Poincaré group is a Lie group (meaning it's both a group and a manifold). The tangent space at the identity element can be given the structure of a Lie algebra (a vector space with a way to "multiply" its members, similar to a cross product or a commutator) called the Poincaré algebra. The members of the Poincaré group are Lorentz transformations by definition, and the members of the Poincaré algebra are generators of Lorentz transformations.

 

[samalkhaiat]

What is the rotation transformation generator?

The set of all rotations:

\bar{r} = R( \theta ) {} r

that keeps r^{2} invariant ( i.e., \bar{r}^{2} = r^{2} ) form a group called the rotation group SO(3) (= Special Orthogonal group in 3 dimension).
For a very small (infinitesimal) rotations, the matrix R will differ from the unit matrix I_{3} by an infinitesimal 3x3 matrix \omega ( \theta ) [infinitesimal matrix is a matrix whose elements are infinitesimal numbers].
Thus;

\bar{r} = \left[ I + \omega ( \theta ) \right] {} r

or, in terms of the components (x_{1},x_{2},x_{3}) of the vector r,

\bar{x}_{i} = \left[ \delta_{ij} + \omega_{ij}( \theta ) \right] {} x_{j}

Now, the invariance condition;

\bar{x}_{i} \bar{x}_{i} = x_{j}x_{j}

forces the matrix \omega to be antisymmetric;

\omega_{ij} = - \omega_{ji}

Therefore, there are only three independent elements in \omega. So, we may write

<br /> \omega = \left( \begin{array}{ccc} 0 &amp; - \theta_{3} &amp; \theta_{2} \\ \theta_{3} &amp; 0 &amp; - \theta_{1} \\ - \theta_{2} &amp; \theta_{1} &amp; 0 \end{array} \right)<br />

We rewrite this as

<br /> \omega = i \theta_{1} J_{1} + i \theta_{2} J_{2} + i \theta_{3} J_{3} \equiv i \theta_{k}J_{k}<br />

where

<br /> J_{1} = \left( \begin{array}{ccc} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; i \\ 0 &amp; -i &amp; 0 \end{array} \right)<br />

<br /> J_{2} = \left( \begin{array}{ccc} 0 &amp; 0 &amp; -i \\ 0 &amp; 0 &amp; 0 \\ i &amp; 0 &amp; 0 \end{array} \right)<br />

and

<br /> J_{3} = \left( \begin{array}{ccc} 0 &amp; i &amp; 0 \\ -i &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{array} \right)<br />

Now it is easy to see that these J's satisfy the following commutation relations

\left[ J_{1}, J_{2} \right] \equiv J_{1}J_{2} - J_{2}J_{1} = i J_{3}

\left[ J_{3}, J_{1} \right] = i J_{2}

and

\left[ J_{2}, J_{3} \right] = i J_{1}

we write these in the compact form;

[ J_{i} , J_{j} ] = i \epsilon_{ijk} J_{k}

In quantum mechanics, the components of the angular momentum operator satisfy the same commutation relations. Therefore, one speaks of the angular momentum operator as the generator of the rotation group SO(3).
Mathematically, the matrix \omega is an element of 3-dimensional vector soace [the Lie algebra of SO(3)] whose basis vectors,J_{i}, satisfy the above-derived commutation relations [Lie brackets]. The components \theta_{i} of the Lie algebra element \omega, play the role of local coordinates of the group manifold. All elements of the group SO(3) can be represented as exponentials of Lie algebra elements, i.e., the matrix R( \theta ) can now be written as

R = \exp (i \theta_{k} J_{k} )

What is the Lorentz group generator?

Doing the same thing with Lotentz transformations leads to six 4x4 matrices (generators).
Try it yourself.


regards

sam

 

[cathalcummins]

The set of all rotations:

\bar{r} = R( \theta ) {} r

that keeps r^{2} invariant ( i.e., \bar{r}^{2} = r^{2} ) form a group called the rotation group SO(3) (= Special Orthogonal group in 3 dimension).
For a very small (infinitesimal) rotations, the matrix R will differ from the unit matrix I_{3} by an infinitesimal 3x3 matrix \omega ( \theta ) [infinitesimal matrix is a matrix whose elements are infinitesimal numbers].
Thus;

\bar{r} = \left[ I + \omega ( \theta ) \right] {} r

or, in terms of the components (x_{1},x_{2},x_{3}) of the vector r,

\bar{x}_{i} = \left[ \delta_{ij} + \omega_{ij}( \theta ) \right] {} x_{j}

Now, the invariance condition;

\bar{x}_{i} \bar{x}_{i} = x_{j}x_{j}

forces the matrix \omega to be antisymmetric;

\omega_{ij} = - \omega_{ji}

Therefore, there are only three independent elements in \omega. So, we may write

<br /> \omega = \left( \begin{array}{ccc} 0 &amp; - \theta_{3} &amp; \theta_{2} \\ \theta_{3} &amp; 0 &amp; - \theta_{1} \\ - \theta_{2} &amp; \theta_{1} &amp; 0 \end{array} \right)<br />

We rewrite this as

<br /> \omega = i \theta_{1} J_{1} + i \theta_{2} J_{2} + i \theta_{3} J_{3} \equiv i \theta_{k}J_{k}<br />

where

<br /> J_{1} = \left( \begin{array}{ccc} 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; i \\ 0 &amp; -i &amp; 0 \end{array} \right)<br />

<br /> J_{2} = \left( \begin{array}{ccc} 0 &amp; 0 &amp; -i \\ 0 &amp; 0 &amp; 0 \\ i &amp; 0 &amp; 0 \end{array} \right)<br />

and

<br /> J_{3} = \left( \begin{array}{ccc} 0 &amp; i &amp; 0 \\ -i &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \end{array} \right)<br />

Now it is easy to see that these J's satisfy the following commutation relations

\left[ J_{1}, J_{2} \right] \equiv J_{1}J_{2} - J_{2}J_{1} = i J_{3}

\left[ J_{3}, J_{1} \right] = i J_{2}

and

\left[ J_{2}, J_{3} \right] = i J_{1}

we write these in the compact form;

[ J_{i} , J_{j} ] = i \epsilon_{ijk} J_{k}

In quantum mechanics, the components of the angular momentum operator satisfy the same commutation relations. Therefore, one speaks of the angular momentum operator as the generator of the rotation group SO(3).
Mathematically, the matrix \omega is an element of 3-dimensional vector soace [the Lie algebra of SO(3)] whose basis vectors,J_{i}, satisfy the above-derived commutation relations [Lie brackets]. The components \theta_{i} of the Lie algebra element \omega, play the role of local coordinates of the group manifold. All elements of the group SO(3) can be represented as exponentials of Lie algebra elements, i.e., the matrix R( \theta ) can now be written as

R = \exp (i \theta_{k} J_{k} )



Doing the same thing with Lotentz transformations leads to six 4x4 matrices (generators).
Try it yourself.


regards

sam

This should be framed. You just tied a lot of things together which were popping about.