Introduction

In nonrelativistic quantum mechanics, spin is already unusual. It is not orbital motion in space, yet it behaves mathematically like angular momentum. In quantum field theory, spin is not an isolated particle label added by hand. It comes from how the field transforms under spatial rotations.

Relativity forces a deeper version of the same idea.

If physical laws are to be the same in all inertial frames, then fields must transform correctly not only under ordinary spatial rotations, but under the full Lorentz group. The Lorentz group contains:

• spatial rotations,
• boosts, which relate inertial frames moving at constant velocity relative to each other.

Relativistic spin is the study of how fields transform under this larger spacetime symmetry group.

The basic story is:

1. Ordinary rotations have generators satisfying angular-momentum commutation relations.
2. Fields can transform under rotations through their spatial arguments and through their internal components.
3. The Lorentz group extends rotations to spacetime by adding boost generators.
4. The rotation and boost generators together obey the Lorentz algebra.
5. The Lorentz algebra can be rewritten as two independent angular-momentum algebras.
6. Lorentz representations are therefore classified by two spin labels, $(s_L,s_R)$.
7. Left-handed and right-handed Lorentz 2-spinors are the fundamental relativistic spin-$\frac12$ objects.

This lesson is the bridge from ordinary spin to relativistic spinors and, ultimately, to the Dirac equation.

Learning Objectives

By the end, you should be able to:

• Explain how fields transform under ordinary spatial rotations.
• Distinguish rotations of a field's spatial argument from rotations of its internal components.
• Derive the orbital angular-momentum generators as differential operators.
• Understand finite-dimensional matrix representations of the rotation algebra.
• Define the Lorentz group as the symmetry group preserving the Minkowski interval.
• Write finite boosts using rapidity.
• Construct boost generators.
• Explain why boost generators alone do not form a closed algebra.
• Write the Lorentz algebra of rotations and boosts.
• Show how the Lorentz algebra splits into two commuting angular-momentum algebras.
• Classify Lorentz representations using the pair $(s_L,s_R)$.
• Define left-handed and right-handed Lorentz 2-spinors.
• Explain why left- and right-handed spinors transform the same under spatial rotations but differently under boosts.

Prerequisite Knowledge

You should already know:

• basic special relativity,
• Minkowski spacetime,
• Lorentz transformations,
• angular momentum in quantum mechanics,
• matrix exponentials,
• field transformations,
• basic Pauli matrix notation.

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1. Review: spatial rotations of fields

Consider a field $\phi(\mathbf r)$. Under a spatial rotation in three dimensions, the spatial argument of the field changes.

Symbolically,

$$\phi(\mathbf r)\rightarrow \phi(\mathbf r'(n,\theta)),$$

where $n$ is the rotation axis and $\theta$ is the rotation angle.

For a rotation about the $z$-axis, cylindrical coordinates make the transformation simple:

$$\phi(r,\varphi,z) \rightarrow \phi(r,\varphi-\theta,z).$$

The minus sign appears because of the usual convention that counterclockwise rotations are positive. The new field value at the new point equals the old field value at the corresponding old point.

Using the Taylor expansion of the exponential differential operator,

$$\phi(r,\varphi-\theta,z) = e^{-\theta \frac{\partial}{\partial \varphi}} \phi(r,\varphi,z),$$

where

$$e^{-\theta \frac{\partial}{\partial \varphi}} = \sum_{n=0}^{\infty} \frac{(-\theta)^n}{n!} \frac{\partial^n}{\partial \varphi^n}.$$

So a finite rotation is generated by the differential operator $\partial/\partial\varphi$.

Because rotations are continuous, a finite rotation can also be viewed as an infinite sequence of infinitesimal rotations:

$$e^{-\theta \frac{\partial}{\partial\varphi}} = \lim_{N\to\infty} \left( e^{-(\theta/N)\frac{\partial}{\partial\varphi}} \right)^N.$$

It is conventional to write this in angular-momentum form:

$$e^{-\theta \frac{\partial}{\partial\varphi}}\phi(r,\varphi,z) = e^{\frac{i}{\hbar}\theta \hat L_z} \phi(r,\varphi,z),$$

with

$$\hat L_z=i\hbar\frac{\partial}{\partial\varphi}.$$

The factor $i\hbar$ is a convention at the classical-field level. After quantization, it matches the usual notation for angular momentum.

Returning to Cartesian coordinates gives the orbital angular-momentum generators:

$$\hat L_x = i\hbar \left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right),$$ $$\hat L_y = i\hbar \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right),$$ $$\hat L_z = i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right).$$

These generators describe how the spatial argument of a field changes under rotations.

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2. The rotation algebra

Rotations are not translations. This shows up in the commutation relations.

The orbital angular-momentum generators satisfy

$$[\hat L_i,\hat L_j] = i\hbar\epsilon_{ijk}\hat L_k.$$

Here $\epsilon_{ijk}$ is the antisymmetric Levi-Civita tensor, and repeated indices are summed.

This algebra is the defining structure of rotations. Anything that genuinely transforms as a rotation must have generators obeying the same commutation relation.

So we now ask:

Can something besides the spatial argument of a field transform under rotation?

Yes. The internal components of a multi-component field can rotate into each other.

For a three-component vector field, the internal component rotations are generated by $3\times 3$ matrices. One useful convention is

$$\hat S_x = i\hbar \begin{pmatrix} 0&0&0\\ 0&0&-1\\ 0&1&0 \end{pmatrix},$$ $$\hat S_y = i\hbar \begin{pmatrix} 0&0&1\\ 0&0&0\\ -1&0&0 \end{pmatrix},$$ $$\hat S_z = i\hbar \begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&0 \end{pmatrix}.$$

These matrices generate rotations among the Cartesian components of a 3-vector, and they also satisfy

$$[\hat S_i,\hat S_j] = i\hbar\epsilon_{ijk}\hat S_k.$$

So a vector field transforms in two ways under rotations:

1. its spatial argument rotates, generated by $\hat L_i$;
2. its internal components rotate, generated by $\hat S_i$.

The total angular momentum is therefore built from orbital plus spin parts.

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3. Diagonalizing $S_z$

Cartesian vector components are intuitive, but they are not always the most useful basis. Since rotations are exponentials of generators, it is often useful to work in a basis of eigenvectors of one generator, usually $S_z$.

For the vector representation, the three eigenvalues of each $\hat S_i$ are

$$+\hbar,\qquad 0,\qquad -\hbar.$$

In the $S_z$-eigenbasis,

$$\hat S_z = \hbar \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}.$$

The corresponding basis vectors are

$$e_0=e_z,$$ $$e_{\pm} = \frac{e_x\mp i e_y}{\sqrt2}.$$

This is still just ordinary three-dimensional rotation algebra. No quantum mechanics is required yet. The notation uses $i\hbar$ because it is convenient and later matches the quantum notation.

The important question is:

Are there other finite-dimensional matrices, larger or smaller, that also satisfy

$$[\hat S_i,\hat S_j] = i\hbar\epsilon_{ijk}\hat S_k?$$

The answer is yes. This is the origin of spin-$\frac12$, spin-1, spin-$\frac32$, and so on.

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4. Finite representations of the rotation algebra

In the vector representation,

$$|\hat S|^2 = \hat S_x^2+\hat S_y^2+\hat S_z^2 = 2\hbar^2 I_{3\times 3}.$$

Because

$$[\hat S_i,\hat S_x^2+\hat S_y^2+\hat S_z^2]=0,$$

we can choose simultaneous eigenvectors of $|\hat S|^2$ and $\hat S_z$.

The ladder operators are

$$\hat S_{\pm} = \frac{\hat S_x\pm i\hat S_y}{\sqrt2}.$$

They raise and lower the $S_z$ eigenvalue:

$$\hat S_z\hat S_{\pm} = \hat S_{\pm}(\hat S_z\pm\hbar).$$

Equivalently, when acting on an eigenvector of $\hat S_z$, $\hat S_+$ increases the $S_z$ eigenvalue by $\hbar$, while $\hat S_-$ decreases it by $\hbar$.

The operator identity used in the finite-representation analysis is

$$\sum_{i=1}^{3}\hat S_i^2 = \hat S_z^2 + 2\hat S_{\pm}\hat S_{\mp} \mp \hbar \hat S_z.$$

At the top and bottom of each finite ladder,

$$\hat S_{\pm}|\psi_\pm\rangle=0.$$

The vector representation has spin $s=1$, because

$$|\hat S|^2 = s(s+1)\hbar^2 I = 2\hbar^2 I \quad\Rightarrow\quad s=1.$$

This review matters because the Lorentz algebra will later reduce to two independent copies of this same angular-momentum algebra.

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5. The Lorentz group

Relativity replaces the Euclidean distance of space with the Minkowski interval

$$ds^2 = dx^2+dy^2+dz^2-c^2dt^2.$$

A Lorentz transformation is any linear transformation that preserves this interval:

$$ds^2 = dx'^2+dy'^2+dz'^2-c^2dt'^2.$$

A standard boost in the $x$-direction is

$$x' = \cosh\beta\,x - \sinh\beta\,ct,$$ $$ct' = \cosh\beta\,ct - \sinh\beta\,x,$$ $$y'=y, \qquad z'=z.$$

The parameter $\beta$ is the rapidity. It is related to velocity by

$$\cosh\beta = \frac{1}{\sqrt{1-v^2/c^2}},$$ $$\sinh\beta = \frac{v/c}{\sqrt{1-v^2/c^2}},$$

and therefore

$$v=c\tanh\beta.$$

A boost is a change to a uniformly moving inertial frame.

The Lorentz group is therefore the group of generalized rotations in four-dimensional spacetime.

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6. Infinitesimal boosts and boost generators

For an infinitesimal boost in the $x$-direction,

$$\cosh\beta\simeq 1, \qquad \sinh\beta\simeq \beta.$$

So

$$x'\simeq x-\beta ct,$$ $$ct'\simeq ct-\beta x,$$

with

$$y'=y, \qquad z'=z.$$

In matrix form,

$$\begin{pmatrix} x'\\ y'\\ z'\\ ct' \end{pmatrix} = \left[ I + \beta \begin{pmatrix} 0&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&0 \end{pmatrix} + O(\beta^2) \right] \begin{pmatrix} x\\ y\\ z\\ ct \end{pmatrix}.$$

We define the boost generator $\hat T_1$ by

$$\begin{pmatrix} x'\\ y'\\ z'\\ ct' \end{pmatrix} = e^{\frac{i}{\hbar}\beta\hat T_1} \begin{pmatrix} x\\ y\\ z\\ ct \end{pmatrix}.$$

This convention gives

$$\hat T_1 = i\hbar \begin{pmatrix} 0&0&0&-1\\ 0&0&0&0\\ 0&0&0&0\\ -1&0&0&0 \end{pmatrix}.$$

Similarly,

$$\hat T_2 = i\hbar \begin{pmatrix} 0&0&0&0\\ 0&0&0&-1\\ 0&0&0&0\\ 0&-1&0&0 \end{pmatrix},$$ $$\hat T_3 = i\hbar \begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&-1\\ 0&0&-1&0 \end{pmatrix}.$$

These are the generators of boosts in the $x$, $y$, and $z$ directions.

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7. Boost generators do not close among themselves

The three boost generators alone do not form a closed algebra.

For example, multiplying the matrices gives a commutator of the form

$$[\hat T_3,\hat T_1] = -i\hbar \hat S_2.$$

This is not another boost generator. It is a spatial rotation generator.

The four-vector rotation generators are the ordinary $3\times 3$ spatial rotation generators extended with an extra row and column of zeros for the $ct$ component:

$$\hat S_1^{(4)} = i\hbar \begin{pmatrix} 0&0&0&0\\ 0&0&-1&0\\ 0&1&0&0\\ 0&0&0&0 \end{pmatrix},$$ $$\hat S_2^{(4)} = i\hbar \begin{pmatrix} 0&0&1&0\\ 0&0&0&0\\ -1&0&0&0\\ 0&0&0&0 \end{pmatrix},$$ $$\hat S_3^{(4)} = i\hbar \begin{pmatrix} 0&-1&0&0\\ 1&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}.$$

Thus two boosts can produce a rotation. This is one of the characteristic features of Lorentz symmetry.

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8. The Lorentz algebra

The rotation generators and boost generators together form a closed algebra.

The rotation-rotation commutators are

$$[\hat S_j,\hat S_k] = i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

The rotation-boost commutators are

$$[\hat S_j,\hat T_k] = i\hbar\epsilon_{jk\ell}\hat T_\ell.$$

This means rotating a boost direction turns it into a boost in the rotated direction.

The boost-boost commutators are

$$[\hat T_j,\hat T_k] = -i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

The minus sign is essential. It comes from the minus sign in the Minkowski metric:

$$ds^2=dx^2+dy^2+dz^2-c^2dt^2.$$

Time is not exactly like space.

Together, the six generators

$$\hat S_1,\hat S_2,\hat S_3, \qquad \hat T_1,\hat T_2,\hat T_3$$

form the Lorentz algebra.

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9. Why Lorentz representations matter

In nonrelativistic physics, fundamental equations must be covariant under spatial rotations. That is why fields are classified as scalars, vectors, spinors, and so on.

For example:

• scalar equations describe scalar fields,
• vector equations describe vector fields,
• spinor equations describe spinor fields.

In relativistic physics, the same idea applies to spacetime rotations. Fundamental equations must be Lorentz-covariant. So we must classify all possible ways fields can transform under the Lorentz group.

Two obvious representations are:

1. Lorentz scalars, which do not change under Lorentz transformations;
2. four-vectors, which transform like $(x,y,z,ct)$.

The crucial question is:

Are there more possibilities?

Yes. The Lorentz algebra has more structure than the rotation algebra, but it is also reducible in a very useful way.

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10. The Lorentz algebra is reducible

Define two new sets of generators:

$$\hat{\mathbf R}_L = \frac12 (\hat{\mathbf S}+i\hat{\mathbf T}),$$ $$\hat{\mathbf R}_R = \frac12 (\hat{\mathbf S}-i\hat{\mathbf T}).$$

Equivalently, component by component,

$$\hat R_{Lj} = \frac12 (\hat S_j+i\hat T_j),$$ $$\hat R_{Rj} = \frac12 (\hat S_j-i\hat T_j).$$

Using the Lorentz algebra, one finds

$$[\hat R_{Lj},\hat R_{Lk}] = i\hbar\epsilon_{jk\ell}\hat R_{L\ell},$$ $$[\hat R_{Rj},\hat R_{Rk}] = i\hbar\epsilon_{jk\ell}\hat R_{R\ell},$$

and

$$[\hat R_{Lj},\hat R_{Rk}] = 0.$$

So $\hat{\mathbf R}_L$ and $\hat{\mathbf R}_R$ each obey an ordinary angular-momentum algebra, and every left-handed generator commutes with every right-handed generator.

This is the central result:

$$\text{Lorentz algebra} \cong \text{left angular momentum algebra} \oplus \text{right angular momentum algebra}.$$

The possible Lorentz transformation laws are therefore just the possible ways of rotating in 3D, but separately for left-handed and right-handed spacetime rotations.

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11. Lorentz representations

Every representation of ordinary angular momentum is labeled by one spin quantum number

$$s=0,\frac12,1,\frac32,\dots$$

For the Lorentz group, every representation is labeled by two spin quantum numbers:

$$(s_L,s_R),$$

where

$$s_L,s_R = 0,\frac12,1,\frac32,\dots$$

independently.

Important examples are:

$$\text{Lorentz scalar}: \qquad (s_L,s_R)=(0,0).$$ $$\text{Four-vector}: \qquad (s_L,s_R)=\left(\frac12,\frac12\right).$$

There are also Lorentz 2-spinors:

$$\text{left-handed Lorentz 2-spinor}: \qquad (s_L,s_R)=\left(\frac12,0\right),$$ $$\text{right-handed Lorentz 2-spinor}: \qquad (s_L,s_R)=\left(0,\frac12\right).$$

These are the basic relativistic spin-$\frac12$ building blocks.

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12. Left-handed Lorentz 2-spinors

A left-handed Lorentz 2-spinor has representation label

$$(s_L,s_R) = \left(\frac12,0\right).$$

It has two complex components:

$$\eta = \begin{pmatrix} \eta_1\\ \eta_2 \end{pmatrix}.$$

Its Lorentz generators are

$$\hat{\mathbf R}_L = \frac{\hbar}{2}\boldsymbol{\sigma}, \qquad \hat{\mathbf R}_R = 0,$$

where

$$\boldsymbol{\sigma} = (\sigma_1,\sigma_2,\sigma_3)$$

are the Pauli matrices.

The ordinary spatial rotation generator is

$$\hat{\mathbf S} = \hat{\mathbf R}_L+\hat{\mathbf R}_R.$$

So for a left-handed spinor,

$$\hat{\mathbf S} = \frac{\hbar}{2}\boldsymbol{\sigma}.$$

Therefore, as far as ordinary 3D rotations are concerned, a left-handed Lorentz 2-spinor behaves like the usual nonrelativistic spin-$\frac12$ two-spinor.

The boost generator is obtained from

$$\hat{\mathbf R}_L-\hat{\mathbf R}_R = i\hat{\mathbf T}.$$

Equivalently,

$$\hat{\mathbf T} = -i(\hat{\mathbf R}_L-\hat{\mathbf R}_R).$$

For a left-handed spinor,

$$\hat{\mathbf T} = -i\frac{\hbar}{2}\boldsymbol{\sigma}.$$

A finite boost in the $z$-direction is

$$\eta' = e^{\frac{i}{\hbar}\beta\hat T_3}\eta.$$

Using

$$\hat T_3 = -i\frac{\hbar}{2}\sigma_3,$$

we get

$$\eta' = e^{\frac{\beta}{2}\sigma_3}\eta.$$

Since

$$\sigma_3 = \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix},$$

this gives

$$\eta' = \begin{pmatrix} e^{\beta/2}&0\\ 0&e^{-\beta/2} \end{pmatrix} \begin{pmatrix} \eta_1\\ \eta_2 \end{pmatrix}.$$

So under boosts, the two components of a left-handed Lorentz 2-spinor change in size.

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13. Right-handed Lorentz 2-spinors

A right-handed Lorentz 2-spinor has representation label

$$(s_L,s_R) = \left(0,\frac12\right).$$

It has two complex components:

$$\xi = \begin{pmatrix} \xi_1\\ \xi_2 \end{pmatrix}.$$

Its Lorentz generators are

$$\hat{\mathbf R}_R = \frac{\hbar}{2}\boldsymbol{\sigma}, \qquad \hat{\mathbf R}_L = 0.$$

Again,

$$\hat{\mathbf S} = \hat{\mathbf R}_L+\hat{\mathbf R}_R = \frac{\hbar}{2}\boldsymbol{\sigma}.$$

So under ordinary spatial rotations, the right-handed Lorentz 2-spinor also behaves like a usual spin-$\frac12$ two-spinor.

But the boost generator is now

$$\hat{\mathbf T} = -i(\hat{\mathbf R}_L-\hat{\mathbf R}_R) = +i\frac{\hbar}{2}\boldsymbol{\sigma}.$$

For a finite boost in the $z$-direction,

$$\xi' = e^{\frac{i}{\hbar}\beta\hat T_3}\xi.$$

Using

$$\hat T_3 = +i\frac{\hbar}{2}\sigma_3,$$

we get

$$\xi' = e^{-\frac{\beta}{2}\sigma_3}\xi.$$

Therefore

$$\xi' = \begin{pmatrix} e^{-\beta/2}&0\\ 0&e^{+\beta/2} \end{pmatrix} \begin{pmatrix} \xi_1\\ \xi_2 \end{pmatrix}.$$

So the two components of a right-handed Lorentz 2-spinor change in size oppositely to the corresponding components of a left-handed Lorentz 2-spinor.

This is the key physical difference:

$$\text{left-handed and right-handed spinors transform the same under rotations,}$$

but

$$\text{they transform oppositely under boosts.}$$

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14. Connection to the Dirac equation

The Dirac equation is built from both left-handed and right-handed Lorentz 2-spinors.

A Dirac spinor can be viewed, in chiral form, as combining one left-handed and one right-handed two-spinor:

$$\Psi = \begin{pmatrix} \eta\\ \xi \end{pmatrix}.$$

The reason this is necessary is that relativity distinguishes left and right under boosts, even though ordinary rotations do not.

So relativistic spin-$\frac12$ theory is not just ordinary spin-$\frac12$ quantum mechanics with time added. It requires the Lorentz transformation behavior of left- and right-handed 2-spinors.

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Worked Examples

Example 1: Why a vector field has spin-1 components

A three-component vector field transforms under spatial rotations by rotating both its spatial argument and its internal components.

In the basis

$$e_0=e_z, \qquad e_\pm=\frac{e_x\mp i e_y}{\sqrt2},$$

the component generator is diagonal:

$$\hat S_z = \hbar \begin{pmatrix} 1&0&0\\ 0&0&0\\ 0&0&-1 \end{pmatrix}.$$

Therefore the three components correspond to spin projections

$$m=+1,\qquad m=0,\qquad m=-1.$$

When the vector field is quantized, creation operators associated with these components create particles with the corresponding spin projection.

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Example 2: Why two boosts can produce a rotation

Boost generators satisfy

$$[\hat T_j,\hat T_k] = -i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

For example,

$$[\hat T_3,\hat T_1] = -i\hbar \hat S_2.$$

So a boost in the $z$-direction and a boost in the $x$-direction do not combine as just another boost. Their noncommutativity produces a rotation around the $y$-axis.

This is a purely relativistic effect and is closely related to the geometry of Minkowski spacetime.

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Example 3: Left-handed versus right-handed boost behavior

For a left-handed spinor,

$$\hat{\mathbf T}_L = -i\frac{\hbar}{2}\boldsymbol{\sigma}.$$

For a right-handed spinor,

$$\hat{\mathbf T}_R = +i\frac{\hbar}{2}\boldsymbol{\sigma}.$$

Under a $z$-boost,

$$\eta' = \begin{pmatrix} e^{\beta/2}&0\\ 0&e^{-\beta/2} \end{pmatrix} \eta,$$

while

$$\xi' = \begin{pmatrix} e^{-\beta/2}&0\\ 0&e^{+\beta/2} \end{pmatrix} \xi.$$

Thus the two spinors look identical under spatial rotations but opposite under boosts.

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Intuition

Ordinary spin comes from how objects transform under rotations in space.

Relativistic spin comes from how objects transform under the full Lorentz group of spacetime.

At first, this looks harder because the Lorentz group has six generators:

$$3\ \text{rotations} + 3\ \text{boosts}.$$

But the Lorentz algebra can be reorganized into two ordinary angular-momentum algebras:

$$\hat{\mathbf R}_L = \frac12(\hat{\mathbf S}+i\hat{\mathbf T}), \qquad \hat{\mathbf R}_R = \frac12(\hat{\mathbf S}-i\hat{\mathbf T}).$$

So relativity does not destroy the angular-momentum picture. It doubles it.

A relativistic field is classified by how much spin it carries in each sector:

$$(s_L,s_R).$$

Examples:

$$(0,0):\ \text{scalar},$$ $$\left(\frac12,\frac12\right):\ \text{four-vector},$$ $$\left(\frac12,0\right):\ \text{left-handed 2-spinor},$$ $$\left(0,\frac12\right):\ \text{right-handed 2-spinor}.$$

The deepest point is that boosts can distinguish left from right. Ordinary spatial rotations cannot.

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Common Mistakes

• Thinking a scalar field does not transform at all under rotation. Its value does not rotate internally, but its spatial argument still changes.
• Confusing orbital angular momentum $\hat L$ with spin angular momentum $\hat S$.
• Thinking spin is added by hand instead of coming from field transformation properties.
• Treating boosts as if they form a closed algebra by themselves.
• Forgetting that the commutator of two boosts gives a rotation.
• Missing the minus sign in

$$[\hat T_j,\hat T_k] = -i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

• Thinking the Lorentz group requires totally new representation theory unrelated to angular momentum.
• Missing the decomposition into left-handed and right-handed angular-momentum algebras.
• Assuming left- and right-handed 2-spinors differ under ordinary rotations.
• Forgetting that left- and right-handed spinors differ specifically under boosts.
• Thinking a Dirac spinor is just one ordinary two-component spinor.

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Short Summary

A field transforms under rotations both through its spatial argument and, if it has multiple components, through internal component rotations. The orbital generators are differential operators,

$$\hat L_x = i\hbar \left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right), \quad \hat L_y = i\hbar \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right), \quad \hat L_z = i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right),$$

and they obey

$$[\hat L_i,\hat L_j] = i\hbar\epsilon_{ijk}\hat L_k.$$

Internal spin generators obey the same algebra:

$$[\hat S_i,\hat S_j] = i\hbar\epsilon_{ijk}\hat S_k.$$

Relativity extends rotations to the Lorentz group, which preserves

$$ds^2=dx^2+dy^2+dz^2-c^2dt^2.$$

The Lorentz group has three rotation generators $\hat S_i$ and three boost generators $\hat T_i$. Their algebra is

$$[\hat S_j,\hat S_k] = i\hbar\epsilon_{jk\ell}\hat S_\ell,$$ $$[\hat S_j,\hat T_k] = i\hbar\epsilon_{jk\ell}\hat T_\ell,$$ $$[\hat T_j,\hat T_k] = -i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

The decisive step is defining

$$\hat{\mathbf R}_L = \frac12(\hat{\mathbf S}+i\hat{\mathbf T}), \qquad \hat{\mathbf R}_R = \frac12(\hat{\mathbf S}-i\hat{\mathbf T}).$$

These obey two independent angular-momentum algebras:

$$[\hat R_{Lj},\hat R_{Lk}] = i\hbar\epsilon_{jk\ell}\hat R_{L\ell},$$ $$[\hat R_{Rj},\hat R_{Rk}] = i\hbar\epsilon_{jk\ell}\hat R_{R\ell},$$ $$[\hat R_{Lj},\hat R_{Rk}] = 0.$$

Therefore Lorentz representations are classified by two spin labels:

$$(s_L,s_R).$$

The scalar is $(0,0)$, the four-vector is $(\frac12,\frac12)$, the left-handed Lorentz 2-spinor is $(\frac12,0)$, and the right-handed Lorentz 2-spinor is $(0,\frac12)$.

Left- and right-handed 2-spinors both transform as ordinary spin-$\frac12$ objects under spatial rotations:

$$\hat{\mathbf S} = \frac{\hbar}{2}\boldsymbol{\sigma}.$$

But they transform oppositely under boosts:

$$\hat{\mathbf T}_L = -i\frac{\hbar}{2}\boldsymbol{\sigma}, \qquad \hat{\mathbf T}_R = +i\frac{\hbar}{2}\boldsymbol{\sigma}.$$

This left-right distinction is the structure behind relativistic spinor theory and the Dirac equation.

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Practice Problems

1. Explain why even a scalar field changes under a spatial rotation.

2. Starting from

$$\phi(r,\varphi,z)\rightarrow \phi(r,\varphi-\theta,z),$$

derive

$$\hat L_z=i\hbar\frac{\partial}{\partial\varphi}.$$

3. What is the difference between orbital angular momentum and spin angular momentum for a field?

4. Why must internal rotation generators obey

$$[\hat S_i,\hat S_j]=i\hbar\epsilon_{ijk}\hat S_k?$$

5. Show that the vector representation has spin $s=1$ using

$$|\hat S|^2=2\hbar^2I.$$

6. What is rapidity $\beta$, and how is it related to velocity $v$?

7. Why does a boost in the $x$-direction mix $x$ and $ct$?

8. Why do boosts alone fail to form a closed algebra?

9. Explain the physical meaning of

$$[\hat T_j,\hat T_k] = -i\hbar\epsilon_{jk\ell}\hat S_\ell.$$

10. Why does the minus sign in the boost-boost commutator come from the Minkowski metric?

11. Derive the commutators of

$$\hat{\mathbf R}_L = \frac12(\hat{\mathbf S}+i\hat{\mathbf T}), \qquad \hat{\mathbf R}_R = \frac12(\hat{\mathbf S}-i\hat{\mathbf T}).$$

12. What do the two labels $(s_L,s_R)$ mean?

13. Why is a four-vector represented by

$$\left(\frac12,\frac12\right)$$

rather than by one spin label?

14. What is the difference between a left-handed and right-handed Lorentz 2-spinor?

15. Why do left- and right-handed 2-spinors transform the same under rotations but differently under boosts?

16. Why is the Dirac equation naturally built from both left-handed and right-handed Lorentz 2-spinors?