What is the Operator for Spin-3/2 Rotation in Fermion Fields?

In summary, The original poster is asking for help finding the spin-3/2 version of the rotation operator for a spin-3/2 fermion field. They have tried using the sum of the spin-1/2 and spin-1 operators, but it does not give the correct eigenvalues. They are looking for clarification on how a spinor-vector transforms under the Lorentz group.
  • #1
tmc
293
1
A long literature search has given me nothing, so I'm turning to this forum for help.

I have a spin-3/2 fermion field, and I want to find its wave functions corresponding to its 4 pure-spin states, +3/2, +1/2, -1/2, -3/2, which is normally done by finding the 4 eigenfunctions of its rotation operator (which has exactly these four eigenvalues). My problem is that I can't find this operator.

For a spin-1/2 particle, the operator is
[tex]
R_{1/2} = \frac{1}{2} \left[
\begin{array}{ c c c c}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array} \right]
[/tex]
which has eigenvalues +1/2 and -1/2. For a spin-1 particle, the operator is
[tex]
R_1 = \left[
\begin{array}{ c c c c}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0
\end{array} \right]
[/tex]
having this time eigenvalues +1, 0 and -1.

I'm looking for the spin-3/2 version of these operators. The original paper by Rarita-Schwinger about the spin-3/2 field states that this operator should be the sum of the above two operators (one would rotate the spinor index, while the other would rotate the vector index), which gives eigenvalues +1/2, -1/2, +sqrt(5)/2 and -sqrt(5)/2; multiplying the second operator by a factor of sqrt(2) would give the correct eigenvalues, but I don't see how I could ever justify that.

The operator diag(3/2,1/2,-1/2,-3/2) would obviously give the right eigenvalues, but I strongly doubt it does what I want it to do, mostly because its equivalent representation for spin-1 is diag(1,0,-1) which does not even have the right dimensions.
 
Last edited:
Physics news on Phys.org
  • #2
Well the Rarita-Schwinger field has a spinor index and a vector index, and they are not usually combined in a single matrix. However, (not having looked at the paper, my connection is quite bad at the moment), I think what R+S mean is that the operator is the [tex]\oplus[/tex] sum of the two operators, just like in quantum mechanics we write [tex]J_x = L_x + S_x[/tex], but the two operators on the RHS act on different spaces.

Thus, the Lorentz transformation acting on such a field would be of the form [tex]\Lambda = e^{i \theta^{ij} L_{ij} + i \omega^{ij} S_{ij}[/tex], where [tex]L_{12}[/tex] is a matrix that acts on Lorentz vectors, and [tex]S_{12}[/tex] is a matrix that acts on Lorentz spinors. The notation [tex]X_{12}[/tex] means a generator that rotates between the 1-2 planes, so in this case, the z axis. Likewise, [tex]X_{01}[/tex] would be a boost. Anyway, I hope the meaning is clear.
 
  • #3
I've been adding them inside the exponential as you said, although I did have them all acting together, basically as matrices multiplying. I'll try to keep my various indices separate.

So basically, I'll be doing:
[tex](\delta_\nu^\mu + \theta L_\nu^\mu)(\delta_b^a + (\omega \cdot \Sigma)^a_b)\Psi_{\mu a} = (1+\lambda) \Psi_{\nu b} [/tex].

Whenever I have some free time I'll plug that through and see what comes out
 
Last edited:
  • #4
After doing some quick calculations, this does not allow the +3/2 eigenvalue, at least not when using

[tex]
L = \left[
\begin{array}{ c c c c}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0
\end{array} \right]
[/tex]

[tex]
\omega \cdot \Sigma = \frac{\theta}{2}\left[
\begin{array}{ c c c c}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array} \right]
[/tex]

Am I doing something wrong here, and if not, do you have any other ideas?

Thanks for your help.
 
  • #6
Great posts in that thread, but I still don't see how [6L] or [6R] transform under the Lorentz group. You go over in some detail on how [2L] transforms, but I don't see how I can use that or a similar approach for [6L].

Would you mind expanding on that, or giving me the name of a textbook or reference which would explain the transformations?
 
  • #7
tmc said:
Great posts in that thread, but I still don't see how [6L] or [6R] transform under the Lorentz group. You go over in some detail on how [2L] transforms, but I don't see how I can use that or a similar approach for [6L].

Would you mind expanding on that, or giving me the name of a textbook or reference which would explain the transformations?

1)

On Minkowski space, we identify the irreducible representation [itex](j_{L},j_{R})[/itex] by a spinorial tensor field

[tex]
\left[(2j_{L}+1)(2j_{R}+1)\right] \equiv \Psi_{l_{1}...l_{2j_{L}}\dot{r}_{1}...\dot{r}_{2j_{R}}}
[/tex]

which is symmetric in the indices [itex](l_{1}l_{2}...l_{2j_{L}})[/itex] and [itex](\dot{r}_{1}\dot{r}_{2}...\dot{r}_{2j_{R}})[/itex]; (otherwise it is reducible).

So, for (1,1/2) we have

[tex][6_{R}] \equiv \Psi_{l_{1}l_{2}\dot{r}} = \Psi_{l_{2}l_{1}\dot{r}}[/tex]

and for (1/2,1) we have

[tex][6_{L}] \equiv \Psi_{l\dot{r}_{1}\dot{r}_{2}} = \Psi_{l\dot{r}_{2}\dot{r}_{1}}[/tex]

Under Lorentz transformation [SL(2,C)], we have

[tex]
\Psi^{'}_{st\dot{r}_{1}} = D^{l_{2}}{}_{s} D^{l_{1}}{}_{t} \bar{D}^{\dot{r}}{}_{\dot{r}_{1}} \Psi_{l_{2}l_{1}\dot{r}}
[/tex]

[tex]
\Psi^{'}_{l_{1}\dot{s}\dot{t}} = \bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} D^{l}{}_{l_{1}} \Psi_{l\dot{r}_{1}\dot{r}_{2}}
[/tex]

The form of [itex]D(\bar{D})[/itex] is given in the other thread.
By contracting the above with [itex](\sigma^{\mu})^{st}[/itex] and [itex](\bar{\sigma}^{\mu})^{\dot{s}\dot{t}}[/itex] respectively, we find

[tex]
(\sigma^{\mu})^{st}\Psi^{'}_{st\dot{r}_{1}} = \left( D^{l_{2}}{}_{s} D^{l_{1}}{}_{t}(\sigma^{\mu})^{st}\right) \bar{D}^{\dot{r}}{}_{\dot{r}_{1}}\Psi_{l_{2}l_{1}\dot{r}}
[/tex]

[tex]
(\bar{\sigma}^{\mu})^{\dot{s}\dot{t}} \Psi^{'}_{l_{1}\dot{s}\dot{t}} = \left( \bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} (\bar{\sigma})^{\dot{s}\dot{t}}\right) D^{l}{}_{l_{1}} \Psi_{l\dot{r}_{1}\dot{r}_{2}}
[/tex]

One can show that [itex]\sigma^{\mu}[/itex] behaves on the one hand as a rank-2 spinor, on the other as a 4-dimensional Lorentz vector, i.e.,

[tex]
D^{l_{1}}{}_{s} D^{l_{2}}{}_{t} (\sigma^{\mu})^{st} = \Lambda(D)^{\mu}{}_{\nu} (\sigma^{\nu})^{l_{1}l_{2}}
[/tex]

[tex]
\bar{D}^{\dot{r}_{1}}{}_{\dot{s}} \bar{D}^{\dot{r}_{2}}{}_{\dot{t}} (\bar{\sigma}^{\mu})^{\dot{s}\dot{t}} = \Lambda(\bar{D})^{\mu}{}_{\nu} (\bar{\sigma}^{\nu})^{\dot{r}_{1}\dot{r}_{2}}
[/tex]

Thus

[tex]
(\Psi^{\mu}_{\dot{s}})^{'} = \bar{D}^{\dot{r}}{}_{\dot{s}} \Lambda(D)^{\mu}{}_{\nu} \Psi^{\nu}_{\dot{r}}
[/tex]

[tex]
(\Psi^{\mu}_{s})^{'} = D^{l}{}_{s} \Lambda(\bar{D})^{\mu}{}_{\nu} \Psi^{\nu}_{l}
[/tex]

where

[tex]
\Psi^{(8)}_{\mu R} \equiv \Psi^{\mu}_{\dot{r}} = \frac{1}{2} (\sigma^{\mu})^{st} \Psi_{st\dot{r}}
[/tex]

[tex]
\Psi^{(8)}_{\mu L} \equiv \Psi^{\mu}_{l} = \frac{1}{2} (\bar{\sigma}^{\mu})^{\dot{r}\dot{s}} \Psi_{l\dot{r}\dot{s}}
[/tex]

The S-R bispinor-vector is given by

[tex]\Psi^{(16)}_{\mu} = \Psi^{(8)}_{\mu L} + \Psi^{(8)}_{\mu R}[/tex]

******
2)

Under Lorentz group, a multi-component field transforms according to

[tex]\Psi^{(16)} \rightarrow \exp ( \frac{i}{2} \omega_{\mu\nu}J^{\mu\nu}) \Psi^{(16)}[/tex]

Now

[tex](J^{\mu\nu})^{(16 \times 16)} \equiv (J^{\mu\nu})^{[4] \otimes ([2] \oplus [2])}[/tex]

can be written as

[tex]J^{[4]}_{\mu\nu} \otimes 1^{[2] \oplus [2]} + 1^{[4]} \otimes J_{\mu\nu}^{[2] \oplus [2]}[/tex]

In components this reads

[tex](J_{\mu\nu})^{\rho a}_{\sigma b} = (J_{\mu\nu})^{\rho}_{\sigma}\delta^{a}_{b} + \delta^{\rho}_{\sigma}(J_{\mu\nu})^{a}_{b}[/tex]

where

[tex](J_{\mu\nu})^{\rho}_{\sigma} = \delta^{\rho}_{\nu} \eta_{\mu\sigma} - \delta^{\rho}_{\mu} \eta_{\nu\sigma}[/tex]

and

[tex](J_{\mu\nu})^{a}_{b} = - \frac{i}{4} [\gamma_{\mu}, \gamma_{\nu}]^{a}_{b}[/tex]

***

3)

You can also use the nice but lengthy formulation of Hayward which leads to 6x6 spin matrix

[tex]\left( \begin{array}{cc} S^{3/2}_{(4 \times 4) & 0_{(4 \times 2)} \\ 0_{(2 \times 2)} & S^{1/2}_{(2 \times 2)} \end{array} \right)[/tex]

See
Haward,R.W.,1976,"The dynamics of fields of higher spins",National Bureau of Standards, Monograph 154, Washington D.C.

On SL(2,C), see

Weinberg,S., in "Lectures on Particles and Field Theory", Brandies Summer Institute, 1964, Vol.2.

Moussa P. & Stora R., in "Methods in Subnuclear Physics", Vol.2, 1968, Gordon and Breach.

Lurie D.,1968, "Particles and Fields", Interscience, N.Y.

Nice and simple treatment can be found in

Jean Hladik,1999,"Spinors in Physics", Springer.
M. Carmeli & S. Malin,2000, "Theory of Spinors", World Scientific.

regards

sam
 

1. What is the purpose of Spin 3/2 Rotation Matrices?

Spin 3/2 Rotation Matrices are used to describe the behavior of particles with spin 3/2, such as baryons and resonances. They are also used in quantum mechanics to represent the rotation of spin 3/2 states.

2. How are Spin 3/2 Rotation Matrices different from Spin 1/2 Rotation Matrices?

Spin 3/2 Rotation Matrices are different from Spin 1/2 Rotation Matrices in that they have a higher dimensionality and more complex mathematical properties. While Spin 1/2 Rotation Matrices can be represented as 2x2 matrices, Spin 3/2 Rotation Matrices are 4x4 matrices.

3. What is the relationship between Spin 3/2 Rotation Matrices and the Pauli matrices?

The Pauli matrices are a set of 2x2 matrices that are used to represent particles with spin 1/2. Spin 3/2 Rotation Matrices can be constructed using the Pauli matrices and other mathematical operations. This relationship allows for a deeper understanding of the properties of spin 3/2 particles.

4. Are Spin 3/2 Rotation Matrices applicable to all particles with spin 3/2?

No, Spin 3/2 Rotation Matrices are only applicable to particles with spin 3/2 that can be described using quantum mechanics. These include baryons and resonances, but not all particles with spin 3/2.

5. Can Spin 3/2 Rotation Matrices be used to describe the rotation of classical objects?

No, Spin 3/2 Rotation Matrices are only applicable to quantum systems. They cannot be used to describe the rotation of classical objects, as classical mechanics and quantum mechanics have different mathematical frameworks.

Similar threads

Replies
7
Views
826
  • Quantum Physics
Replies
1
Views
218
Replies
12
Views
1K
  • Quantum Physics
Replies
3
Views
1K
Replies
6
Views
1K
Replies
1
Views
718
Replies
22
Views
1K
Replies
4
Views
1K
Replies
2
Views
425
Replies
6
Views
1K
Back
Top