# Question about the helicity operator

• jdstokes
In summary, the conversation discusses the use of Dirac spinors in relativistic quantum field theory and how they can be chosen to be eigenstates of the helicity operator. The speaker wants to show that the helicity operator commutes with the Dirac Hamiltonian if the direction of the helicity is the same as the momentum vector. They also mention the use of the angular momentum operator and suggest looking into the Jacob and Wick formalism.
jdstokes
In relativistic quantum field theory the Dirac spinors can be chosen to be eigenstates of the helicity operator $\vec{\Sigma}\cdot \vec{p} /|\vec{p}|$.

I want to show that $\vec{\Sigma}\cdot \vec{a}$ commutes with the Dirac Hamiltonian only if $\vec{a}\propto \vec{p}$. As usual I'm using Einstein summation everywhere.

$[\vec{\Sigma}\cdot \vec{a}, H_\mathrm{Dirac}]\psi=[\vec{\Sigma}\cdot \vec{a}, -i\hbar c \gamma^i\partial_i + mc^2]\psi =-i\hbar c[\Sigma^ia_i, \gamma^i \partial_i]\psi$

$[\Sigma^i a_i, \gamma^l \partial_l]\psi=[\frac{i}{2}\epsilon^{ijk}\gamma_j\gamma_k a_i, \gamma^l \partial_l]\psi= \frac{i}{2}\epsilon_{ijk}[\gamma^j\gamma^k a^i, \gamma^l \partial_l]\psi = \frac{i}{2}\epsilon_{ijk}(\gamma^j\gamma^k a^i \gamma^l \partial_l-\gamma^l \partial_l\gamma^j\gamma^k a^i ]\psi$

since $\epsilon^{ijk} = \epsilon_{ijk}(-1)^3$ and $\gamma_i = -\gamma^i,\, a_i = -a^i$ in flat spacetime with (+,-,-,-) signature.

Now suppose $a^i = \partial^i$. Then using equality of mixed partials and dividing out any constants gives

$\epsilon_{ijk}(\gamma^j\gamma^k\gamma^l \partial_l\partial^i\psi - \gamma^l\gamma^j\gamma^k \partial_l\partial^i \psi)$

Using the relation $\{ \gamma^\mu,\gamma^\nu \} = 2\eta^{\mu\nu}$ twice and the fact that $\eta^{ij}\partial_i = \partial^j$ in flat spacetime gives

$\epsilon_{ijk}(\gamma^k \partial^j - \gamma^j\partial^k)\partial^i \psi$

Now, in the sum over k,j the term in brackets is anti-symmetric but so is $\epsilon_{ijk}$ so I don't see why this should vanish.

Any help would be appreciated.

Last edited:
Woops $\epsilon_{ijk}$ is symmetric in j,k since $\epsilon^{ijk}$ is anti-symmetric.

Edit: Hang on, $\epsilon_{ijk}$ should still be anti-symmetric, just differing in sign from $\epsilon^{ijk}$. I still don't see why this should vanish?

Last edited:
Presumably $$\partial^j \partial k^k$$ is symmetric in $$j,k$$.

Sorry I didn't understand what you wrote. Are you suggesting something along these lines?

$\epsilon_{ijk}\gamma^k\partial^j = -\epsilon_{ikj}\gamma^k\partial^j = -\epsilon_{ijk}\gamma^j\partial^k$.

Therefore

$\epsilon_{ijk}(\gamma^k\partial^j - \gamma^j\partial^k)\partial^i\psi = -4\epsilon_{ijk}\gamma^j\partial^k\partial^i\psi$.

But

$\epsilon_{ijk}$ is anti-symmetric in i,k whereas $\partial^i\partial^k$ is symmetric in i,k. Therefore each term in the sum over j vanishes.

Last edited:
Um, yes... each term actually vanishes separately, the gamma matrix is simply along for the ride.

You'll find it much easier to work directly with standard Dirac solutions in two component form, which are built around the helicity operator. Or look up the angular momentum operator -- see, for example, F. Gross's Relativistic Quantum Mechanics and Field Theory.

Helicity, with a fixed direction of course, commutes with free Hamiltonians for any spin. This is the basis for the elegant Jacob and Wick formalism -- something every particle physicist should master.
Regards,
Reilly Atkinson

## 1. What is the helicity operator?

The helicity operator is a mathematical operator used in quantum mechanics to describe the spin of a particle. It is represented by the Greek letter "lambda" (λ) and measures the projection of the particle's spin onto its direction of motion.

## 2. How is the helicity operator related to angular momentum?

The helicity operator is directly related to the angular momentum of a particle. In quantum mechanics, angular momentum is quantized and can only take on certain discrete values. The helicity operator helps determine the allowed values of angular momentum for a particle.

## 3. Can the helicity of a particle change?

No, the helicity of a particle is a conserved quantity and cannot change. This means that the projection of a particle's spin onto its direction of motion remains constant even as the particle moves and interacts with other particles.

## 4. What are the applications of the helicity operator?

The helicity operator is commonly used in quantum mechanics to study the spin properties of particles. It is also used in particle physics to analyze the behavior of subatomic particles and in nuclear magnetic resonance (NMR) spectroscopy to determine the structure of molecules.

## 5. How is the helicity operator different from the spin operator?

The helicity operator and the spin operator are related but not the same. The spin operator measures the total angular momentum of a particle, while the helicity operator measures the projection of that spin onto a specific direction. Additionally, the spin operator is a vector while the helicity operator is a scalar.

• Quantum Physics
Replies
1
Views
853
• Quantum Physics
Replies
4
Views
999
• Quantum Physics
Replies
6
Views
811
• Quantum Physics
Replies
1
Views
542
• Quantum Physics
Replies
1
Views
615
• Quantum Physics
Replies
56
Views
3K
• Quantum Physics
Replies
10
Views
2K
Replies
20
Views
2K
• Quantum Physics
Replies
1
Views
893
• Quantum Physics
Replies
4
Views
803