Is \(|p,\lambda\rangle\) an Eigenstate of the Helicity Operator?

Click For Summary
SUMMARY

The discussion focuses on whether the state \(|p,\lambda\rangle\) is an eigenstate of the helicity operator \(\frac{\vec{p}}{|\vec{p}|}\cdot\vec{J}\) for massless particles. The state is defined using the Lorentz transformation \(L(p)\) applied to the reference state \(|p_R,\lambda\rangle\), which is an eigenstate of the operator \(J_3\). The key to solving the problem lies in utilizing the properties of unitary transformations and the commutation relationships of angular momentum operators. The suggested approach emphasizes transforming the problem back to the reference vector state to simplify the eigenvalue question.

PREREQUISITES
  • Understanding of Lorentz transformations and their representation in quantum mechanics
  • Familiarity with angular momentum operators and their commutation relations
  • Knowledge of eigenstates and eigenvalues in quantum mechanics
  • Basic proficiency in manipulating exponential operator forms
NEXT STEPS
  • Study the properties of unitary transformations in quantum mechanics
  • Learn about the Lorentz group and its representations
  • Explore the algebra of angular momentum operators in quantum mechanics
  • Investigate the method of transforming problems in quantum mechanics to simplify calculations
USEFUL FOR

Quantum physicists, students studying quantum mechanics, and researchers working on particle physics or angular momentum theory will benefit from this discussion.

CharlieCW
Messages
53
Reaction score
5

Homework Statement



For massless particles, we can take as reference the vector ##p^{\mu}_R=(1,0,0,1)## and note that any vector ##p## can be written as ##p^{\mu}=L(p)^{\mu}_{\nu}p^{\nu}_R##, where ##L(p)## is the Lorentz transform of the form

$$L(p)=exp(i\phi J^{(21)})exp(i\theta J^{(13)})exp(i\alpha J^{(30)})$$

Where ##(\theta,\phi)## are the spherical coordinates of ##\vec{p}## and ##\alpha=sinh^{-1}(\frac{1}{2}(p^0-1/p^0))##. This allows to define the general state for the massless particle as:

$$|p,\lambda\rangle=U(L(p))|p_R,\lambda\rangle$$

Where ##|p_R,\lambda\rangle## is an eigenstate with value ##\lambda## of the operator ##J_3##. Show that ##|p,\lambda\rangle## is an eigenstate of the helicity operator ##\frac{\vec{p}}{|\vec{p}|}\cdot\vec{J}##.

Homework Equations



$$J_3|p_R,\lambda\rangle=\lambda|p_R,\lambda\rangle$$

$$\vec{p}=|\vec{p}|(sin\theta cos\phi, sin\theta sin\phi, cos\theta )$$

$$U(\Lambda_a)U(\Lambda_b)=U(\Lambda_a \Lambda_b)$$

The Attempt at a Solution



For the last week, I've been trying to verify this last statement by expanding the exponentials or using commutators. For example, by using the commutation relationship

$$[J_i,J_k]=i\epsilon_{ijk}J_k$$

But I only end with non-reducible expressions. I also tried expanding the exponentials of the operators using the relationship

$$e^{A}=1+A+\frac{1}{2}A^2+\frac{1}{6}A^3+...$$

Without arriving at a result. Particulary, I don't understand how to act using the unitary transformations, as when I even try to start by calculating:

$$|p,\lambda\rangle=U(L(p))|p_R,\lambda\rangle)=U(exp(i\phi J^{(21)})exp(i\theta J^{(13)})exp(i\alpha J^{(30)}))|p_R,\lambda\rangle$$

Or even the direct calculation:

$$(\frac{\vec{p}}{|\vec{p}|}\cdot\vec{J})|p_R,\lambda\rangle=(\frac{\vec{p}}{|\vec{p}|}\cdot\vec{J})U(L(p))|p_R,\lambda\rangle)$$

I don't know how to reduce terms. Do you have any suggestions?
 

Attachments

  • upload_2018-9-5_14-2-33.png
    upload_2018-9-5_14-2-33.png
    67.5 KB · Views: 535
Physics news on Phys.org
I think your approach is going to need to be a bit more algebraic. Rather than trying to expand those exponential representation operators, you'll want to map the question back to a question about the reference vector, and take advantage of the fact that we're talking here about the Lorentz group (with inverses and everything!).

Given:
|p,\lambda\rangle=U(L(p))|p_R,\lambda\rangle
then
U(L(p))^{-1}|p,\lambda\rangle=|p_R,\lambda\rangle
now try to frame the eigen-value question you're being asked to the transformed question on \lvert p_R, \lambda\rangle.

This is the basic transform--solve--transform-back method.
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
12K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K