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How to label two particle states using orbital angular momentum?

+ 4 like - 0 dislike
77 views

Weinberg claims in his QFT text(section 3.7):

It is often convenient to work with the S-matrix in a basis of free-particle states in which all variables are discrete, except for the total momentum and energy. This is possible because the components of the momenta $\vec p_1,..,\vec p_n$ in an n-particle state of definite total momentum p and total energy $E$ form a $(3n - 4)$-dimensional compact space; for instance, for $n = 2$ particles in the center-of-mass frame with $\vec p = 0$, this space is a two-dimensional spherical surface. Any function on such a compact space may be expanded in a series of generalized 'partial waves', such as the spherical harmonics that are commonly used in representing functions on the two-sphere. We may thus define a basis for these n-particle states that apart from the continuous variables $p$ and $E$ is discrete: we label the free-particle states in such a basis as $\Phi_{E \vec p N}$, with the index $N$ incorporating all spin and species labels as well as whatever indices are use to label the generalized partial waves.

For two massless particles, these basis states are related to the tensor product basis by the relation:
$$(\Phi_{\vec p_1\sigma_1n_1;\vec p_2\sigma_2n_2},\Phi_{E,\vec p,l,m,\sigma_1',\sigma_2',n_1',n_2'})$$

$$=\sqrt{\frac{E}{|\vec p_1|E_1E_2}}\delta^{(3)}(\vec p-\vec p_1-\vec p_2)\delta(E-\sqrt{m_1^2+\vec p_1^2}-\sqrt{m_2^2+\vec p_2^2})$$

$$Y^{*m}_l(\hat p_1)\delta_{\sigma_1\sigma_1'}\delta_{\sigma_2\sigma_2'}\delta_{n_1n_1'}\delta_{n_2n_2'}$$

My questions are:

1. Can I define $\Phi_{E,\vec p,l,m,\sigma_1',\sigma_2',n_1',n_2'}$ as eigenstates of energy, momentum and some "orbital angular momentum" operator and obtain this relation? If not, then:

2. How do I derive this formula? The text says:

the states may be labelled by their total momentum $\vec p = \vec p_1 + \vec p_2$, the energy $E$, the species labels $n_1, n_2$, the spin z-components $\sigma_1, \sigma_2,$ and a pair of integers $l,m$ (with $|m| < l$) that specify the dependence of the state on the directions of, say, $p_1$. 

How do i formally justify this?

3.  Can it be done in 1-particle states also? Is there a labelling like $\Phi_{E,l,m}$ such that$$(\Phi_{E,l,m,\sigma},\Phi_{\vec p,\sigma'})=\delta(E-\sqrt{M^2+\vec p^2})Y^m_l(\hat p)\delta_{\sigma \sigma'}?$$

4. How does a general Lorentz transformation act on the states labelled this way? The total angular momentum must be eigenvalues of the $J^{ij}$ generators of Lorentz group. So, for single particle states if we club together the "spin" and "orbital" angular momenta using clebsch-gordan coefficients for $SU(2)$ and relabel the states, the group action should be derived using the commutation relations. Am I right? What about the 2-particle states? In particular, I'm interested in finding out how the parity operator $P$ acts on these states because I want to know how "states with orbital angular momentum $l$ has parity eigenvalues $(-1) ^l$" works out for positronium bound states in relativistic quantum mechanics.

asked May 23 in Theoretical Physics by Rajat Mandal (20 points) [ revision history ]
edited May 29 by Rajat Mandal

1 Answer

+ 3 like - 0 dislike

Given a total energy $E$ and a total momentum $\vec{p}$ for a two particles system, the possible momenta of the individual particles are  $\vec{p_1}$ and $\vec{p}- \vec{p_1}$ where $\vec{p_1}$ is a solution of the equation

$E=\sqrt{\vec{p_1}^2+m_1^2}+\sqrt{(\vec{p}-\vec{p_1})^2+m_2^2}$.

The set of solutions of this equation is a compact surface $M$ in $\mathbb{R}^3$. The idea is that rather than to parametrize states by a given value of $\vec{p_1}$, i.e. by a point on $M$, one can parametrize them using any basis of the space of functions on $M$ (not just the "basis" of delta functions). The surface $M$ can be parametrize by a 2-sphere and so a natural basis of functions on $M$ comes from the spherical harmonics $Y_l^m$  on the 2-sphere.

For a one-particle state, there is no analogue of the manifold $M$: total energy and momentum are the same as individual energy and momentum.

answered May 28 by 40227 (4,660 points) [ revision history ]

For a one-particle state, there is no analogue of the manifold $M$: total energy and momentum are the same as individual energy and momentum.

Note that for 1-patticle states, I omitted the total momentum. The state labels are $\Phi_{E,l,m}$, not $\Phi_{E,\vec p,l,m}$. The point precisely is to say that here the analogue of $M$ is the total momentum $\vec p$ which, for a given energy $E$, lies on a 2-sphere because it is constrained by $$\vec p^2=E^2-m^2>0.$$Now can I trade the $\vec p$ label for some discrete variables $(l,m)$? I guess I can, but there is another difficulty. I'll add it as a fourth question.

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