B Questions about Parity: Learn How Intrinsic Parity Relates to State Vector

[hgandh]

A pair of non-interacting particles can be described by the state vector:
\begin{equation}
\Psi_{p_1,\sigma_1,p_2,\sigma_1, t_1, T_1, t_2, T_2}
\end{equation}

Where T is the isospin and t is the 3rd-component. The parity of this state is the product of the intrinsic parities of the two particles. Now, we do a change of basis:
\begin{equation}
\Psi_{p_1,\sigma_1,p_2,\sigma_1, t_1, T_1, t_2, T_2} \Rightarrow \Psi_{E, p, j, \sigma, l, t, s, T}
\end{equation}
Where E is the total energy, p is the total momentum, j is the total angular momentum, l is the orbital angular momentum, sigma is the total angular momentum 3-component, s is the total spin, and t, T are the total isospin. My question is, how would we describe parity in this basis? Obviously the it must equal the product of the intrinsic parities of the two particles. Would the parity in this basis be
\begin{equation}
(-1)^{l}\eta_T
\end{equation}
Where \eta_T is the intrinsic parity corresponding to isospin T? Or is this completely wrong?

 

[PeterDonis]

Obviously the it must equal the product of the intrinsic parities of the two particles.

But in the new basis you introduced, there aren't "two particles"; there is just one quantum system. Its parity is $\eta_T$, the parity of the total isospin of the system.

The fact that the parity of the state of the total system must be the product of the intrinsic parities of the two particles is expressed as $\eta_T = \eta_{T_1} \eta_{T_2}$. But this fact is basis independent, and there's no way to write it down entirely in terms of quantities in a single basis, since in one basis you only have the isospins of the two individual particles, and in the other basis you only have the isospin of the total system.

 

[hgandh]

But in the new basis you introduced, there aren't "two particles"; there is just one quantum system. Its parity is $\eta_T$, the parity of the total isospin of the system.

The fact that the parity of the state of the total system must be the product of the intrinsic parities of the two particles is expressed as $\eta_T = \eta_{T_1} \eta_{T_2}$. But this fact is basis independent, and there's no way to write it down entirely in terms of quantities in a single basis, since in one basis you only have the isospins of the two individual particles, and in the other basis you only have the isospin of the total system.

I think using an example would help me understand better. Let's say we have two non-interacting pions with $T_ 1=T_2=1$ and total parity of $\eta_{1}\eta_{2}=+1$. This state can be expressed as a superposition of states in the new basis as
\begin{equation}
\sum_{j,\sigma,l,t,s,T}\Psi_{E,p,l,j,\sigma,l,t,s,T}
\end{equation}
Where $T= 0, 1, 2$. The parity of these states, $\eta_T$ must be +1 right? So does this restrict the allowable values of the orbital angular momentum for a given T?

 

[PeterDonis]

The parity of these states,$\eta_T$ must be +1 right?

The parity of the superposition as a whole must be $+ 1$, since it's the same overall state in a different basis. That does not necessarily mean that the parity of each individual term in the superposition must be $+ 1$.