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So I know that the total exchange operator is the product of the coordinate, spin and isospin exchange operators, as,

\begin{equation} P_{12} = P_{12}^r P_{12}^\sigma P_{12}^\tau \end{equation}

The coordinate $P_{12}^r$, spin $P_{12}^\sigma$ and isospin $P_{12}^\tau$ exchange operators are introduced according to,

\begin{align} P_{12}^r |r_1 \sigma_1 q_1, r_2 \sigma_2 q_2 \rangle = |r_2 \sigma_1 q_1, r_1 \sigma_2 q_2 \rangle \\ P_{12}^\sigma |r_1 \sigma_1 q_1, r_2 \sigma_2 q_2 \rangle = |r_1 \sigma_2 q_1, r_2 \sigma_1 q_2 \rangle \\ P_{12}^\tau |r_1 \sigma_1 q_1, r_2 \sigma_2 q_2 \rangle = |r_1 \sigma_1 q_2, r_2 \sigma_2 q_1 \rangle \end{align}

Furthermore, the spin and isospin exchange operators can be expressed as,

\begin{align} P_{12}^\sigma = \frac{1}{2} (1 + \sigma_1 \cdot \sigma_2 )\\ P_{12}^\tau = \frac{1}{2} (1 + \tau_1 \cdot \tau_2) \end{align}

However, I don't know the expression of the coordinate exchange operator. How can it be expressed?

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    $\begingroup$ The expression will never be as simple, because it acts on an infinite dimensional space. $\endgroup$
    – Javier
    Commented Nov 25, 2021 at 16:07

1 Answer 1

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Just a formal wisecrack: you know how to Lagrange-shift functions, $$ e^{a\partial_x + b\partial_y} f(x,y)=f(x+a,y+b), $$ a mere rewriting of the Taylor expansion.

You may further perform two evaluations in an important order, $$ f(x+a,y+b) \Big|^{x=0}_{y=0}= f(a,b), $$ and $$ f(a,b) \Big|^{a=y}_{b=x}= f(y,x), $$ so that $$ P^r_{12} f(x,y)= \left( \left((e^{a\partial_x + b\partial_y} f(x,y) ) \Big|^{x=0}_{y=0} \right ) \Bigg|^{a=y}_{b=x} \right) =f(y,x) $$ so, hyper-formally, $$ P^r_{12}=\Big|^{a=y}_{b=x} \circ \Big|^{x=0}_{y=0} \circ e^{a\partial_x + b\partial_y} . $$ This is an infinite-dimensional representation, so its exchange operator is predictably messier than for the other two finite-dimensional representations you exchanged. Not clear to me what you might do with it...

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