# Why must the deuteron wavefunction be antisymmetric?

Wikipedia article on deuterium says this:

The deuteron wavefunction must be antisymmetric if the isospin representation is used (since a proton and a neutron are not identical particles, the wavefunction need not be antisymmetric in general).

I wonder why does the wave function need to be antisymmetric when isospin representation is used. I assume that if two somehow different particles are exchanged the total wavefunction changes sign. Is it so? Why?

Thanks

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Nope, liberias, these two conditions may look similar but they're not the same thing. The spatial inversion is just the spatial part of the exchange of the two nucleons. The full condition is $\psi(i_1,s_1,x_1;i_2,s_2,x_2)=-\psi(i_2,s_2,x_2;i_1,s_1,x_1)$ where $i,s,x$ are all isospin-, spin-, and space-related quantum numbers of the 1st and 2nd particle. If you imagine that the three types of degrees of freedom are independent - not quite true - then $\psi(i_1,s_1,x_1;i_2,s_2,x_2)=\psi_i(i_1,i_2)\psi_s(s_1,s_2)\psi_x(x_1,x_2)$. Each of the three factors is either symmetric or antisymmetric. – Luboš Motl Feb 8 '11 at 7:02
One more comment: if you consider parity, i.e. flipping left hand and right hand (as opposed to a rotation by 180 degrees, for example, that may also exchange them), that would exchange the particles, but it could also change the "internal wave function" by a sign, the eigenvalue of the $P$ operator. It just happens that the electron and positron have the opposite parity, so you get another minus sign from parity. All these minus signs are "totally invisible" to classical physics - because the probabilities are squared amplitudes. But all those signs play a big role in quantum mechanics. – Luboš Motl Feb 8 '11 at 8:46