# 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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## 1 Answer

The neutron and proton may be viewed as the same particle - sometimes referred to as the nucleon. A proton is a "nucleon with the isospin up" and the neutron is a "nucleon with the isospin down". With this qualification, it's still true that nucleons are identical fermions, so their total wave function has to be antisymmetric.

In the simplest Ansatz, the total wave function is the tensor product of the isospin wave function; spin wave function; and orbital (spatial) wave function. The odd number of those has to be antisymmetric for the result to be antisymmetric. The (anti)symmetry of each factor is governed by the total isospin (0 antisymmetric or 1 symmetric); total spin (0 antisymmetric or 1 symmetric); orbital angular momentum (even means symmetric and odd means antisymmetric).

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So because nucleons are fermions their wavefunction has to be antisymmetric. But the deuteron is a boson, what is the argument for the antisymmetry of the total wavefunction? – liberias Feb 7 '11 at 20:47
Dear liberias, if there is only 1 particle - such as 1 boson called the deuteron - the condition of symmetry or antisymmetry is absolutely vacuous. The symmetry or antisymmetry is a nontrivial condition only because it says that the wave function remains the same or changes the sign if you exchange two particles of the same kind. To do so, you need at least two particles of the same kind. You don't have any two bosons, so it's nonsense to talk about the relationship of the bosonic character and the (anti)symmetry. We're exchanging fermions, so the relevant statistics is the Fermi statistics. – Luboš Motl Feb 7 '11 at 21:06
Can one say that this exchange of coordinates of two same particles is actually the same as spatial inversion of the relative coordinate r=r_1-r_2 in case of symmetric potential? I assume that if this holds, the parity of the deuteron wave function would be negative, which is of course not true. – liberias Feb 7 '11 at 21:19
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