Wave function of a system of two identical fermions In N. Zettili's 'Quantum Mechanics Concepts and Applications' [chapter 8, solved problem 8.3], we have to find wave function and ground state energy of a system having two identical fermions and in the solution it is mentioned that the wave function of the system must be anti-symmetric.
But if we have a system of two identical fermions(say electrons), the net spin of the system is then either 1 or 0, hence integral. So the wave function must be symmetric!
Is this right or am I missing something because it is repeatedly written in Zettili in subsequent problems as well?
 A: The total wave function needs to be antisymmetric under particle interchange, e.g.:
$$\Psi(x_1, x_2) = - \Psi(x_2, x_1) $$
where the total wave function is factored into space, spin, and maybe color, and so on:
$$\Psi(x_1, x_2) = \psi(x_1, x_2)\chi_{1, 2}C_{1, 2} $$
where $C=1$ for colorless electrons.
As you point out, the spin can be $S=0$ or $S=1$. Spin zero is antisymmetric:
$$ \chi_{1, 2}=\frac 1{\sqrt 2}(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle) = -\chi_{2, 1} = -[\frac 1{\sqrt 2}(|\downarrow\uparrow\rangle-|\uparrow\downarrow\rangle)]$$
while spin one are the three symmetric combinations, such as:
$$ \chi_{1,2}=|\uparrow\uparrow\rangle=+\chi_{2,1}=+[|\uparrow\uparrow\rangle]$$
For the total wave function to be antisymmetric, the spatial part of the wave function needs to have opposite symmetry, so that if the ground state is spin zero:
$$ \psi(x_1, x_2) = \psi(x_2, x_1) $$
so that:
$$ \Psi(x_1, x_2) = \psi(x_1, x_2)\chi_{1, 2} = [+\psi(x_2, x_1)][-\chi_{2, 1}] = -\Psi(x_2, x_1)$$
Particle interchange means interchanging all coordinates: space, spin, color, flavor, and whatever else may be relevant.
Now the comment that the total spin of the system is 0 or 1 and hence is symmetric is interesting. Yes, the 2 particles form a composite boson, but there is only one boson, and hence, nothing to interchange.
The best example of this is $^4$He. It is obviously a boson, as it forms a condensate--superfluid helium, and indeed, the superfluid wave function is symmetric under interchange of any two helium atoms. Nevertheless, the 2 electrons, 2 protons, and 2 neutrons in each helium atom are antisymmetric under interchange.
A: The book is right; you are wrong. An antisymmetric wave function of two electrons can have spin 0 or spin 1.
The wave function has a spatial part and a spin part. The overall wave function for two fermions must be antisymmetric.
If the spatial part is symmetric and the spin part is antisymmetric, you have total spin 0.
If the spatial part is antisymmetric and the spin part is symmetric, you have total spin 1.
A: A bound state of two fermions is a boson, whether the total spin is $0$ or $1$. The wavefunction of the constituent fermions is still antisymmetric, but the wavefunction of the bosons (regarded as elementary objects) is symmetric. By "wavefunction" here, I mean the whole thing, considering both spatial and spin degrees of freedom. This is clarified below.
Suppose we have four fermions arranged in the form of two integer-spin bound states. The "wavefunction" of that two-boson system must be symmetric with respect to exchange of the two bosons (the two bound states, each regarded as a unit), even though it is still antisymmetric with respect to the interchange of any two of the fermions. This can be seen in a formulation using creation/annihilation operators. Suppose that $a_n^\dagger$ are operators that create fermions, where $n$ labels different possible modes that a fermion can be in — different combinations of momentum and spin-orientation. To simplify the notation, I'll pretend that there is only a discrete list of possible modes. These operators satisfy the anticommutation relations
$$
  a_n^\dagger a_m^\dagger = - a_m^\dagger a_n^\dagger.
\tag{1}
$$
An operator 
$$
  \sum_{n,m}f(n,m)a_n^\dagger a_m^\dagger
\tag{2}
$$
creates a pair of fermions, and we can choose the complex-valued "wavefunction" $f(n,m)$ so that this pair corresponds to a bound state. The anticommutation relation (1) implies that the wavefunction $f(n,m)$ might as well be antisymmetric; the symmetric part doesn't contribute to the sum (2) at all.
Now consider the operators
\begin{align}
  b_j^\dagger\equiv \sum_{n,m}f_j(n,m)a_n^\dagger a_m^\dagger,
\tag{3}
\end{align}
where the index $j$ can be used, say, to label bound states in different locations or with different total momenta. The anticommutation relations (1) imply that these new operators commute with each other:
$$
 b_j^\dagger b_k^\dagger =b_k^\dagger b_j^\dagger.
\tag{4}
$$
Therefore, in an operator that creates two of these objects, such as
$$
  \sum_{j,k}g(j,k)b_j^\dagger b_k^\dagger,
\tag{5}
$$
the "wavefunction" $g(j,k)$ might as well be symmetric, because the commutaiton relation (4) implies that the antisymmetric part does not contribute to the sum (5) at all.
For this reason, a bound state of two fermions is a boson.
