# Writing wave functions with spin of a system of particles

Suppose I have 2 fermions in a potential $V(x)$. Both particles are moving in one dimension: the $x$ axis. Then, neglecting the interaction between the particles, the spatial wave function of the system would be of the form $$\psi_{n_{1}}(x_{1})\psi_{n_{2}}(x_{2})$$

Now, if I'm considering particles with spin 1/2, the notation $\alpha(1)$ indicates that the particle 1 has spin up, and $\beta(2)$ denotes the particle 2 having spin down.

Now, I want to write the complete wave function, a function of the form $$\psi_{n_{1}n_{2}s_{1}s_{2}}(x_{1},x_{2},s_{1},s_{2})=\psi_{n_{1}}(x_{1})\psi_{n_{2}}(x_{2})F(\alpha,\beta)$$ where $F(\alpha,\beta)$ is a function of the spin of the system.

To this end, I have that the only physically possible functions $F(\alpha,\beta)$ are:

Symmetric: $\chi_{\alpha}:=\alpha(1)\alpha(2),\quad\chi_{\beta}:=\beta(1)\beta(2),\quad\chi_{+}:=\frac{1}{\sqrt{2}}\left[\alpha(1)\beta(2)+\alpha(2)\beta(1)\right]$

Antisymmetric: $\chi_{-}:=\frac{1}{\sqrt{2}}\left[\alpha(1)\beta(2)-\alpha(2)\beta(1)\right]$

In order to write down the complete wave function with spin, I understand I have to consider the energy levels. For example the ground state: $\psi_{1}(x_{1})\psi_{1}(x_{2})$.

If $\psi_{1}(x_{1})\psi_{1}(x_{2})$ is symmetric (as I understand it is), then I must multiply this function times the antisymmetric function $\chi_{-}$ (in order to get an antisymmetric wave function, for two fermions).

If $\psi_{1}(x_{1})\psi_{1}(x_{2})$ is antisymmetric (and I understand this is impossible, since the ground state is not degenerate), then I'd have 3 wave functions, obtained by multiplying $\psi_{1}(x_{1})\psi_{1}(x_{2})$ times $\chi_{\alpha}$, $\chi_{\beta}$ and $\chi_{+}$.

Now, for the 1st excited level, say $\psi_{2}(x_{1})\psi_{1}(x_{2})$, my question is, what happens when this function is not symmetric neither antisymmetric?

I mean, I could built a symmetric

$$f_S=\frac{1}{\sqrt{2}}\left[\psi_{2}(x_{1})\psi_{1}(x_{2})+\psi_{2}(x_{2})\psi_{1}(x_{1})\right]$$

or an antisymmetric

$$f_A=\frac{1}{\sqrt{2}}\left[\psi_{2}(x_{1})\psi_{1}(x_{2})-\psi_{2}(x_{2})\psi_{1}(x_{1})\right]$$ wave function. But which one of these must I choose? Or, must I calculate the resultant complete wave functions with both? Then, when I count the states with the 1st excited energy, I'd have 4 instead of 1 or 3.

• I'm not sure if you understand what the terms 'symmetric' and 'antisymmetric' are. the function $\Psi(x_1,x_2) = \psi_1(x_1) \psi_1(x_2)$ is symmetric under exchange of $x_1 \leftrightarrow x_2$ because $\Psi(x_1,x_2) = \Psi(x_2,x_1)$. There is no clause 'if $\psi_1(x_1) \psi_1(x_2)$ is antisymmetric'... because it just isn't antisymmetric, by definition. It has nothing to do with it being impossible since the ground state is not degenerate. Why are there so many upvotes to this qn ?? Commented Jun 27, 2013 at 12:38

If $\psi_1(x_1)\psi_1(x_2)$ is antisymmetric (and I understand this is impossible, since the ground state is not degenerate)

The ground state is degenerate, since both particles have the same $n$ (principal) quantum number and thus the same energy. In general, for $N$ particles, the symmetric and antisymmetric wavefunction may be constructed as \begin{align}\psi_{_S}&\equiv\sqrt{\frac{N_1!\cdots{N}_k!}{N!}}\sum_P\hat{P}\,\phi_{n_1}(\zeta_1)\phi_{n_2}(\zeta_2)\ldots\phi_{n_N}(\zeta_N)\\[0.1in]\psi_{_A}&\equiv\sqrt{\frac{N_1!\cdots{N}_k!}{N!}}\begin{vmatrix}\phi_{n_1}(\zeta_1)&\cdots&\phi_{n_1}(\zeta_N)\\\vdots&&\vdots\\\phi_{n_N}(\zeta_1)&\cdots&\phi_{n_N}(\zeta_N)\end{vmatrix}\end{align} respectively, where $\zeta_i$ are the internal degrees of freedom and $N_i$ is the degeneracy of the $i$-th set of degenerated particles (for the antisymmetric part, most usually $N_1!\cdots{N}_k!=1$). In your case (given that you can always write the wavefunction as a product of the spatial and spin parts), $$\psi_{_A}=\begin{vmatrix}\psi_1(x_1)&\psi_1(x_2)\\\psi_1(x_1)&\psi_1(x_2)\end{vmatrix}=0$$ which is why the spatial antisymmetric part is impossible for the ground state. For fermions this is a natural consequence of the Pauli exclusion principle, since you would allow the possibility of two particles being in the same state, given that the spin part would be symmetric.

Now, for the first excited level there is no restriction about considering both the symmetric and antisymmetric parts, in fact you must consider them both. Just as when you must consider the three posibilities from the triplet spin state $$\chi_{_T}=\begin{cases}\chi_\alpha\\\chi_\beta\\\chi_+\end{cases}$$ you may consider both solutions (4 in total, as you say), $$\psi=\begin{cases}\frac{1}{\sqrt{2}}\left[\psi_1(x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1)\right]\chi_-\\\frac{1}{\sqrt{2}}\left[\psi_1(x_1)\psi_2(x_2)-\psi_1(x_2)\psi_2(x_1)\right]\chi_{_T}\end{cases}$$ the thing is that this is a set of possible solutions, just as what you found out for the spin part with the triplet state, the particles may have this or that state, usually when dealing with fermions the only restriction to take care of is Pauli exclusion principle. You may thus consider them all to construct the total wavefunction.

Now, the thing Trimok says,

Note that, mathematically, you can have a total antisymmetric wave function, without having a specific symmetry in the spatial part or in the spin part, for instance : $$\psi_1(x_1)\psi_2(x_2) \alpha(s_1)\beta(s_2) - \psi_2(x_1)\psi_1(x_2) \beta(s_1)\alpha(s_2)$$

could be misleading. This can be seen if you construct the first excited state from the Slater determinant (the general expression for $\psi_{_A}$), say, $n_1=1$, $n_2=2$, $\alpha(1)$, $\beta(2)$, i.e. $$\mathcal{S}_1\equiv\frac{1}{\sqrt{2}}\begin{vmatrix}\psi_1(x_1)\alpha(1)&\psi_1(x_2)\alpha(2)\\\psi_2(x_1)\beta(1)&\psi_2(x_2)\beta(2)\end{vmatrix}$$ which is the expression given by Trimok, but the thing, again, is that you must consider all possible solutions for this state, meaning that for $n_1=1$, $n_2=2$, you can have \begin{align}\alpha(1),&\,\alpha(2)\\\beta(1),&\,\beta(2)\\\beta(1),&\,\alpha(2)\\\alpha(1),&\,\beta(2)\end{align} you can interchange $n_1,\,n_2$ also if you please (there's no new information). For the first, the Slater determinant pops out the antisymmetric spatial solution times $\chi_\alpha$, the second, the antisymmetric spatial part times $\chi_\beta$, but you may take the third, $$\mathcal{S}_2\equiv\frac{1}{2}\begin{vmatrix}\psi_1(x_1)\beta(1)&\psi_1(x_2)\beta(2)\\\psi_2(x_1)\alpha(1)&\psi_2(x_2)\alpha(2)\end{vmatrix}$$ and the fourth to build the antisymmetric part times $\chi_+$ as $\mathcal{S}_1+\mathcal{S}_2$ and to build the symmetric part times $\chi_-$ as $\mathcal{S}_1-\mathcal{S}_2$. Here both must be taken in count because of the indistinguishability of particles, considering $\mathcal{S}_1$ or $\mathcal{S}_2$ alone is just insufficient. As I showed first, all this is taken care of if you just factor the spatial and spin parts of the wavefunction and treat each one apart, as you were doing.

• Trimok is actually right. The wave function does not have to be symmetric/antisymmetric for the spatial/spin parts of the function or the opposite. It does not need to be factorized either. Actually this is only the case for 2 particles when you want the spin part to be an eigenstate of the ${\sf S}^2$ and ${\sf S}_z$ operators. But this is not necessarily the best or unique option always. For instance, in molecules ${\sf S}^2$ is not conserved. Commented Jul 1, 2013 at 8:26
• Yes, I did not say he was wrong; as I show, the expression given by Trimok comes out naturally from the Slater determinant for one possible configuration of the first excited state $\mathcal{S}_1$, which is indistinguishable from $\mathcal{S}_2$. The only thing that need to be antisymmetric is the total wavefunction, but one must consider all possible solutions for a given state, that seems to be the core of the question, and as I said, considering $\mathcal{S}_1$ or $\mathcal{S}_2$ alone is insufficient.
– user24999
Commented Jul 1, 2013 at 16:29

The choice depends on the fermionic physical problem.

Take for instance ferromagnetism.

We want to minimize electrostatic energy between 2 contiguous electrons, and to do that, we will have to maximize their mean distance.

One may show, that due to the Pauli Exclusion principle, the mean distance is greater, when the spatial part of the wave function is antisymmetric (see for instance https://physics.stackexchange.com/a/69267/6316).

But the whole wave function have to be antisymmetric, so if the spatial part of the wave function is antisymmetric, the spin part of the wave function is symmetric.

Practically, in this problem, the spin are all up, or all down. And this is a symmetric configuration for the spin part of the wave function. So this is coherent.

But, for an other physical problem, with fermions,you may have to minimize another kind of energy, which maybe demands a symmetric spatial part of the wave function. If so, you have to choose an antisymmetric spin part for the wave function.

Note that, mathematically, you can have a total antisymmetric wave function, without having a specific symmetry in the spatial part or in the spin part, for instance :

$$\psi_1(x_1)\psi_2(x_2) \alpha(s_1)\beta(s_2) - \psi_2(x_1)\psi_1(x_2) \beta(s_1)\alpha(s_2)$$.

• Separation into space and spin part is possible when $[H, \textbf{S}^2]=0$, and in the simultaneous eigenbasis, space and spin part will have definite symmetry. For example, two electrons in the 1-D box. @Trimok
– SRS
Commented Apr 23, 2018 at 19:49

If your particles are fermions you sould use an antisymmetric wavefunction function to describe them, if they are bosons the wavefunction must be symmetric.

• 0 down vote If your particles are fermions you sould use an antisymmetric wavefunction to describe them, if they are bosons the wavefunction must be symmetric . Commented Jun 27, 2013 at 11:40
• I understand that. My question is, what happens when the spatial part of the wave function is not symmetric neither antisymmetric? I can built a symmetric or an antisymmetric wave function, but which one of these two should I use? Commented Jun 27, 2013 at 21:44