# 2 Fermions in a 1D infinite potential well

I had a brief question regarding 2 fermions (with spin) in a 1D infinite potential well (going from $$-L/2$$ to $$+L/2$$, where $$L$$ is the width of the well).

To start off with the question, "what would be the wave function (both space and spin) for the 2 fermions in the ground state?"

Here's what I've got so far:

From the previous lectures in my intro to QM class, I do know how to calculate the wave functions of a 1D infinite potential well. Those being:

$$\psi(x)=\sqrt{2/L}\cos(k(n)x),$$ where $$k(n)=n\pi/L$$ and $$n=1,3,5,...$$

or $$\psi(x)=\sqrt{2/L}\sin(k(n)x),$$ where $$k(n)=n\pi/L$$ and n=2,4,6,... thus, the ground state wave function would be: $$\psi(x)=\sqrt{2/L}\cos(πx/L)$$ However, my main confusion seems to stem from how I would write this (mainly for the space eigenfunctions, as the spin ones make relative sense to me) for the 2-fermion ground-state system.

Per my understanding, the net wave function of the 2-fermion system must be antisymmetic and since they posses the same n-value (since they are in the ground state: n=1), they would have a symmetric space eigenfunction and an antisymmetric spin one (singlet).

However, while in my textbook (Quantum Physics by Robert Eisberg and Robert Resnick) the symmetric space eigenfunction is given as $$\psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1)$$ on the internet, I keep seeing space eigenfunctions given simply as $$\psi(x_1,x_2)=\psi_1(x_1)\psi_2(x_2)$$ (such as here: Fermions in a well)

Why is that so? What am I missing, or where am I going wrong?

I'd greatly appreciate any guidance.

• In the situation of the linked answer, as well as in yours, the two particles are in the same state. Hence the product is already symmetric under exchange of the particles.
– lcv
Nov 24, 2023 at 20:40

So in the well the states are labeled by two quantum numbers: $$(n,\sigma)$$, where $$n$$ is the integer related to the wavenumber and $$\sigma=\uparrow,\downarrow$$ is the quantum number of the projection of the spin along a given axis. States with different $$\sigma$$ but same $$n$$ are degenerate. Moreover, the spin part and the spatial part can be decoupled, namely $$\Psi_{n\sigma}(x) = \psi_n(x) \chi_{\sigma}$$, where for example we can choose $$\chi_{\sigma}$$ as the canonical basis set for the Hilbert space $$\mathbb{C}^2$$ of a single-spin $$\chi_{\uparrow}=(1, 0)^T$$ and $$\chi_{\downarrow} = (0,1)^T$$. Notice that now the Hilbert space of two particles is the tensor product of the two Hilbert spaces, so in particular the spin part is the tensor product $$\mathbb{C}^2 \otimes \mathbb{C}^2$$. Thus we will need to distinguish $$\chi_{\sigma}(1)$$ and $$\chi_{\sigma}(2)$$, the basis sets of the two $$\mathbb{C}^2$$ Hilbert spaces that are "multiplied" via tensor product.
Now the two states that we are interested in are 1: $$(n,\sigma=\uparrow)$$ and 2: $$(n,\sigma=\downarrow)$$, and the elctronic wave function is a Slater determinant with respect to these states: $$\Psi(x_1, x_2) = \frac{1}{\sqrt{2}} \left[ \Psi_1(x_1) \Psi_2(x_2) - \Psi_1(x_2) \Psi_2(x_1) \right]$$ $$= \frac{1}{\sqrt{2}} \left[ \psi_n(x_1) \chi_{\uparrow}(1) \psi_n(x_2) \chi_{\downarrow}(2) - \psi_n(x_2) \chi_{\uparrow}(2) \psi_n(x_1)(1) \chi_{\downarrow} \right]$$ Notice that here the spatial part can be collected: $$\Psi(x_1,x_2) = \psi_n(x_1)\psi_n(x_2) \frac{1}{\sqrt{2}} \left(\chi_{\uparrow}(1)\chi_{\downarrow}(2) - \chi_{\downarrow}(2)\chi_{\uparrow}(1) \right)$$ so the spatial part is in fact just the product of the two $$\psi_n(x_1)\psi_n(x_2)$$ (which is symmetric under exchange $$x_1 \leftrightarrow x_2$$), and the spin part is the singlet.
• The full fermionic wave function $\Psi(x_1, x_2)$ must be antisymmetric with respect to the particle exchange $1 \leftrightarrow 2$, right? But then as you can see in the end the spatial part is in fact symmetric $\psi_n(x_1)\psi_n(x_2)$, and the spin part is antisymmetric (singlet) Apr 27, 2023 at 22:46
• In fact it is the same: you can rewrite $\psi_n(x_1)\psi_n(x_2)$, as ${1/2}(\psi_n(x_1)\psi_n(x_2)+\psi_n(x_2)\psi_n(x_1))$, which is what your book says, and which is maybe more generalizable to situations where the two electrons have different $n$ Apr 27, 2023 at 23:28