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.

  • $\begingroup$ 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. $\endgroup$
    – lcv
    Nov 24, 2023 at 20:40

1 Answer 1


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.

  • $\begingroup$ pardon me for asking, but why is your wavefunction expressed as a linear difference of the product of individual wavefunctions, instead of a sum. Isn't the ground state of the system, when both fermions occupy the same spacial quantum state and, thus, must have a singlet spin state? $\endgroup$ Apr 27, 2023 at 22:33
  • $\begingroup$ 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) $\endgroup$
    – Matteo
    Apr 27, 2023 at 22:46
  • $\begingroup$ Thank you for your response! Ah, I do now understand why your final spacial part is symmetric, but what I can't understand is why can't your spacial part (also) be described as such: Ψ(𝑥1,𝑥2)=(√1/2)[Ψ1(𝑥1)Ψ2(𝑥2)+Ψ1(𝑥2)Ψ2(𝑥1)] (I'm mainly wondering that as that is what is listed in my textbook and I am trying to connect that to what you have kindly presented to me!) $\endgroup$ Apr 27, 2023 at 23:08
  • $\begingroup$ 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 $n$ $\endgroup$
    – Matteo
    Apr 27, 2023 at 23:28
  • $\begingroup$ Oh, that makes much more sense actually. Thank you very much for the clarification @Matteo $\endgroup$ Apr 27, 2023 at 23:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.