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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.

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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