# Exclusion Principle for Two $e^-$ in Harmonic Oscillator Ground State

Suppose we have two $e^-$ in a one Dimensional Harmonic Oscillator with total spin $1$. I am looking for the ground state (and I've read this question Ground State Wavefunction of Two Particles in a Harmonic Oscillator Potential ). We know that $$H=\frac{p_{1}^2}{2m}+\frac{1}{2}m\omega^2 x_1^2+\frac{p^2_{2}}{2m}+\frac{1}{2}m\omega^2x_2^2 = H_1 + H_2$$

where $[H_1,H_2]=0$ and we obviously know the solution to these identical hamiltonians giving us eigenvalues $E_{ni} =\hbar ω (n_i+1/2)$.Thus we'll have energy $E_{n_1,n_2}=\hbar ω(n_1 +n_2 +1)$ and we'll have to deal with linear combinations of $$|n_1,n_2,s=1,m_s \rangle = |n_1,n_2 \rangle \otimes |s=1,m_s \rangle$$

(I know how to right the proper antisymmetric state-as seen in the linked question-that's not what I'm interested in here)

Obviously the lowest energy we can have is with $n_1 = n_2 =0$, but will this violate the exclusion principle? Here's what I'm getting confused with:

We say that we have total spin $1$. Adding two $1/2$ spins we get 3 states with total spin $1$ : $$|s=1,m_s =1 \rangle = |+ +\rangle$$

$$|1,0 \rangle = \frac{1}{\sqrt{2}} ( |+-\rangle +|-+\rangle)$$

$$|1,-1\rangle = |--\rangle$$

So saying that we have total spin $1$ means that the spin-part of our states will be a linear combination of the above three states,right? If yes, then the electrons are considered to be in the same spin state or not? If yes then the state with $n_1=n_2=0$ violates the exclusion principle and the ground state is $E_{10}$, right?

• it's not that they are in the same spin state, it's that their state is symmetric under particle interchange---so the spatial wavefunction has to be antisymmetric under particle interchange.
– JEB
Commented May 28, 2018 at 15:20
• I see and we can't make it antisymmetric with $n_1 =n_2 =0$, thanks. Commented May 28, 2018 at 15:45
• @Dimiris Imma make the comment an answer, with an explanation of where I think the confusion arises.
– JEB
Commented May 28, 2018 at 16:48

It's not that they are in the same spin state, it's that their state is symmetric under particle interchange---so the spatial wavefunction has to be antisymmetric under particle interchange.

I think this causes confusion for the following reason: $x_1$ and $x_2$ are different coordinates and the 2 particles are in some external potential. Contrast this to 2 identical particle interacting with each other. I that case, the problem factors into 2 coordinates:

The center of mass motion: $$x_{CM} = x_1 + x_2$$

The distance between the particles: $$x = x_1 - x_2$$

We ignore the COM motion and solve the problem in $x$. Because of this, the spatial parity of $\psi(x)$ becomes a proxy for particle interchange symmetry:

$$-x = -(x_1 -x_2) = (x_2 - x_1) \equiv x_{1\rightarrow 2} - x_{2 \rightarrow 1}$$

The SHO solutions are spatial parity eigenstates (sine and cosine), but in this case, that is not relevant, e.g.:

$$f(x_1,x_2) = \sin x_1\sin x_2$$

and

$$g(x_1, x_2) = \cos x_1\cos x_2$$

are composed of opposite spatial parity functions, but have the same parity (+) under interchange.

• Hi JEB, I have difficulty following the explanation. Could you fill it out with some more explanation? In particular, using $x_1$ and $x_2$ in the same dimension confuses me. Typically, these are said to be separate dimensions in configuration space. Commented Jun 28 at 10:07
• good question. It would be better to say electron 1 is in $x$ and electron 2 is in $y$, then a 2 particle wave function $\psi(x, y) = f(x)g(y) \pm g(x)f(y)$ would be totally antisymmetric wave functions for spin singlet (triplet), for arbitrary $f$ and $g$.
– JEB
Commented Jun 29 at 5:07