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Question:

Two identical, non-interacting spin-$1/2$ particles are in a 1D Harmonic Oscillator Potential. Their Hamiltonian is given by

$$H=\frac{p_{1x}^2}{2m}+\frac{1}{2}m\omega^2 x_1^2+\frac{p^2_{2x}}{2m}+\frac{1}{2}m\omega^2x_2^2 $$

What is the ground state wave function for the two particles in the singlet and triplet states; i.e., when $S=0$ and $S=1$, respectively.

Attempt: I believe that, for non-interacting indistinguishable particles, we have

$$\psi=\frac{1}{\sqrt{2}}\left \{\psi_1 (x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1)\right \}$$

As well, the ground state of a single particle in a 1D Harmonic Oscillator Potential is

$$\psi_0(x)=\left ( \frac{m\omega}{\pi \hbar}\right )^{1/4}\exp \left \{-\frac{m\omega}{2\hbar}x^2 \right\}$$

Therefore, would our $\psi$ for the two particle system just be

$$\psi=\frac{2}{\sqrt{2}}\left ( \frac{m\omega}{\pi \hbar}\right )^{1/2} \left (\exp \left \{-\frac{m\omega}{2\hbar}(x_1^2+x_2^2) \right\} \right )$$ I feel like I'm missing something. Also, how do i account for the $S=0$ and $S=1$ cases? I'm very confused how I incorporate them into my general case above, which does not consider spin. Any help would be appreciated.

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From the expression of the Hamiltonian you can see that $\hat{H}=\hat{H}_{1}+\hat{H}_{2}$. Also you have the following commutation relation $[\hat{H}_{1},\hat{H}_{2}]=0$. Thus, you can label the eigenstates as $|n_{1}n_{2},SM_{s}\rangle$. Where $|SM_{s}\rangle$ is the two particle total spin state. Now, the ket corresponding to the ground state is $|n_{1}n_{2},SM_{s}\rangle=|00,00\rangle$. To see where the spin comes into play, we look at the first excited state. The first excited energy is $E_{1}=E(10)=E(01)$ with four possible kets:

For the singlet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|00\rangle$

For the triplet state $\left(\frac{1}{\sqrt{2}}|10\rangle+\frac{1}{\sqrt{2}}|01\rangle\right)|S=1,M_{s}=0,\pm1\rangle$

Having fermions, the antisymmetric wave function is

$$\psi=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$

(there's a plus in your wave function and that is for integer spin particles). This wave function can be split into a spatial and spin part. Being an antisymmetric wave function, when the spatial part is symmetric the spin part is antisymmetric and vice versa.

$$\psi(x_{1},x_{2},M_{1},M_{2})=\left\{ \begin{array}{ll} \psi^{S}(x_{1},x_{2})\chi^{A}(M_{1},M_{2})\\ \psi^{A}(x_{1},x_{2})\chi^{S}(M_{1},M_{2}) \end{array}\right.$$

where the spatial part is

$$\psi^{S}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})+\psi_{1}(x_{2})\psi_{2}(x_{1}))$$

$$\psi^{A}(x_{1},x_{2})=\frac{1}{\sqrt{2}}(\psi_{1}(x_{1})\psi_{2}(x_{2})-\psi_{1}(x_{2})\psi_{2}(x_{1}))$$

and the spin part

$$\chi^{A}(M_{1},M_{2})=\frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}-\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=}0,\hspace{5mm} S=0$$

$$\chi^{S}(M_{1},M_{2})=\left\{\begin{array}{ll} \uparrow_{1}\uparrow_{2},\hspace{29mm} M_{s}=1\\ \frac{1}{\sqrt{2}}(\uparrow_{1}\downarrow_{2}+\uparrow_{2}\downarrow_{2}) \hspace{5mm} M_{s=0}\\ \downarrow_{1}\downarrow_{2} \hspace{29mm} M_{s}=-1 \end{array}\right \} , S=1$$

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  • $\begingroup$ Wouldn't the wave function for the ground state then be zero in the case of an antisymmetric spatial part? $\endgroup$
    – Bronzeclocksofbenin
    Commented Nov 20, 2013 at 17:13
  • $\begingroup$ My notation was a little sloppy. I should have wrote for the one particle wave function something like this $\psi_{n}^{(1)}(x_{1})$. Thus, the antisymmetric spatial part would be $\psi_{nn'}^{A}(x_{1},x_{2})\propto(\psi_{n}^{(1)}(x_{1})\psi_{n'}^{(2)}(x_{2})-\psi_{n'}^{(1)}(x_{2})\psi_{n}^{(2)}(x_{1}))$. For $n=n'$ the ground state would be zero, but taking into account Pauli's exclusion principle, you can't have two fermions in the same energy state. $\endgroup$
    – nijankowski
    Commented Nov 20, 2013 at 18:37
  • $\begingroup$ @nijankowski In your formula for $\chi_S$ for the middle case do you mean $\uparrow_1 \downarrow_2 + \uparrow_2 \downarrow_\bf{1}$? $\endgroup$
    – jim
    Commented Mar 11, 2021 at 15:33

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