3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ Wouldn't the wave function for the ground state then be zero in the case of an antisymmetric spatial part? $\endgroup$ – Bronzeclocksofbenin Nov 20 '13 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 Nov 20 '13 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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