Spin-##\frac{1}{2}## particles in infinite square well

[Alex145]

Homework Statement

Construct the four lowest-energy configurations for particles of spin-$\frac{1}{2}$ in the infinite square well, and specify their energies and their degeneracies. Suggestion: use the notation $\psi_{n_1,n_2}(x_1, x_2) |s,m>$. The notation is defined in the textbook.

Homework Equations

$$\psi_{n_1, n_2}(x_1, x_2) = \psi_{n_1}(x_1)\psi_{n_2}(x_2)$$
$$\psi_{n_1, n_2}(x_1, x_2) = \frac{1}{A}(\psi_{n_1}(x_1)\psi_{n_2}(x_2) + \psi_{n_1}(x_2)\psi_{n_2}(x_1))$$
$$\psi_{n_1, n_2}(x_1, x_2) = \frac{1}{A}(\psi_{n_1}(x_1)\psi_{n_2}(x_2) - \psi_{n_1}(x_2)\psi_{n_2}(x_1))$$
$$\psi_n(x) = \sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})$$
$$E = (n_1^2+n_2^2)k$$

The Attempt at a Solution

Ground State:
The position wave function can be symmetric so as to produce the lowest energy (E = 2k). Since the particles are fermions they must then be in anti-symmetric spin states. I imagine then that I would take the second equation in the relevant equations. However, the textbook is very explicit in saying that it is reserved for identical bosons only. In the textbook they ignore spin all together in related questions, so it is a bit confusing. My question is if I am allowed to use the mentioned equation when I take spin into account? I would then have

$$\psi_{1, 1}(x_1, x_2)|0, 0> = \frac{1}{A}(\frac{4}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a}))|0,0>$$

Where A = 2 as determined in the text when the two wave functions are equal.

After I am confident in my answer for the ground state I imagine I would then use anti-symmetric position states and symmetric spin states for the second energy level and switching back for the third.

 

[DrClaude]

The total wave function must have a definite symmetry. For fermions, you can thus have a symmetric spatial wf with an anti-symmetric spin state, or an anti-symmetric spatial wf with a symmetric spin state.

 

[Alex145]

The total wave function must have a definite symmetry. For fermions, you can thus have a symmetric spatial wf with an anti-symmetric spin state, or an anti-symmetric spatial wf with a symmetric spin state.

Thats what I thought. So what I did above is reasonable for the ground state?

 

[DrClaude]

Thats what I thought. So what I did above is reasonable for the ground state?

Apart from the normalisation factor, which is a bit strange, the approach is correct.

 

[Alex145]

Apart from the normalisation factor, which is a bit strange, the approach is correct.

Thank you very much, I'll have a look into the normalization. The spin part of the wave function also has a factor of $\frac{1}{\sqrt{2}}$ so overall it will be reduced by that factor regardless.

 

[DrClaude]

The spin part of the wave function also has a factor of $\frac{1}{\sqrt{2}}$

Are you sure about that?

 

[Alex145]

Well for two spin-$\frac{1}{2}$ particles they can have total spin states, using the notation of the text

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

Where $(\uparrow) = |\frac{1}{2},\frac{1}{2}>$ and $(\downarrow) = |\frac{1}{2},\frac{-1}{2}>$ for single particles

For the ground state, since I am confining the spatial wave functions to be symmetric, I have to choose the last spin state since it's the only one that is anti-symmetric. Unless I am missing something, it carries the factor $\frac{1}{\sqrt{2}}$. The overall wavefunction for the ground state is then

$$\psi_{1,1}(x_1, x_2) = \frac{\sqrt{2}}{a}sin(\frac{\pi x_1}{a})sin(\frac{\pi x_2}{a})(\uparrow \downarrow-\downarrow \uparrow)$$

Which I think is normalized as well.

 

[DrClaude]

Ok. I thought you meant to add a $1/\sqrt{2}$ in front of $| 0,0\rangle$.

 

[Alex145]

Ok. I thought you meant to add a $1/\sqrt{2}$ in front of $| 0,0\rangle$.

Sorry for the confusion. I figured out the overall states. Does including spin affect the total energy of the states? I previously assumed it wouldn't but am now unsure and I can't find anything on that. In the text the energy of the spinless states is just

$$E = k(n_1^2+n_2^2) , k = \frac{\hbar^2 \pi^2}{2 m a^2} $$

 

[DrClaude]

As no details are given, I would assume that spin-spin interactions are to be neglected. And there is no external field. Therefore, I would take the energy to be only the infinite-square-well energy.