Identical Particles - Silly question

[IHateMayonnaise]

Identical Particles -- Silly question

Homework Statement

Reviewing for final, can someone check this really quick?

Two non-interacting particles are in an infinite cube, each side of length L. Determine the two-particle wave function and also the energy of the ground state and the first excited state:

a) The particles are distinguishable

b) The particles are identical bosons of spin 0

c) The particles are identical fermions of spin 1/2 (also: identify the singlet and triplet states)

Homework Equations

For a 3-D infinite potential box:

$$\psi(x,y,z)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_x \pi x}{L}\right)Sin\left(\frac{n_y \pi y}{L}\right)Sin\left(\frac{n_z \pi z}{L}\right)$$

$$E_{n_xn_yn_z}=\frac{\hbar^2 \pi^2}{2 m L^2} (n_x^2+n_y^2+n_z^2)$$

And also:

$$\psi(\vec{r_1},\vec{r_2})_{\pm}=A[\psi_a(\vec{r_1})\psi_b(\vec{r_2})\pm\psi_a(\vec{r_2})\psi_b(\vec{r_1})]$$

And don't forget about spin:

$$\psi(x)=\psi(x)\chi^{\pm}$$

The Attempt at a Solution

a) For distinguishable

Ground state:
$$E_{111}=\frac{3\hbar^2 \pi^2}{2 m L^2}$$

First excited state (degeneracy exists):
$$E_{112}=E_{121}=E_{211}=\frac{3\hbar^2 \pi^2}{ m L^2}$$

$$\psi(\vec{r_1},\vec{r_2})=\Psi_a(x_1,y_1,z_1)\Psi_b(x_2,y_2,z_2)$$

Where
$$\psi_a(x_1,y_1,z_1)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_1} \pi x_1}{L}\right)Sin\left(\frac{n_{y_1} \pi y_1}{L}\right)Sin\left(\frac{n_{z_1} \pi z_1}{L}\right)$$

$$\psi_b(x_2,y_2,z_2)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_2} \pi x_2}{L}\right)Sin\left(\frac{n_{y_2} \pi y_2}{L}\right)Sin\left(\frac{n_{z_2} \pi z_2}{L}\right)$$

b)For identical bosons of spin 0

Ground state:
$$E_{111}=\frac{3\hbar^2 \pi^2}{2 m L^2}$$

First excited state (no degeneracy)
$$E_{112}=\frac{3\hbar^2 \pi^2}{ m L^2}$$

$$\psi(\vec{r_1},\vec{r_2})_{+}=A[\psi_a(x_1,y_1,z_1)\psi_b(x_2,y_2,z_2)+\psi_a(x_2,y_2,z_2)\psi_b(x_1,y_1,z_1)]$$

Where

$$\psi_a(x_1,y_1,z_1)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_1} \pi x_1}{L}\right)Sin\left(\frac{n_{y_1} \pi y_1}{L}\right)Sin\left(\frac{n_{z_1} \pi z_1}{L}\right)$$

$$\psi_b(x_2,y_2,z_2)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_2} \pi x_2}{L}\right)Sin\left(\frac{n_{y_2} \pi y_2}{L}\right)Sin\left(\frac{n_{z_2} \pi z_2}{L}\right)$$

$$\psi_a(x_2,y_2,z_2)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_1} \pi x_2}{L}\right)Sin\left(\frac{n_{y_1} \pi y_2}{L}\right)Sin\left(\frac{n_{z_1} \pi z_2}{L}\right)$$

$$\psi_b(x_1,y_1,z_1)=\left(\frac{2}{L}\right)^{\frac{2}{3}}Sin\left(\frac{n_{x_2} \pi x_1}{L}\right)Sin\left(\frac{n_{y_2} \pi y_1}{L}\right)Sin\left(\frac{n_{z_2} \pi z_1}{L}\right)$$

c) Identical fermions of spin 1/2

Ground state:
$$E_{112}=\frac{3\hbar^2 \pi^2}{m L^2}$$

First excited state:
$$E_{122}=\frac{9\hbar^2 \pi^2}{2 m L^2}$$

$$\psi(\vec{r_1},\vec{r_2})_{-}=A[\psi_a(x_1,y_1,z_1)\chi^{+}\psi_b(x_2,y_2,z_2)\chi^{+}-\psi_a(x_2,y_2,z_2)\chi^{+}\psi_b(x_1,y_1,z_1)\chi^{+}]$$

Wave functions are the same as in part b).

How do I distinguish between a singlet and triplet state? I know singlet is S=0, and triplet is S=1, but not quite sure what it wants.

 

[kuruman]

You just can't multiply the spatial wavefunctions arbitrarily by χ⁺ like that. The spinors themselves must have "1" and "2" subscripts.

It is easier to write the total fermion wavefunction as the product of two separate wavefunctions, one for the spatial part and one for the spin part:

ψ = (spatial)*(spin)

Because we are talking about fermions, the total wavefunction must be antisymmetric under particle exchange. This means the if the spatial part is symmetric, the spin part must be antisymmetric and vice-versa.

I note that the spin triplet state is symmetric under particle exchange while the spin singlet state is antisymmetric.

 

[IHateMayonnaise]

You just can't multiply the spatial wavefunctions arbitrarily by χ⁺ like that. The spinors themselves must have "1" and "2" subscripts.

It is easier to write the total fermion wavefunction as the product of two separate wavefunctions, one for the spatial part and one for the spin part:

ψ = (spatial)*(spin)

Because we are talking about fermions, the total wavefunction must be antisymmetric under particle exchange. This means the if the spatial part is symmetric, the spin part must be antisymmetric and vice-versa.

I note that the spin triplet state is symmetric under particle exchange while the spin singlet state is antisymmetric.

By "1" and "2" subscripts do you mean $\chi^\pm$?

 

[kuruman]

No. In your notation, the spin singlet state is

$$|0,0>=\frac{\chi^{+}_{1}\chi^{-}_{2}-\chi^{-}_{1}\chi^{+}_{2}}{\sqrt{2}}$$

and the spin triplet has states

$$|1,1>=\chi^{+}_{1}\chi^{+}_{2}$$

$$|1,0>=\frac{\chi^{+}_{1}\chi^{-}_{2}+\chi^{-}_{1}\chi^{+}_{2}}{\sqrt{2}}$$

$$|1,-1>=\chi^{-}_{1}\chi^{-}_{2}$$

Check out their properties under particle exchange. You may also wish to look them up in wikipedia.