# Identical Particles - Silly question

• IHateMayonnaise
In summary, the conversation discusses determining the two-particle wave function and energy states for two non-interacting particles in an infinite cube with different properties (distinguishable, identical bosons, and identical fermions). The conversation also covers the use of spinors and how to distinguish between a singlet and triplet state in the case of identical fermions. The key takeaway is that the total wavefunction must be antisymmetric under particle exchange for fermions.
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:

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

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

## 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.

Last edited:

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.

kuruman said:
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 $\chi^\pm$?

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.

## What are identical particles?

Identical particles are particles that have the same intrinsic properties, such as mass, spin, and charge. These properties are indistinguishable from one another, making it impossible to tell them apart.

## Can identical particles be in the same place at the same time?

Yes, identical particles can occupy the same quantum state and be in the same place at the same time. This is known as the Pauli exclusion principle, which states that no two identical fermions (particles with half-integer spin) can occupy the same quantum state at the same time.

## Do identical particles have the same behavior?

Yes, identical particles have the same behavior because they have the same intrinsic properties. This means they will interact with their surroundings in the same way and have the same probability of being found in a particular location.

## Can identical particles be distinguished by their position or motion?

No, identical particles cannot be distinguished by their position or motion. This is because their properties are indistinguishable, making it impossible to track their individual movements.

## How do identical particles affect the behavior of matter?

Identical particles play a crucial role in determining the behavior of matter. For example, the behavior of gases is affected by the identical particles (molecules) that make up the gas. The Pauli exclusion principle also plays a role in determining the electron configuration of atoms, which ultimately affects the properties of matter.

Replies
26
Views
4K
Replies
3
Views
1K
Replies
9
Views
2K
Replies
1
Views
1K
Replies
9
Views
3K
Replies
16
Views
1K
Replies
3
Views
1K
Replies
1
Views
1K
Replies
7
Views
2K
Replies
0
Views
964