Fermions, Bosons, and nonidentical particles in a 1-d oscillator

In summary, I'm having a hard time understanding how to treat fermions, bosons, and distinguishable particles differently for this problem.
  • #1
MaestroBach
40
3
Homework Statement
Consider two noninteracting particles of mass m in the harmonic oscillator potential well. For
the case with one particle in the single-particle state |n> and the other in state |k> with n not equal to k,
calculate the expectation value of the squared interparticle spacing <(x1-x2)^2>, assuming (a)
the particles are distinguishable, (b) the particles are identical spin-0 bosons, and (c) the particles
are identical spin-1/ 2 fermions in a spin triplet state. Use bra-ket notation as far as you
can, but you will have to do some integrals.
Relevant Equations
N/A
I'm having a hard time understanding how to treat fermions, bosons, and distinguishable particles differently for this problem.

To the best of my understanding, I know that my overall state for bosons must be symmetric, and because they're spin-0, this means there's only one coupled spin state available for them, ie ##\ket{S m_s} = \ket{0 0}##, and since their spin is symmetric then their space representation must also be symmetric, giving me ##\psi_{nk} = \psi_n(x_1)\psi_k(x_2) + \psi_n(x_2)\psi_n(x_1)##, from which I can calculate ##<(x_1-x_2)^2>##.

However, my confusion comes in when I'm dealing with fermions and distinguishable particles.

For fermions, as far as I understand, the spin could be both asymetric or symmetric, but that would change whether my space representation is asymetric or symmetric. How do I decide which to use? This is especially confusing for me, given that the spin does not play a part in my calculation of ##<(x_1-x_2)^2>##, but choosing an asymetric or symmetric spin changes the space representation which to me seems like it would change
##<(x_1-x_2)^2>##, giving me two different possible answers.

As for distinguishable particles, I'm not even sure where to begin...

I appreciate any help!
 
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  • #2
MaestroBach said:
To the best of my understanding, I know that my overall state for bosons must be symmetric, and because they're spin-0, this means there's only one coupled spin state available for them, ie ##\ket{S m_s} = \ket{0 0}##, and since their spin is symmetric then their space representation must also be symmetric, giving me ##\psi_{nk} = \psi_n(x_1)\psi_k(x_2) + \psi_n(x_2)\psi_n(x_1)##, from which I can calculate ##<(x_1-x_2)^2>##.
Sounds good. You might want to include a normalization constant for ##\psi_{nk}##

However, my confusion comes in when I'm dealing with fermions and distinguishable particles.
You are given that the spin state for the fermions is the triplet state.

For the distinguishable particles, you can distinguish the particle in state n from the particle in state k. Let ##x_1## denote the position variable for the particle in state n and let ##x_2## denote the position variable for the particle in state k. How would your express the wavefunction ##\psi_{nk}##?
 
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  • #3
TSny said:
You are given that the spin state for the fermions is the triplet state.

Wow. I can't believe I missed that. Thanks a ton.

TSny said:
For the distinguishable particles, you can distinguish the particle in state n from the particle in state k. Let ##x_1## denote the position variable for the particle in state n and let ##x_2## denote the position variable for the particle in state k. How would your express the wavefunction ##\psi_{nk}##?

So for this, would the wavefunction ##\psi_{nk}## just be ##\psi_{nk} = \psi_n(x_1)\psi_k(x_2)##?

For distinguishable particles I don't even care about spin right? Just that they're distinguishable?
 
  • #4
MaestroBach said:
So for this, would the wavefunction ##\psi_{nk}## just be ##\psi_{nk} = \psi_n(x_1)\psi_k(x_2)##?

For distinguishable particles I don't even care about spin right? Just that they're distinguishable?
Yes, that's right.
 
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  • #5
TSny said:
Yes, that's right.

Really appreciate the help!
 
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Related to Fermions, Bosons, and nonidentical particles in a 1-d oscillator

What are fermions, bosons, and nonidentical particles?

Fermions and bosons are two types of particles that make up the building blocks of matter. Fermions, such as electrons and protons, have half-integer spin and follow the Pauli exclusion principle, meaning that no two fermions can occupy the same quantum state. Bosons, such as photons and gluons, have integer spin and do not follow the exclusion principle. Nonidentical particles refer to particles that are not identical in terms of their mass, charge, or other properties.

What is a 1-d oscillator?

A 1-d oscillator is a theoretical model used in physics to describe the motion of a particle in one dimension. It is often used to study the behavior of particles in a potential energy well, such as an atom or a molecule.

How do fermions and bosons behave in a 1-d oscillator?

In a 1-d oscillator, fermions and bosons behave differently due to their different spin properties. Fermions follow the Pauli exclusion principle and cannot occupy the same quantum state, while bosons do not follow this rule and can occupy the same state. This leads to different energy levels and behavior for fermions and bosons in a 1-d oscillator.

What is the significance of studying fermions, bosons, and nonidentical particles in a 1-d oscillator?

Studying these particles in a 1-d oscillator allows us to better understand their behavior and interactions in different physical systems. This knowledge is crucial in fields such as quantum mechanics, condensed matter physics, and atomic and molecular physics.

What are some real-world applications of studying fermions, bosons, and nonidentical particles in a 1-d oscillator?

One potential application is in developing new materials with unique properties, such as superconductors, which rely on the behavior of fermions in a 1-d oscillator. Additionally, understanding the behavior of particles in a 1-d oscillator can also aid in the development of new technologies, such as quantum computers and sensors.

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