Idential particles, postion wavefunction for fermions.

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SUMMARY

The discussion centers on the treatment of identical particles, specifically fermions, in Griffiths' textbook on quantum mechanics. The wavefunction for two identical particles is expressed as ##\Psi(\vec{r_{1}},\vec{r_{2}})=\pm \Psi(\vec{r_{2}},\vec{r_{1}})##, with the sign depending on whether the particles are bosons or fermions. In the case of two electrons in a spin singlet state, the position wavefunction remains unchanged upon particle exchange, while the spin component does change sign. This indicates that Griffiths initially ignores spin when discussing the identity of identical particles, as noted in the first edition of his book.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wavefunctions
  • Familiarity with the concept of identical particles in quantum theory
  • Knowledge of spin states and their implications for fermions
  • Experience with Griffiths' "Introduction to Quantum Mechanics" textbook
NEXT STEPS
  • Review Griffiths' "Introduction to Quantum Mechanics" for detailed explanations on identical particles
  • Study the implications of spin in quantum mechanics, focusing on fermions and bosons
  • Explore the mathematical formulation of wavefunctions for identical particles
  • Investigate the role of spin in quantum entanglement and its effects on particle exchange
USEFUL FOR

Students and educators in quantum mechanics, particularly those studying the behavior of identical fermions and the implications of spin in wavefunctions.

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1. In griffiths the following is written down in the chapter of identical particles:

##\Psi(\vec{r_{1}},\vec{r_{2}})=\pm \Psi(\vec{r_{2}},\vec{r_{2}})##

Where it's + for bosons and - for fermions.

However in class we have seen that for two electrons in the spin singlet situation the POSITION part of their wavefunction doesn't change signs upon swapping the particles. It's the spin part that changes sign in that case.

How to concile these two results? Does Griffiths take spin into account when writing the mentioned identity above? Because at least it looks like he means the position parts of the wavefunctions at first glance.
 
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Check to see if Griffiths is ignoring spin at this point of the text. In the first edition, he says at the beginning of the chapter, "we'll ignore spin for the moment". Quite a bit later in the chapter, he brings in spin.
 
TSny said:
Check to see if Griffiths is ignoring spin at this point of the text. In the first edition, he says at the beginning of the chapter, "we'll ignore spin for the moment". Quite a bit later in the chapter, he brings in spin.

Oh yes, you are entirely correct. I should have searched a bit longer in the book before posting. Thanks.
 

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