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Idential particles, postion wavefunction for fermions.

  1. Jan 24, 2015 #1
    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.
     
  2. jcsd
  3. Jan 24, 2015 #2

    TSny

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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.
     
  4. Jan 24, 2015 #3
    Oh yes, you are entirely correct. I should have searched a bit longer in the book before posting. Thanks.
     
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