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Correct understanding of symmetrization requirement?

  1. Jan 23, 2013 #1
    Hi all,

    I want to make sure I have the right understanding of the symmetrization requirement...in particular what is discussed in section 5.1 in Griffiths' Introduction to Quantum Mechanics 2nd edition. When we have a two-electron state, the wave function (described by the product both its position wave function and spinor) has to be antisymmetric with exchange of the electron. Since the singlet spinor function is antisymmetric, it must be combined with a symmetric spatial function in order for it to properly describe a fermion (in this case an electron).

    I am just really confused because Griffiths seems to derive that just the position wave function is antisymmetric for fermions and symmetric for bosons on pages 203-205.

    However, on page 210 he seems to pull out of thin air the fact that the whole wave function has to be antisymmetric for fermions and symmetric for bosons and the position wave function for a fermion can be BOTH symmetric or antisymmetric depending on the corresponding spinor.

    Any help with the clarification would be greatly appreciated.
    Thanks,
    D
     
  2. jcsd
  3. Jan 23, 2013 #2
    of course,it should be the total wave function for fermion to be antisymmetric when pauli principle is taken into account.
     
  4. Jan 23, 2013 #3
    yes. But Griffiths derives Pauli exclusion principle by showing that when the positional wave functions are antisymmetric with exchange, then the wave function cancels when the particle states are the same.
     
  5. Jan 23, 2013 #4
    you will have to multiply the positional wave function by those spin wave function to obtain the total wave function.
     
  6. Jan 23, 2013 #5

    vanhees71

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    For fermions the N-body Hilbert space of indistinguishable particles is constructed by the totally antisymmetrized tensor products of N single-particle states. Let's discuss non-relativistic quantum mechanics, where spin and momentum (or spin and position) are compatible observables.

    An electron has spin 1/2, and a nice single-particle basis is given by the position-spin basis, [itex]|\vec{x},\sigma \rangle[/itex], where [itex]\vec{x} \in \mathbb{R}^3[/itex] and [itex]\sigma \in \{-1/2,+1/2 \}[/itex].

    Now let's look at the Hilbert space for two electrons. It consists of all superpositions of antisymmetrized single-particle states, i.e.,
    [tex]|\Psi \rangle=\frac{1}{\sqrt{2}} (|\psi_1 \rangle \otimes |\psi_2 \rangle - |\psi_2 \rangle \otimes |\psi_1 \rangle).[/tex]
    The wave two-particle wave function is given by
    [tex]\Psi(\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2) = \frac{1}{\sqrt{2}} \left [\psi_1(\vec{x}_1,\sigma_1) \psi_2(\vec{x}_2,\sigma_2)-\psi_2(\vec{x}_1,\sigma_1) \psi_1(\vec{x}_2,\sigma_2) \right].[/tex]
    Here
    [tex]\psi_j(\vec{x},\sigma)=\langle \vec{x},\sigma|\psi_j \rangle[/tex]
    are the single-electron wave functions.

    Of course you can also use a basis with good total spin [itex]S[/itex]. For two spin-1/2 particles the total spin can be either 0 or 1. There is only one state with total spin 0:
    [tex]|\vec{x}_1,\vec{x}_2,S=0,\Sigma=0 \rangle = \frac{1}{\sqrt{2}} (|\vec{x}_1, 1/2 \rangle \otimes |\vec{x}_2,-1/2 \rangle - | \vec{x}_1,-1/2 \rangle \otimes |\vec{x}_2,-1/2 \rangle.[/tex]
    The [itex]S=0[/itex] part of the above antisymmetrized product state is thus
    [tex]\Psi(\vec{x}_1,\vec{x}_2,S=0,\Sigma=0)=\frac{1}{2} \left [\psi_1(\vec{x}_1,1/2) \psi_2(\vec{x}_2,-1/2)-\psi_2(\vec{x}_1,1/2) \psi_1(\vec{x}_2,-1/2) - \psi_1(\vec{x}_1,-1/2) \psi_2(\vec{x}_2,1/2)+\psi_2(\vec{x}_1,-1/2) \psi_1(\vec{x}_2,1/2) \right].[/tex]
    It's obviously antisymmetric, when interchanging the single-particle spins and symmetric, when interchanging the single-particle positions. Alltogether it's still totally antisymmetric.

    The same you can do for the [itex]S=1[/itex] basis,
    [tex]|\vec{x}_1,\vec{x}_2,S=1,\Sigma=1 \rangle=|\vec{x}_1,1/2 \rangle \otimes \vec{x}_2,1/2 \rangle,[/tex]
    [tex]|\vec{x}_1,\vec{x}_2,S=1,\Sigma=0 \rangle = \frac{1}{\sqrt{2}} \left (|\vec{x}_1,1/2 \rangle \otimes |\vec{x}_2,-1/2 \rangle + |\vec{x}_1,-1/2 \rangle \otimes |\vec{x}_2,1/2 \rangle \right ),[/tex]
    [tex]|\vec{x}_1,\vec{x}_2,S=1,\Sigma=-1 \rangle=|\vec{x}_1,-1/2 \rangle \otimes \vec{x}_2,-1/2 \rangle.[/tex]
    The two-particle wave function for all three states is of course symmetric under exchange of the single-particle spins and thus necessarily antisymmetric under exchange of the single-particle positions. In total it's antisymmetric under single-particle-state exchange.

    Any other two-body state is given as a superposition of several such antisymmetrized two-particle product states.
     
  7. Jan 27, 2013 #6
    dracobook, this excerpt from Townsend's excellent QM book should help clarify things.
     
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