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I read in Landau's quantum mechanics(section 62) that if the Hamiltonian does not involve spin, the total wave function can be written $$ \psi ( x_1, x_2, ...) \chi ( \sigma_1 , \sigma_2 ,...)$$ in which coordinate and spin dependence are separated.

Does this mean that for such Hamiltonian for a system of identical particles, there is always a complete set of wavefunctions of form $$ \psi ( x_1, x_2, ...) \chi ( \sigma_1 , \sigma_2 ,...)$$ which are either symmetric or antisymmetric under interchange of labels?

If that is true, why exactly are the spin part and the coordinate part be neither symmetric nor antisymmetric in general, according to Landau(section 63)?

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  • $\begingroup$ If I understand the problem correctly, the reason is that you can put spins in whichever direction you like, but for 1/2-spin particle the spatial part must be antisymmetric to ensure that you don't have the same spins at the same location with the same spin state. $\endgroup$ – Phoenix87 Dec 29 '14 at 13:01
  • $\begingroup$ @Doris - could you clarify the question a little? At present it's not obvious whether your claims about (anti)symmetry apply to both $\chi$ and $\psi$ or whether they apply to either $\chi$ or $\psi$. Also are the particles involved general, or are they bosons/fermions? If you gave a page reference in Landau's book, that might help people answer! $\endgroup$ – Edward Hughes Dec 29 '14 at 13:35
  • $\begingroup$ The question is not formulated clearly. $\endgroup$ – Sofia Dec 29 '14 at 14:04
  • $\begingroup$ @Doris , In the total Hilbert space comprising space variables and spin variables, you can define an orthonormal base. Among the possible bases, you can choose one in which the functions are factorized as you say, with the space part separated from the spin part. Such a factorization holds if the Hamiltonian doesn't comprise terms of interaction between the spins and the space-behavior. $\endgroup$ – Sofia Dec 29 '14 at 14:23
  • $\begingroup$ @Doris (continuation) However, about symmetry or antisymmetry, you have probably a typo mistake. You say "neither symmetric or antisymmetric". No, its "either symmetric or antisymmetric". The total wave-function of a collection of identical fermions has to be anti-symmetrical in any two fermions, i.e. either the space-part, or the spin part, but not both, has to change sign if you interchange them. For identical bosons the wave-function has to be symmetrical at such an interchange. Either both space and spin parts, change sign at the interchange, or both remain unchanged. $\endgroup$ – Sofia Dec 29 '14 at 14:27
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This is actually a very insightful question.

You are correct to say that a general wavefunction for a Hamiltonian without a spin interaction may be written as

$$\psi(x_1,x_2,\dots)\chi(\sigma_1,\sigma_2,\dots)$$

where $\psi$ is the spatial wavefunction and $\chi$ the spin wavefunction.

For a system of bosons, the overall wavefunction must be invariant under all permutations, in other words

$$P (\psi \chi) = \psi \chi$$

where $P$ is the parity operator. We call such overall wavefunctions symmetric.

For a system of fermions, the overall wavefunction must change sign under an odd permutation, in other words

$$ P (\psi \chi) = \textrm{sign}(P)\psi \chi$$

We call such overall wavefunctions antisymmetric.

We must now examine what restrictions these conditions place on the spatial and spin wavefunctions $\psi$ and $\chi$.

Suppose we have a system of 2 identical bosons. Then we must have either

  1. both $\psi$ and $\chi$ are symmetric under interchange
  2. both $\psi$ and $\chi$ are antisymmetric under interchange

In both cases this produces a symmetric overall wavefunction, since $(-1)^2 = 1$.

Now more generally, suppose we have $n$ identical bosons. Then there is more freedom in choosing $\psi$ and $\chi$ such that you get something overall symmetric. If I choose $\psi$ and $\chi$ carefully enough, they don't have to be completely symmetric or antisymmetric themselves!

This comes down to the fact that for several particles, there are more representations of the symmetric group than just the totally symmetric or totally antisymmetric ones. In fact you get one representation for each Young diagram you can draw.

$\psi$ and $\chi$ have definite linear transformations under particle permutation, so must lie in representations of the symmetric group. We just need to know which Young diagrams match up correctly to give the right overall symmetry behaviour.

For bosons, both $\psi$ and $\chi$ must be represented by the same Young diagram in order to get overall symmetry. For fermions, the Young diagrams for $\psi$ and $\chi$ must be related by swapping rows and columns in order to get overall antisymmetry.

Feel free to ask some more questions if you are still confused. I've tried to elucidate what Landau was talking about in section 63 - reread his excellent account and see whether you understand it better now!

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    $\begingroup$ Nice overview. The prime example would be $He^4$ atom at low temperatures, which is just a bose, made out of fermions. By the way there's a nice Feynman chapter on this (identical particles), specially section 4-6 onward. Here's the link. $\endgroup$ – Phonon Jan 6 '15 at 2:02
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Editted: total wave function of Fermion system = anti-symmetric --> either spatial or spin function is anti-symmetric

total wave function of boson system = symmetric --> both symmetric

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  • $\begingroup$ This is too simplistic unfortunately (you are only treating the case of two particles)! $\endgroup$ – Edward Hughes Dec 30 '14 at 23:36
  • $\begingroup$ @EdwardHughes: it's the simple form of course but there is nothing more deeper i think. $\endgroup$ – P.A.M Dec 31 '14 at 10:16
  • $\begingroup$ See my answer - when you have more than 2 particles, the possible representations of the symmetric group are more complicated! $\endgroup$ – Edward Hughes Dec 31 '14 at 12:25

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