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As far as I know, fermions are the particles which exhibit antisymmetric states: $\hat{P}\left|n_1\right>\otimes\left|n_2\right> = -\left|n_2\right>\otimes\left|n_1\right>$. Often times, we decompose the state $\left|n_1\right>\otimes\left|n_2\right>$ into its spatial and spin parts: $$\psi(x_1,x_2) \chi_{1,2} = \Psi = \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \left< x_1,m_{s1} |n_1\right>\otimes\left< x_2,m_{s2} |n_2\right> $$ (see Kardar eq. 7.31) since then we can conclude that, if $\psi$ is symmetric, then $\chi$ must be antisymmetric and vice-versa.

What prohibits us from having neither $\psi$ nor $\chi$ symmetric/antisymmetric. For example, I believe that the following would maintain our desired antisymmetry of $\left|n_1\right>\otimes\left|n_2\right>$: $$ \psi(x_1,x_2)=i\,\psi(x_2,x_1) \quad \text{and} \quad \chi_{1,2}=i\, \chi_{2,1}$$ since then $$ \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \bigg(\left< x_1,m_{s1} \right|\otimes\left< x_2,m_{s2}\right|\bigg) \hat{P}\bigg(\left| n_1\right> \otimes\left|n_2\right>\bigg) = \sum_{m_{s1}}\sum_{m_{s2}} C(m_{s1},m_{s2}) \left< x_1,m_{s1} |n_2\right>\otimes\left< x_2,m_{s2} |n_1\right> = \psi(x_2,x_1)\chi_{2,1} = -\psi(x_1,x_2)\chi_{1,2} $$ as desired.

Is this just allowed mathematically but doesn't occur in reality? Is there a process prohibiting it? Did I make a mistake?

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  • $\begingroup$ can you write down a function of 2 variables, $f(x,y)$, that satisfies $f(y,x) = if(x,y)$? I have been unable to get past $f(x,y)=0$. $\endgroup$
    – JEB
    Nov 3, 2020 at 20:15
  • $\begingroup$ I believe that I can give you the general form of such an $f$! If we want $f(x,y)=if(y,x)$, then all of the information in the function occurs in the half plane $y<x$ (or one could choose $y>x$); the other half plane will be given by symmetry. If $\theta(x)$ is the standard Heaviside theta function, then $1-\theta(y-x)$ is equal to $1$ for $y<x$ and $0$ otherwise. Similarly, $1-\theta(x-y)$ is equal to $1$ for $y>x$ and $0$ otherwise. Thus, we can construct $f$ from any function $g(x,y)$ as $f(x,y)=[(1-\theta(y-x))+i(\theta(x-y)]g(x,y)$. $\endgroup$ Nov 3, 2020 at 21:30

1 Answer 1

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The permutation $P_{12}$ satisfies $P_{12}^2=\hat{\mathbb{I}}$ and your transformation does not satisfy this since it multiples the spatial or the spin part (which are separate objects) by $i$. In other words you have $P_{12}^2\psi(x_1,x_2)=-\psi(x_1,x_2)$ which is not possible for a transposition.

A more sophisticated answer is that there are only irreducible representations of $S_2$; one multiplies by $-1$ the other by $+1$; note that repeating the permutation twice then gives the identity, as expected.

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  • $\begingroup$ How is it proper to apply $\hat{P}$ to $\psi(x_1,x_2)$? I (implicitly) defined $\hat{P}$ as an endomorphism of $\mathcal{H}_1\otimes\mathcal{H}_2$ where $\mathcal{H}_1$ is the vector space for particle $1$ and $\mathcal{H}_2$ is the vector space for particle $2$ (the vector spaces can be identical). I do not believe that function $\psi$ lives in $\mathcal{H}_1\otimes\mathcal{H}_2$ so I don't see how I can apply $\hat{P}$ to it. $\endgroup$ Nov 3, 2020 at 18:18
  • $\begingroup$ Edit to above comment: I misspoke. Not only can $\mathcal{H}_1=\mathcal{H}_2$, but I believe that we must have that. $\endgroup$ Nov 3, 2020 at 18:42
  • $\begingroup$ If you’re thinking of endomorphism this is complicated. Of course you need the 2 copies of the Hilbert spaces to be identical else the particles cannot be identical. $\psi(x_1,x_2)$ is already a tensor product space so $P_{12}[\phi_a(x_1)\otimes \phi_b(x_2)]= \phi_a(x_2)\otimes \phi_b(x_1)$. $\endgroup$ Nov 3, 2020 at 21:39
  • $\begingroup$ Yes - I mis-implied that $\mathcal{H}_1$ could be different than $\mathcal{H}_2$. I do not understand what you mean by $\psi(x_1,x_2)$ is already a tensor product space, however. As far as I know, $\psi$ is just a function of two real variables $x_1,x_2\in\mathbb{R}^3$ and thus $\psi$ is not an element of $\mathcal{H}_1\otimes\mathcal{H}_1$ and thus not in the domain of $\hat{P}$. While I can see a natural definition of a permutation operator on $\psi$, I do not believe that $\hat{P}$ is said operator. $\endgroup$ Nov 3, 2020 at 21:47
  • $\begingroup$ $\psi(x_1,x_2)$ can be expanded in a basis of product wave functions since the set of all such products is complete; by linearity you may thus work on the product states. Moreover non-interacting particles have product states they serve as basis set. $\endgroup$ Nov 3, 2020 at 22:49

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