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?

  • $\begingroup$ 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
    Commented 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 $f(x,y)=[(1-\theta(y-x))+i(\theta(x-y)]g(x,y)$. $\endgroup$ Commented Nov 3, 2020 at 21:30

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ Commented 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. $\endgroup$ Commented 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 $P_{12}[\phi_a(x_1)\otimes \phi_b(x_2)]= \phi_a(x_2)\otimes \phi_b(x_1)$. $\endgroup$ Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented Nov 3, 2020 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.