In Weinberg's textbook on QFT(google book preview), he discussed the phase acquired after interchanging particle labels in the last paragraph of page 171 and the footnote of page 172. It seems he's suggesting interchanging particles of same species but different spin states will only bring a phase determined by convention, that is, the phase does not have to be $\pm1$. I'm having a hard time understanding this. Is Weinberg actually treating particles of same species but different spin states as distinguishable particles? If so can I take it further and conclude particles of same species, same spin states but different momenta are also distinguishable, so that the interchange phase is also conventional?

EDIT: I did a calculation of the commutation relation of $a'_m$ and $a'_n$ as Lubos suggested and verified his conclusion. Recently I did the same calculation again but could not reproduce the result I had, here is my recent calculation:

$a'_ma'_n=(-1)^{N_{m',m'<m}} a_{m'}(-1)^{N_{n',n'<n}}a_{n'}=\pm (-1)^{N_{m',m'<m}}(-1)^{N_{n',n'<n}} a_{m'}a_{n'}$

where $+$ is taken when $m'\neq n'$, $-$ is taken when $m'=n'$. Also

$a'_na'_m=(-1)^{N_{n',n'<n}}a_{n'}(-1)^{N_{m',m'<m}} a_{m'}=\pm (-1)^{N_{m',m'<m}}(-1)^{N_{n',n'<n}} a_{n'}a_{m'}$

again $+$ is taken when $m'\neq n'$, $-$ is taken when $m'=n'$. So $a'_m$ and $a'_n$ should obey the same commutation relations as those of $a_m$ and $a_n$.

From another persepctive, it seems just from Pauli exclusion principle we can deduce the antisymmetry of fermion wavefunctions:

Exclusion principle tells us the square of a fermion creation or annihilation operator must be zero, so $a^2_m=0,\ a^2_n=0,\ (a_m+a_n)^2=0$(assuming no superselection rule on $m,\ n$ so that the 3rd operator is well defined), and we can easily see this implies $a_ma_n+a_na_m=0$.

So, I'm again puzzled by this question.


1 Answer 1


Yes, you may consider different $j_z$ polarizations of a single particle species to be "distinguishable particle species"; and you may even consider different $\vec k$ modes of a field to produce distinguishable particles even at the same $j_z$. However, the formalism needed for such general phases will not be manifestly rotationally symmetric, local, and/or it will be unnatural from various viewpoints.

But you may always redefine annihilation operators e.g. in this way: $$ a_m \to a'_m (-1)^{N_{m',m\lt m}} $$ The labels $m,m'$ refer to values of $j_z$.

Instead of a power of $(-1)$, I could put a power of any phase $\exp(i\alpha)$. The creation operator is still the Hermitian conjugate to the operator above and all operators that are bilinear in $a_m$ (only for the same values of $m$, twice) will be invariant under the addition of the primes.

But with this redefinition, $a'_m$ and $a'_{m'}$ will anticommute if the unprimed $a$'s commuted, and vice versa. It's because for $m'\lt m$, $a'_{m'}$ changes the $N$ in the exponent by $\pm 1$, so it matters how you order it. In other words, $$ (-1)^{N_{m',m\lt m}} a_{m'} = -a_{m'} (-1)^{N_{m',m\lt m}} $$ Note the minus sign on the right hand side. However, operators such as the Hamiltonian typically contain lots of polynomial terms in $a_m$ with mixed values of the indices $m$, and those terms will have to be remapped to include additional explicit powers of $(-1)$ similar to the power of $(-1)$ above.

Similar treatment may be done with the values of momenta $\vec k$; we may need to "order the possible values of $\vec k$ in some way", however.

Let me mention one example of a situation in which the statement "the different polarizations are distinguishable particles, after all" is familiar and uncontroversial. The proton and the neutron are two components of an "isodoublet". They're inherently the same particle, nucleon, in two states $I_z=\pm 1/2$ that are analogous to $j_z=\pm 1/2$. We may consider them to be the same particle or two distinguishable particle species, depending on the perspective.

Of course, because they're fermions, we always prefer the wave function to switch the sign when a proton is exchanged with a neutron. That's guaranteed if they're considered polarizations of an isodoublet but it's more natural even if it is not. However, as I mentioned above, if you don't care about various unnatural powers of $(-1)$ or phases appearing in the Hamiltonian and elsewhere, you may always redefine the operators in such a way that the phase associated with the exchange of two distinguishable particle species – or even two polarizations or momentum modes etc. – is anything you want.

  • $\begingroup$ It's getting late here so I'll read it more carefully tomorrow, before I go to bed can I just ask what is the meaning of your N? $\endgroup$
    – Jia Yiyang
    Nov 22, 2012 at 18:49
  • $\begingroup$ The number of particles in a given state, $a^\dagger a$. $\endgroup$ Nov 23, 2012 at 17:09
  • $\begingroup$ I'm having trouble making sense of your $a_m \to a'_m (-1)^{N_{m',m\lt m}}$, do you actually mean something like $a_m \to \sum\limits_{m'} a_{m'} (-1)^{N_{m',m'\lt m}}$? Also you mentioned "However, the formalism needed for such general phases will not be manifestly rotationally symmetric, local, and/or it will be unnatural from various viewpoints." Could you elaborate this? Any reference will also be appreciated. $\endgroup$
    – Jia Yiyang
    Nov 23, 2012 at 18:04
  • $\begingroup$ Right, the (self-explanatory) $N$ used in the particular exponent was the total $N$ of all excitations with $m'\lt m$, so it's equal to the sum over $m'$ that you wrote. I don't understand what you don't understand about the quoted sentence, sorry. A manifest rotational symmetry is the rotational symmetry of equations or actions that may be easily seen because all the vector indices are "nicely contracted". For example, $\vec p\cdot \vec p$ is manifestly rotationally symmetric. With some field redefinition, the theory may still have the symmetry but it may be obscured by the formulae. $\endgroup$ Nov 24, 2012 at 8:52
  • $\begingroup$ I do understand what it means by "manifestly rotationally symmetric", but I can't see clearly why the chosen convention is particularly nice for it. $\endgroup$
    – Jia Yiyang
    Nov 24, 2012 at 15:57

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