Argument about identical particles I am reading Schwartz, Quantum field theory and the standard model, p.207, 12.1 Identical Particles and some question arises (I think that I am beginner for quantum field theory and please understand):



Why the underlined statements are true?

*

*Why $| \cdots s_1\vec{p_1}n \cdots s_2\vec{p_2}n \cdots > = \alpha | \cdots s_2\vec{p_2}n \cdots s_1\vec{p_1}n \cdots >$ for some $\alpha=e^{i\phi}$ is true?


*How can we argue to show the below underlined statement (12.9)?
Can anyone help?
 A: 

*

*Why $| \cdots s_1\vec{p}_1n \cdots s_2\vec{p}_2n \cdots \rangle = \alpha | \cdots s_2\vec{p}_2n \cdots s_1\vec{p}_2n \cdots \rangle$ for some $\alpha=e^{i\phi}$ is true?


As has been mentioned in the text, this is because we have fixed the normalisation of the states. Physically speaking, since the particles are indistinguishable, each state must yield the same probabilities and thus can only differ by a phase.



*How can we argue to show the below underlined statement (12.9)?


What happens when $\vec{p}_1 = \vec{p}_2$ in $\langle \vec{p}_1 | \vec{p}_2 \rangle$? Use this and (12.5) to determine $[a_{\vec{p}_1 s_1 n}, a^{\dagger}_{\vec{p}_2 s_2 n}]$. Start by first seeing how the terms in the commutator act on a general state $|\psi\rangle$ (similar to (12.7) in the text) to arrive at (12.9).
Hope these hints help. Feel free to ask for clarifications in the comments :)
A: First, Schwartz is technically talking about indistinguishable rather than identical particles.  Two electrons are identical but if one is on Mars and the other on Earth they are distinguishable: see this question.
If the particles are indistinguishable, then the labelling of the particles cannot affect the outcome of any measurement.  Thus, interchanging any two particles $(i,j)$ in $\vert\psi\rangle$ will give $\vert\psi'\rangle$ so that
$$
\langle \psi\vert \hat{\cal O}\vert\psi\rangle= \langle \psi'\vert \hat{\cal O}\vert\psi'\rangle
$$
for any operator $\hat{\cal O}$.   The only way for this to work is if $\vert\psi'\rangle=e^{i\phi}\vert\psi\rangle$.
In fact, Kaplan in

Kaplan, Inna G. "The exclusion principle and indistinguishability of identical particles in quantum mechanics." Soviet Physics Uspekhi 18, no. 12 (1975): 988

provides an example where $P_{ij}\vert\psi\rangle$ is not a multiple of $\vert\psi\rangle$ and shows that the resulting average value depends on the labelling of the particles.  Basically the argument boils down to requiring the state of a collection of many indistinguishable particles to transform by a 1-dimensional representation of the permutation group.
Having argued that $\vert\psi'\rangle$ must be a multiple $\alpha$ of $\vert\psi\rangle$ if the states are to give the same average values for any operator, it follows that they can only differ by a phase: the multiple must be of the form $\alpha=e^{i\phi}$ so both states are properly normalized.  Since $P_{ij}P_{ij}=1$ for any transposition, if follows that $\alpha^2=1$ and thus $\alpha=\pm 1$.
This (non-relativistic) argument relies on using permutation (rather than braiding transformations) to interchange particles, so the topology of the ambient space is largely irrelevant, although quasi-particles with fractional statistics (anyons) are possible in 2d.
