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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):

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Why the underlined statements are true?

  1. 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?

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

Can anyone help?

Qmechanic
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Plantation
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  • Are you confused about the typo in equation (12.5)? (It should be $s_1 \vec{p}_1 n$ in the second instance on the right-hand side, not $s_1 \vec{p}_2 n$.) Or are you confused about the idea that the two states have to differ only by a phase? – Michael Seifert Jan 09 '23 at 15:30
  • The one that makes me wonder is the second~ – Plantation Jan 10 '23 at 07:24

2 Answers2

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  1. 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.

  1. 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 :)

Song of Physics
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  • Thank you. To mimic the (12.7) to drive the (12.9), it seems that it will be good if we understand the (12.7) certainly. Looking at it again, I don't know why (12.7) holds. We know that (12.2) and (12.4) as in the above image. And how can we deduce from these to (12.7)? Is there an any relation between the ground state $|0>$ and general state $|\psi>$? And for the (12.9), how/where can we use the fact $<\vec{p_1}|\vec{p_2}>=2\omega_1(2\pi)^{3}\delta^{3}(\vec{p_1}-\vec{p_2})$? – Plantation Jan 10 '23 at 07:44
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    @Plantation $| \psi \rangle$ is of course, a linear combination of the basis states, which can be constructed from the vacuum state $| 0 \rangle$ as shown in (12.2) and (12.4). Immediately (12.7) follows from (12.5) by definition, which you can see writing everything down explicitly. As for how to use the $\langle \vec{p}_1 | \vec{p}_2 \rangle$, it would require an elementary understanding of how the Dirac$-\delta$ works, which you can review in a standard QM text. I highly recommend you do so since many such doubts are bound to come up further up in the text. – Song of Physics Jan 10 '23 at 14:06
  • Uhm..Thank you. In fact, I'm still struggling to make a formal proof. I don't know where the $<\vec{p_1} | \vec{p_2}>$ will pop up in the process of argument. Perhaps can you provide more kind hint or step-e.g. what is key point -in detail for formal proof? – Plantation Jan 12 '23 at 06:11
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    @Plantation for the details as I mentioned, you ought to refer to a standard QM text. You'll essentially need the $\langle \vec{p}_1 | \vec{p}_2 \rangle$ to determine the commutator for the case where $\vec{p}_1 = \vec{p}_2$ and unify it with the case $\vec{p}_1 \neq \vec{p}_2$. – Song of Physics Jan 12 '23 at 07:14
  • O.K. Anyway thank you ~~ – Plantation Jan 12 '23 at 07:22
  • @Plantation kindly mark the answer correct if you're satisfied. – Song of Physics Jan 12 '23 at 11:36
  • It seems that a separate associated study is required. :) – Plantation Jan 15 '23 at 06:49
  • It's a little late. I am still don't understand how can we derive the (12.9). I think that I forgot something in the QM text. A thing that makes me annoying is, the objects $\vec{p_1} , s_1 , \vec{p_2},s_2$ are positioned as subscripts of $a$, $a^{\dagger}$. I don't know how can we deal the commutator involving more than two subscripts. What should I catch? Which relationship can we use? Can you give me a hint? – Plantation Oct 21 '23 at 03:16
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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.

ZeroTheHero
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