# Indistinguishable particles in quantum mechanics

If you have two particles of the same species , Quantum mechanics says that $\Phi_{m_{1},x_{1},p_{1},m_{2},x_{2},p_{2}}=\alpha\Phi_{m_{2},x_{2},p_{2},m_{1},x_{1},p_{1}}$ But I don't understand why $\alpha$ doesn't depend on $x$ , $p$ . If $\alpha$ depends on $SO(3)$ invariants as $x^2 , x.p , p^2$ etc then it will be the same on all reference frame why does one require that it doesn't depend on these variables? Even if it depends on $p ,x$ $\alpha$ is a phase factor so it doesn't affect anything why should this be important?

EDIT : I figured out the answer to the second question , for $\alpha$ is a complex number that carries no indices so it cannot change

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+1 for figuring out your own answer! Re the first question, if $\alpha$ depended on those variables, then when you start differentiating wavefunctions to do your Schroedinger equation, I guess things would get in a mess? – twistor59 Mar 31 '13 at 14:17
Well, as I see it, if $\alpha$ depended on $x_{1}$ (not operator, say we are in the position representation), say, it would change the derivative of $\Phi$ with respect of $x_{1}$, and thus would change the $p_{1}$ of the wavefunction. – guillefix May 15 '14 at 13:53

## 2 Answers

Well, $\alpha$ could not depend on $x$ or any function of it since the space-time is homogenius. $p_{12}^2 = m^2$ is a constant (for given identical particles), so the only possiblity is that $\alpha$ depends on 4-invariant $p_1^\mu p_{2 \mu}$ which is identical to $p_2^\mu p_{1 \mu}$. So interchanging particles one more time will still lead (as it is usually derived in textbooks) to $\alpha^2 = 1$ hence $\alpha = \pm 1$. So $p_1 p_2$ dependence cannot exist too.

However, you've asked a good question - there are some more possibilities that you're missing, so I advice you have a look at Weinberg's Quantum Field Theory, Vol 1, Chapter 4.

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Let us step back for a moment to answer your question.

We consider a system of $n$ indistinguishable particles. What does that mean ? Let $S_n$ be the set of all permutations of $n$ elements, and let $\sigma \in S_n$. Then if $P(\sigma)$ is the (unitary) operator representing $\sigma$ on the $n$-particles Hilbert space $\cal H$, the property of "indistinguishability" means that the two vectors $|\psi\rangle$ and $P(\sigma)|\psi\rangle$ represent the same physical state, and this should be true for any state $|\psi\rangle \in {\cal H}$. In other words, we must have $$P(\sigma) = e^{i \chi(\sigma)} {\mathbf 1}$$ where $\chi(\sigma)$ is some real number.

If I understand your question correctly, what you ask is why couldn't we have the number $\chi(\sigma)$ to be an operator depending on the momentum operator $\hat {\vec P}$ or the position operator $\hat {\vec R}$ (for example). But obviously $P(\sigma)$ would not be proportional to the identity operator $\mathbf 1$, violating the above conclusion.

I hope this help !

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