I am an undergraduate student in Physics and Mathematics. I am now preaparing for my final exam in Statistical Mechanics and I would like some help in a particular point. So here it goes:

In the introductory course of Quantum Mechanics we have discussed how a global phase does not affect the physical state of the system (to be specific $|\psi\rangle$ and $-|\psi\rangle$ are the same state).

However, in the course of Statistical Mechanics, we predicted the existence of fermions and bosons by applying an elementary permutation of two particles and seeing that the only possibilities are $\hat{P}|\psi\rangle = \pm |\psi\rangle$

What I have trouble understanding is that if the state is the same after the permutation, why do the particles have such a different behaviour.

Attempt at solution:

In my head, the way this could work is that indeed $|\psi\rangle$ and $-|\psi\rangle$ are the same state but the behaviour of the wave function (symmetric or antisymmetric with respect to particle interchange) is what determines if the system is a fermionic or a bosonic.

I am aware that the question is a bit wishy-washy, but I have been looking around in different sources and no one seems to adress this particular concern.

Any help or comments are appreciated

(PS: This is my first question ever, so any tips in question formulation are also welcome)

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    $\begingroup$ Be careful in thinking of phase as arbitrary. There is only one arbitrary phase in an entire system. In systems of more than one particle the second and subsequent particles have a meaningful phase difference relative the first one. $\endgroup$ – dmckee Jun 7 at 18:40
  • $\begingroup$ @dmckee That seems like an answer, rather than a comment. :-) $\endgroup$ – rob Jun 8 at 17:53

The main difference is more visible if you apply the permutation to some coordinates, say $x_1, x_2$ of the particles. These might be positions or spins or whatever you want, a property of the two particles where $x_1$ is the property of the particle $1$ and $x_2$ of the particle $2$. Let's work with the wave function $\langle x_1, x_2|\psi\rangle$ where we know, $\hat{P}$ being self adjoint $$\pm\langle x_1, x_2|\psi\rangle=\langle x_1, x_2|\hat{P}|\psi\rangle=\langle x_2, x_1|\psi\rangle$$ Now consider the fermonic case, the one with a minus sign$$\langle x_1, x_2|\psi\rangle=-\langle x_2, x_1|\psi\rangle$$ and look at the case $x_1=x_2=x$

$$\langle x, x|\psi\rangle=-\langle x, x|\psi\rangle$$

This clearly implies $\langle x, x|\psi\rangle=0$, lo and behold, two fermions can't be in the same quantum state! Bosons have no such problem, as there is no minus sign. This is the Pauli exclusion principle and it's the main "tangible" difference between bosons and fermions.

  • $\begingroup$ I see, it truly is clearer if you work with wave functions. So the main difference is actually that the wave function that represents a system of fermions is different from the wave function representing a system of fermions (the first one being antisymmetric under permutation of 2 coordinates). However, the physical state of the system before and after the permutation is the same, as it should be, because the particles are identical! $\endgroup$ – Hightide Jun 7 at 10:25
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    $\begingroup$ Yes, to put it more formally, choosing whether your wave functions are symmetric or antisymmetric divides your Hilbert space into two different 1D representations of the permutation group. Basically you "select" all the wave functions that are symmetric and you call them bosonic, same for antisymmetric and fermionic, but both these sets of wave functions have the property that their norm squared doesn't change under $S_N$. All the wave functions that don't belong in either category, you throw them away. $\endgroup$ – user2723984 Jun 7 at 11:10
  • $\begingroup$ That was exactly the answer I needed. Thank you very much! $\endgroup$ – Hightide Jun 7 at 11:21

(to be specific |ψ⟩ and −|ψ⟩ are the same state).

Err... yes and no. As analogy, consider the numbers $i$ and $-i$. Whenever you have a function expressed in terms of real numbers, $i$ and $-i$ are indistinguishable. For instance, $x^2+1=0$ is true for both, $x^3+1=0$ is false for both. When we look at what properties they have through the real numbers, they are the "same" number. Any attempt to create something that identifies one, using just real numbers, will also identify the other. And yet clearly they are different numbers.

Similarly, for any linear operator $H$, we have that $\langle-\psi |H(-\psi)\rangle = \langle\psi |H\psi\rangle$. So any observable will give the same result for both states, and the states are in this sense "indistinguishable" and thus "the same". But clearly they're not actually the same state. One is $\psi$ and the other is $-\psi$. One of them, when added to $\psi$, yields $2\psi$, and the other one yields $0$. They are equivalent states (given a particular definition of "equivalent"), but they are not the same state.

As another analogy, if a copy of Earth out of anti-matter, then (putting aside symmetry violations), that planet would look like Earth. People would think the same way and do the same things as on the original Earth. These two Earths would, in terms of all these observations, be "the same". And yet they would not in fact be the same thing: put two actually identical Earths next to each other, and you have two identical Earths. Put an Earth and an anti-matter Earth next to each other, and you have a huge explosion.


$\newcommand{\Ket}[1]{\left | #1\right>}$ You studied that if $|\psi\rangle$ is a physical wavefunction, it must have permutation symmetry. This means that, if $\hat{P}$ is the permutation operator, then $|\psi\rangle$ must be an eigenfunction of the operator. So, $\hat{P}|\psi\rangle = \pm |\psi\rangle $. We define the wavefunctions that have the +1 eigenvalue as Bosons and that of -1 eigenvalue as Fermions.

So, the wavefunctions that have + 1 eigenvalue are different from that of -1 eigenvalue.

The wavefunction of a system determines all the properties of it, so naturally fermionic and bosonic systems would have different properties.

Let me know if you want to know something specific that I haven't mentioned !

  • $\begingroup$ Thank you for your answer! I am pretty sure I understand it now $\endgroup$ – Hightide Jun 7 at 14:46

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