1
$\begingroup$

I am reading Quantum Statistics from 'Fundamentals of Statistical and Thermal Physics' by Frederick Reif. I have questions in two places.
I understand the following paragraph:

Particles with half-integral spin (Fermi-Dirac statistics): This is applicable when each particle has a total spin angular momentum (measured in units of h) which is half-integral, i.e., $\frac{1}{2}$, $\frac{3}{2}$, . . . (examples might be electrons or He3 atoms). Then the fundamental quantum-mechanical symmetry requirement is that the total wave function $\Psi$ be antisymmetric (i.e., that it changes sign) under interchange of any two particles. In symbols $$\Psi(\cdot\cdot\cdot Q_j \cdot\cdot\cdot Q_i\cdot\cdot\cdot) =-\Psi(\cdot\cdot\cdot Q_i\cdot\cdot\cdot Q_j\cdot\cdot\cdot)\tag{1}$$ Once again, interchange of two particles does not lead to a new state of the gas. Hence the particles must again be considered as genuinely indistinguishable in enumerating the distinct states of the gas. But the change of sign in $(1)$ does imply one additional consequence: Suppose that two particles $i$ and $j$ both in the same single-particle state $s$, are interchanged. In that case one obviously has $$\Psi(\cdot\cdot\cdot Q_j \cdot\cdot\cdot Q_i\cdot\cdot\cdot) =\Psi(\cdot\cdot\cdot Q_i\cdot\cdot\cdot Q_j\cdot\cdot\cdot)\tag{2}$$ But since the fundamental symmetry requirement $(1)$ must also be valid, $(1)$ and $(2)$ together imply that $$\Psi = 0\tag{3}$$ when particles $i$ and $j$ are in the same state $s$ Thus in the Fermi-Dirac case there exists no state of the whole gas for which two or more particles are in the same single-particle state. This is the so-called "Pauli exclusion principle."

Now, I am thinking that if Fermions are indistinguishable particles, then exchanging any two particles from two different states should also keep the wave function same: I am imagining this like I have a quantum system consisting of several energy levels in which I am putting Fermions. So, if I exchange any two of them from any two levels, then also the system looks quite the same as it looks same if we try to put two Fermions in the same level and interchange them. In the former, i.e., my case, then also the wave function should follow $(2)$ and hence be $0$. But it is not the case. I think I am doing some serious mistake in imagining the entire setup. Where am I wrong?

$\endgroup$
  • $\begingroup$ HI The Smart Turtle: I removed your last question. Please only ask 1 question per post. $\endgroup$ – Qmechanic Jun 8 at 7:47
  • $\begingroup$ the -1 sign is a global phase that has no consequence on measurements. Is a gauge $\endgroup$ – lurscher Jun 8 at 11:00
1
$\begingroup$

You don’t need the wavefunctions to be the same for all observables to remain the same. That is because all observables are expectation values. $$\langle O\rangle=\langle\psi|O|\psi\rangle=\big(-\langle\psi|\big)O\big(-|\psi\rangle\big) $$

Rather what you need to preserve are the probabilities, ie $$|\psi|^2=\langle\psi|\psi\rangle$$ And note that any transformation of the kind $|\psi\rangle\to e^{i\theta}|\psi\rangle$ preserves that. That’s why symmetries (changes that preserve the physics of the system) are all unitary transformations. And the pair-exchange operator for a fermion gives a negative sign (or $\theta=\pi$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But it's written in the book that the wavefunction is same. ''for all observables to remain the same ''- where are they remaining same? I am not measuring anything, I am just exchanging two particles. $\endgroup$ – The Smart Turtle Jun 8 at 8:20
  • $\begingroup$ They are talking about whatever measurement you can potentially make, they must remain the same before and after exchange. Note that all relevant physical observables are always expectation values. Thus what you need to preserve is the expectation value. Not the wavefunction itself. $\endgroup$ – Superfast Jellyfish Jun 8 at 8:35
  • $\begingroup$ If that is the case then how will you explain the derivation of Pauli Exclusion Principle in terms of expectation values of observables. Here it is shown through the wave function remaining intact. $\endgroup$ – The Smart Turtle Jun 8 at 8:56
  • $\begingroup$ Consider the simplest case where you have two electrons in the same state. The overall wavefunction is the product of the two which is the individual function squared. The rule says if you exchange electrons you need to have a minus sign. A square that is its own negative . Thus the total wavefunction must equal it’s own negative. Only if it’s zero that’s possible. $\endgroup$ – Superfast Jellyfish Jun 8 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.