0
$\begingroup$

In the second edition of "Introduction to Quantum Mechanics" of David J. Griffiths (2016) there is an example that says (p.209, example 5.1)

Suppose we have two noninteracting --they pass right through one another ... -- particles, both of mass $m$, in the infinite square well. The one-particle states are $$ \psi_n (x) = \sqrt{\dfrac{2}{a}} \sin \left(\dfrac{n\pi }{a} x \right), \qquad E_n = n^2 K$$ (where $K\equiv \pi^2 \hbar^2 /2ma^2$, for convenience). If the particles are $\textit{distinguishable}$, with #1 in state $n_1$ and #2 in state $n_2$, the composite wave function is a simple product: $$ \psi_{n_1 n_2} (x_1, x_2) = \psi_{n_1} (x_1) \psi (x_2), \qquad E_{n_1 n_2} = (n_1^2+n_2^2) K.$$ For example, the ground state is $$ \psi_{11} = \frac{2}{a} \sin (\pi x_1/a) \sin (\pi x_2/a), \qquad E_{11} = 2K$$ [...]

and a problem 5.6. (p-214)

Imagine two noninteracting particles, each of mass $m$, in the infinite square well. If one is in the state $\psi_n$, and the other in state $\psi_m (m \neq n)$, calculate $<(x_1-x_2)^2>$, assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions.

If they are distinguishable particles: $$ <(x_1-x_2)^2>\equiv g_{nm}(a).$$ If they are identical bosons: $$ <(x_1-x_2)^2> = g_{nm}(a) -|I_{nm}|^2.$$ If they are identical fermions: $$ <(x_1-x_2)^2> = g_{nm}(a) + |I_{mn}|^2$$ where $$ g_{nm}(a) = a^2 \left[ \dfrac{1}{6} -\dfrac{1}{2\pi^2} \left( \dfrac{1}{n^2} +\dfrac{1}{m^2}\right)\right], \qquad n,m \in \mathbb{N}$$ and $$ I_{mn} =\dfrac{4a}{\pi^2} \dfrac{mn}{(n^2-m^2)^2} \left\{ (-1)^{m+n} -1 \right\}, \qquad n,m \in \mathbb{N}.$$ So, for bosons and fermions if $I_{nm}=0$ they can be considered as distinguishable particles. This occurs when the wave functions have the same parity on $n,m$. What is the physical interpretation of the last sentence?

If the sentence "they can be considered as distinguishable particles" is false, what is the physical interpretation of $|I_{mn}|=0$? What expectation values are needed to prove that the bosons or the fermions can be considered as distinguished particles?

$\endgroup$
1
$\begingroup$

Just because $\langle (x_1 - x_2)^2 \rangle$ is unaffected by the particle statistics doesn't mean that "they can be considered as distinguishable particles." Other expectation values will be affected by the statistics.

$\endgroup$
  • $\begingroup$ Thank you @tparker. And what is the physical interpretation of $I_{mn}=0$? $\endgroup$ – Clare Francis Mar 3 '17 at 18:48
  • $\begingroup$ @JonathanEstévez-Fernández It just means that the net "exchange forces" due to the particle statistics happen to balance out to zero for that particular expectation value for those particular states. I don't think there's anything very fundamental there, though - I suspect that the higher separation moments, like $\langle (x_1 - x_2)^4 \rangle$ or $\langle (x_1 - x_2)^6 \rangle$, are affected by the particles' statistics. $\endgroup$ – tparker Mar 3 '17 at 23:03
  • $\begingroup$ Thank you again @tparker for your answer. So, is it imposible to distinguish between two identical bosons/fermions with this potential even if $|x_1-x_2| \rightarrow \infty$ (if one particle is in Chicago and the other one in Seattle, for example)? Can it be mathematically proven ? $\endgroup$ – Clare Francis Mar 4 '17 at 9:51
  • $\begingroup$ @JonathanEstévez-Fernández In principle, yes, every single particle in the universe needs to be (anti-)symmetrized with every other particle of that same species, no matter how far away they are. You can't prove that, it's just part of the definition of bosons and fermions. But it turns out that the "exchange force" effects of the (anti-)symmetrizion become extremely difficult to detect if the particles are very far away from each other, as explained in Griffiths pg. 209 or Shankar pg. 273. $\endgroup$ – tparker Mar 6 '17 at 0:45
  • $\begingroup$ @JonathanEstévez-Fernández At first, it might seem disappointing that the cool prediction of the (anti-)symmetrization of faraway particles is so difficult to test in practice. But as Griffiths and Shankar explain, it's actually a very good thing, because if every particle exerted an important influence on every other particle in the universe (no matter how far away), then quantum mechanics would be intractable complicated, because you'd know to know about every particle in the universe in order to make any predictions at all! $\endgroup$ – tparker Mar 6 '17 at 0:49

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.