0
$\begingroup$

In case of superposition of identical particles, we usually just add their amplitudes. For example, if we have several particles having the amplitudes of being in a particular quantum state $\psi_1, \psi_2, \psi_3 ...$, after superposition, we can say, the amplitude of finding one or more particle in that state is $\psi_1 + \psi_2 + \psi_3 ...$, and the probability is $|\psi_1 + \psi_2 + \psi_3 ...|^2$.

Now, my question is, does this superposition work for fermions? I mean, Pauli exclusion principle states that, more than one identical fermions cannot be in the same state at the same time together. Should the amplitude of finding one particle in that state still remain $\psi_1 + \psi_2 + \psi_3 ...$? Or it will be something less than that?

Moreover, since bosons 'like' to stick together, should that change the total amplitude in case of bosonic superposition? Should the total amplitude of finding one or more bosons in a particular state be more than $\psi_1 + \psi_2 + \psi_3 ...$?

Thanks in advance.

$\endgroup$

closed as unclear what you're asking by Aaron Stevens, GiorgioP, user191954, Jon Custer, Chris Apr 30 at 20:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Where are you finding this method? I have never seen this where you just add wavefunctions together. Also your terminology is confusing. I think by superposition of identical particle you mean to say a system of multiple identical particles $\endgroup$ – Aaron Stevens Apr 28 at 11:10
1
$\begingroup$

If we have 2 identical particles $a$ and $b$, and 2 states 1 and 2, the amplitude of finding a particle at $a$ and another at $b$ is

$\psi_a(1) \psi_b(2) ± \psi_a(2) \psi_b(1)$

Now, if we generalize the scenario for 10 states and 2 particles, the amplitude of finding particle $a$ at state 1 and particle $b$ in any state is:

$\psi_a(1)\Bigl(\psi_b(1) + \psi_b(2) + \psi_b(3) + …\Bigr)$

Similarly, the amplitude of finding particle $b$ in state 1 and $a$ in any state is:

$\psi_b(1) \Bigl(\psi_a(1) + \psi_a(2) + \psi_a(3) + …\Bigr)$

If we plus minus the 2 cases, we get the overall amplitude of finding one or more particles in state 1.

$\psi_a(1)\Bigl(\psi_b(1) + \psi_b(2) + \psi_b(3) + …\Bigr) ± \psi_b(1) \Bigl(\psi_a(1) + \psi_a(2) + \psi_a(3) + …\Bigr)$

$\endgroup$
0
$\begingroup$

It's a bit more complicated than that. When you have identical particles and there is an interaction between them, the whole system must be described by only one wave function $\Psi$, not each individual particle, since they are identical and we cannot keep track of which one is which.

Now the exchange between two identical particles would leave the probability if finding the system in a given state invatiant, that is $|\Psi|^2 $ does not change if two particles are interchanged. For fermions that means that the phase changes by a factor -1, as a result of spin statistics theorem, but that is a pretty advanced topic. The result of this is Pauli exclusion principle, since if both exchanged particles were in the exact same state, the wave function would be changed, and that wouldn't be consistent. I personally like the explanation of "Concepts in thermal physics" by Blundell & Blundell.

To be consistent with spin statistics theorem, the wave function of the system must be antisymmetrized for fermions and symmetrized for bosons. You should check Wikipedia article on identical particles.

As a result of this symmetrization and antisymmetrization, it is indeed true that the expected value of the distance between particles is increased for fermions and reduced for bosons.

You can find a lot of information about the topic, here you can see some of the results presented step by step M. Safronova lecture notes.

$\endgroup$
  • $\begingroup$ Hi Alex, thank you for your kind answer. Now, let us consider a system where we have 2 identical particles A and B and 10 different states 1, 2, 3, .... If I want to calculate the probability of finding one or more particles at state 1, several cases arise. 1. Particle A is in state 1, and particle B is in some state. 2. Particle B is in state 1 and particle A is in some state. If I understood it properly, we just add the 2 cases in cases in case of bosons, and subtract them in case of fermions right? $\endgroup$ – Nayeem Hossain Apr 28 at 15:09
  • $\begingroup$ @NayeemHossain Not exactly, as I said you have to symmetrize (bosons) or antisymmetrize (fermions) the wave function. That means that if the particles are really indistinguishable (not only in their intrinsec properties, but are also sufficiently close for their wave function to overlap), you cannot really tell which is in which state, so you would have for the boson example |1,3> + |3,1>, that is the superposition of A is in 1 and B in 3, and A in 3 and B in 1. For the fermions it works similarly, you will have |1,3> - |3,1>. (Sorry for the late answer) $\endgroup$ – Álex De La Calzada Apr 29 at 0:45
  • $\begingroup$ Can you please check the answer I posted below? Thanks in advance. $\endgroup$ – Nayeem Hossain Apr 29 at 3:09
  • 1
    $\begingroup$ @NayeemHossain It makes sense to me, very nicely explained $\endgroup$ – Álex De La Calzada Apr 29 at 7:49

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