1
$\begingroup$

If we are to take two Hydrogen atoms and subject them to the same potential, then wouldn't both Hydrogen atoms be in the same exact quantum state? This bother me because no two identical fermions can be in the same quantum state! This seems to contradict the principle. This applies to any two elements or molecules that are subjected to the same potential.

Say these two Hydrogen atoms are located 1m from each other, then would the only way to distinguish them would be their spatial location?

What is the technical term for two seemingly identical things to be distinguishable by their location?

Improve this question
$\endgroup$
5
  • $\begingroup$ Have you thought that a hydrogen atom is not a fermion? It has an equal number of protons and electrons so the minimum spin is 0, and it is a boson anyway. An unlimited number of bosons can fit in the same energy level/state. They are indistinguishable. Two hydrogen atoms one meter apart will be meeting different potentials and energy levels anyway. Meters are in the realm of classical physics. $\endgroup$
    – anna v
    Mar 30, 2013 at 16:19
  • $\begingroup$ I did not think of the hydrogen as a boson! I guess my question would be if I picked an element that is a fermion, say Lithium atoms? $\endgroup$
    – QEntanglement
    Mar 30, 2013 at 16:29
  • $\begingroup$ Because atoms are neutra, i.e. equal numbers of electrons and protons, they all are bosons. An ion, an electron missing, becomes a fermion. $\endgroup$
    – anna v
    Mar 30, 2013 at 16:46
  • $\begingroup$ Okay, say I have two ions then. Can their remaining electrons be in the same orbital, same energy state, and same spin state? Can this happen if the two ions are separated by a large distance? $\endgroup$
    – QEntanglement
    Mar 31, 2013 at 2:26
  • 1
    $\begingroup$ Did you not see Lubos' answer? The first paragraph covers these changed conditions. As long as there is a different (x,y,z) coordinate the eigen states are different.( BTW an odd number of baryons would also make an atom have at least spin 1/2) $\endgroup$
    – anna v
    Mar 31, 2013 at 4:32

1 Answer 1

Reset to default
4
$\begingroup$

The two hydrogen atoms aren't in the same state in the situation you described: they differ by the location which is an observable specifying the quantum state. Because it's different, the states are different.

Moreover, hydrogen-1 isn't a fermion.

If you talk about two electrons in the potential of two hydrogen nuclei, then you do have two fermions, but otherwise the situation is identical. These two fermions belong to two different states because the location distinguishes (the "in which hole" information) the states.

These things – the fact that with extra "space" or "volume", we obtain new states where electrons may choose to live – are obvious once we learn what the states actually mean and what the Pauli exclusion principle demands, so there doesn't have to exist and there doesn't exist any special term for that.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Okay, so to solidify my understanding, if we consider two electrons in the potential of two hydrogen nuclei, then we have two electrons that are distinguishable because of the fact that the two protons are located at a different locations. Everything technically is distinguishable because of space time itself? With the exception of Bosons? Ex: two photons with same energy that are overlapped traveling the same direction and in phase, or Bose Einstein condensates with atoms all collapsed in the same location with ground state energy? $\endgroup$
    – QEntanglement
    Mar 31, 2013 at 5:13
  • $\begingroup$ Dear QEntanglement, sorry, I don't understand these extra questions. If we consider 2 electrons in the potential created by 2 distant nuclei, the approximate Hilbert space resembles two copies of the bound state found in a single nucleus. This is an approximation that becomes good when the distance of nuclei goes to infinity. In this approximation, the basis is doubled and the "here/there" binary extra information "near which nucleus" plays the role of an extra quantum number for all the bound states which allows one to insert more fermions. There are no "exceptions", why exceptions? $\endgroup$
    – Luboš Motl
    Mar 31, 2013 at 6:19
  • $\begingroup$ I am basically stating, "We can distinguish two Hydrogen Atoms from the fact that they are located at different locations." If the two Hydrogen Atoms were to overlap their wave functions, then we can't tell them apart. The exceptions I was talking about was about how Bosons are indistinguishable. I am trying to understand what we can say is different or the same in the most fundamental way possible. $\endgroup$
    – QEntanglement
    Apr 1, 2013 at 17:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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