5
$\begingroup$

Suppose we have two hydrogen atoms in the ground state with spin of both electrons pointing upwards. Then the two electrons are in the same state. This should be against the exclusion principle. Now suppose we have 1 mole of hydrogen atoms in a chamber. Certainly, most of them will be in the ground state (at sufficiently low temperature), and among any three of those in the ground state, at least two will have spin in the same directions, hence the two electrons are in the same state. How is the exclusion principle valid for those two electrons?

My doubt is mostly about which parameters determine a "state". Suppose two different hydrogen atoms having the same quantum numbers are in different points in space. Are the two electrons in the same state?

Improve this question
$\endgroup$
1
  • $\begingroup$ I have edited the question as it was being marked "too broad". $\endgroup$
    – Archisman Panigrahi
    Commented May 6, 2019 at 16:44

4 Answers 4

Reset to default
17
$\begingroup$

A quantum state includes the information about a particle's position. Two particles with the same quantum numbers at different locations are in different states, so are allowed by the exclusion principle.

Improve this answer
$\endgroup$
18
  • 2
    $\begingroup$ Since there is the uncertainty principle, can we really ever say that the two particles are in the same position? $\endgroup$
    – Archisman Panigrahi
    Commented May 5, 2019 at 19:34
  • 1
    $\begingroup$ No, but then you can't say they have the same momentum either. The point is that the quantum state includes all th information about the particle and it must all be the same to be disallowed by the Pauli Exclusion Principle $\endgroup$
    – By Symmetry
    Commented May 5, 2019 at 19:44
  • 3
    $\begingroup$ Remember that the wavefunction for $n$ identical fermions isn't $n$ different functions defined on $\mathbb{R}^3$ - it's a single function defined on the configuration space $\mathbb{R}^{3n}$, not on real space. The exclusion principle says that the joint probability (density) for two particles to both be at the exact position simultaneously is zero. However, if the wavefunction can be expressed as a Slater determinant of $n$ different single-particle wavefunctions (which is not always the case), then WLOG the single-particle states can be chosen to be orthogonal, but they are ... $\endgroup$
    – tparker
    Commented May 5, 2019 at 23:22
  • 3
    $\begingroup$ ... allowed to overlap. Roughly speaking, you can have a nonzero probability of measuring particle $A$ at position $x$, and also a nonzero probability of measuring particle $B$ at position $x$, as long as the probability of simultaneously measuring particles $A$ and $B$ to both be at position $x$ is zero (neglecting spin, etc.). @ArchismanPanigrahi $\endgroup$
    – tparker
    Commented May 5, 2019 at 23:26
  • 1
    $\begingroup$ @candied_orange That's why I put "(density)" in parentheses in my comment above. Remember that the wavefunction gives you probability densities, not probabilities, and these can be either zero or nonzero. $\endgroup$
    – tparker
    Commented May 6, 2019 at 13:18
4
$\begingroup$
  1. You can't create two electrons with the same momentum, because you can't create even a single electron with a particular, exact momentum. You can create an electron whose spatial wavefunction contains an arbitrarily narrow distribution of momenta, but then the distribution over spatial locations will be very broad. Independent of this, if they are localized in different regions of space, then their spatial states are different.
  2. Assuming the electrons in different H atoms are bound to distinct nucleii, their states will be distinct because of this. In principle, though, if we ignore the nucleii and just put a lot of electrons in a box at low T, we can get a degenerate Fermi gas in which the exclusion principle does matter. The situation is more complicated when nucleii are involved.
  3. The spatial state of any particle is part of its state with regard to the exclusion principle, so no, two electrons in two different atoms are never in the same state. Often we focus on their atomic (orbital+spin) states, and people often just call these the "states," but with regard to the Pauli exclusion principle, the spatial state definitely also matters.
Improve this answer
$\endgroup$
1
$\begingroup$

If you can say they are in different parts of the Universe, then that means you do have position information, even if only a small amount, which means that there is also less momentum information, even if such may be quite ample. It also means that they thus cannot be attributed the same quantum state. Hence, Pauli does not forbid it.

Two electrons with perfect momentum information would, indeed, have no position information at all, and thus would be totally unconnected to any sense of place within the Universe, totally harring all throughout. And if those maximal-information momenta were equal, indeed, Pauli would exclude that.

Improve this answer
$\endgroup$
0
$\begingroup$

The spatial part of the wave function is part of the state.

Two electrons in the same isolated atom have the same spatial part of their state, hence can be subject to the exclusion principle. Two in well-separated atoms don’t have the same spatial wave function, hence can’t be in the same state, hence are not subject to the exclusion principle.

It can get complicated with overlapping cases, I.e. in chemical bonds, and when the electrons have a historical connection leading to entanglement. But in the simple cases in the question, separated electrons have different spatial parts to their state, hence are not in the same state.

Improve this answer
$\endgroup$

Your Answer

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

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