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Based on what we know about atomic structure all atoms have an electron density function which describes the uncertainty in the position of an electron. Thus theoretically 2 atoms of the same element can be different in terms of positions of their respective elections.

why is it then that no matter where they are found 2 atoms of the same element tend to have the exactly same physical and chemical properties ?

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All bound states (groups that are not decaying away, they're more "orbiting" in some way) of nuclei and electrons may be written as a (linear combination of) energy eigenstates – eigenstates of the Hamiltonian $H$.

The Hamiltonian (energy operator) is always the same – it captures all the information about the laws of Nature and there are the same. The possible eigenstates and eigenvalues of this operator are "discrete", i.e. sharply separated from each other, so their set is countable. We may uniquely identify these solutions – eigenstates – by labels, e.g. by integers.

If two atoms have the same label, they are exactly the same because they are exactly the same eigenstates of the same Hamiltonian which defines the laws of physics and those are also the same everywhere.

Incidentally, two atoms are also perfectly identical in a stronger sense: if we exchange them, the wave function remains exactly the same (up to the sign flip, if they are fermions). It has to be so because the atoms' dynamics is described by "field theory" where the fields are either bosonic or fermionic, and two or more atoms are achieved by the action of two or more "creation operators" and those either commute or anticommute with each other which guarantees the symmetry or antisymmetry of the wave function under the permutations of the atoms.

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  • $\begingroup$ While it's certainly true that for all practical purposes two atoms in steady state are interchangeable, would you really go so far as state that the nucleus and electron cloud are technically identical between the atoms? After all, the simple "orbital" Hamiltonian could be modulated in a possibly non-interchangeable way if you would ever set up an experiment to test this. In other words, the atoms electrons will only seem similar as long as you measure them in a way that allows the hamiltonian to stay time-independent (for example) as the poster alludes to I think. $\endgroup$
    – BjornW
    Feb 7, 2015 at 9:18
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    $\begingroup$ Dear @BjornW, all electrons in the Universe - those in atoms and those outside atoms - are perfectly interchangeable, identical, and the wave function is totally antisymmetric in them. Similarly for protons. For helium nuclei, they are identical bosons. It's not going "far". It's what the principle about the indistinguishability says. If it weren't true, the principle would be (completely) wrong. $\endgroup$
    – Luboš Motl
    Feb 7, 2015 at 13:57
  • $\begingroup$ You may use approximate Hamiltonians or treatments where the perfect antisymmetry or perfect symmetry is violated but they're always at most approximations or fudges. The exact treatment doesn't allow any deviation from the perfect symmetry or antisymmetry. $\endgroup$
    – Luboš Motl
    Feb 7, 2015 at 13:58
  • $\begingroup$ What I mean is that an atom with say a dozen electrons undergoing absorption of a photon for example, is not interchangeable with a similar atom drifting around in free space. The atom as a whole including all the electrons is only approximately the same as another such atom due to dynamics, even if all the constituents are indistinguishable. I think John's answer contains pretty much what I wanted to add :) But then again, these distinctions are less related to the chemical properties the OP asks about.. $\endgroup$
    – BjornW
    Feb 7, 2015 at 14:42
  • $\begingroup$ Dear @BjornW, if you have two magnesium atoms, each of them has 12 electrons and the accurate description of the situation is in terms of a wave function of positions of 24 electrons (plus two nuclei) and this wave function is perfectly antisymmetric under the $S_{24}$ permutation group of the 24 electrons. Alternatively, you may overlook the exact internal structure of the atoms. Then you have a wave function that is a function of locations of 2 atoms. $\endgroup$
    – Luboš Motl
    Feb 8, 2015 at 7:50
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I think you've misunderstood what the uncertainty principle tells us.

The electrons in an atom do not have a position because they are delocalised over the whole atom. So two atoms can't behave differently because their electrons are in different positions - all atoms of the same element/isotope have thir electrons delocalised in an identical way.

We can temporarily localise an electron by measuring its position. If we do this the uncertainty principle tells us that if we measure the electron position to a standard deviation of $\Delta x$ then the momentum will have an uncertainty $\Delta p$ given by the usual uncertainty equation:

$$ \Delta x \Delta p \ge \frac{\hbar}{2} $$

But as soon as we've done the measurement, and localised the electron, it will delocalise again. All we would have achieved is to temporarily localise the electron to some region of space.

The obvious question is whether the atom would behave differently during the very short time that we are doing the measurement, and the answer is that yes it would. But this shouldn't be surprising. When you make a measurement on an atom you interact with it and the interaction perturbs both the atom and the measuring equipment.

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  • $\begingroup$ I think your and Lubos answer could be combined in some way :) $\endgroup$
    – BjornW
    Feb 7, 2015 at 9:19

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