3 ms·

This article is extremely misleading. The author is a professor at the University of Strasbourg, so he knows what he's talking about; however, he words things

This article is extremely misleading. The author is a professor at the University of Strasbourg, so he knows what he's talking about; however, he words things in a weird way in an attempt to confirm some kind of overarching point that experimental chemistry is still necessary (which seems kind of obvious IMO). Quantum mechanics can and does predict every element on the periodic table. Whether we can calculate what it predicts at the present time is a different matter entirely. In the words of Peter Gill, "The field of quantum chemistry has become an applied mathematics problem". Yes, it's well-known that naïve quantum chemistry fails to predict experimental results for a variety of reasons, but all of these reasons come back to the fact that we had to use approximations to solve difficult equations that otherwise could not be solved. If our predictions failed in some way fundamental to QM/QFT, every theoretical physicist alive would be shocked — it would be the biggest news in the field in over a century. As better and better numerical approximation schemes are developed, the current issues with quantum chemistry will begin to disappear. Some people argue there are fundamental limits on what a classical computer can discover about quantum chemical systems, but then others argue that quantum computing will surmount these barriers (and others argue that it won't).

danbruc 10y ago
Some people argue there are fundamental limits on what a classical computer can discover about quantum chemical systems, but then others argue that quantum computing will surmount these barriers (and others argue that it won't). There is nothing a quantum Turing machine can compute that can not be computed by a classical Turing machine. However quantum Turing machines may be able to solve problems faster than classical Turing machines but, as far as I know, it is not known whether this is indeed the case. It is known that the speedup is at most exponential.
- Xcelerate 10y ago
  This is what I meant by "fundamental limits". I should have clarified that meant a practical computation time.
jpmattia 10y ago
> so he knows what he's talking about; I'm not so sure. For example: > The commutator of the Hamiltonian with the angular momentum of an electron does not vanish.8 Its eigenvalue is, therefore, not a constant of motion. In fact, the Hamiltonian is invariant under rotations (the electron does not care about which way the protons in the nucleus are rotated), which is why the angular momentum even exists. (From the fact that the Lagrangian is invariant under rotation: Noether's theorem tells you that there is a conserved quantity associated with that invariance. It turns out that rotational invariance gives rise to angular momentum.) I haven't checked his reference 8, but "Boston Studies Series in the Philosophy of Science" is not the first place I'd go to understand constants of the motion and how they relate to things that commute with the Hamiltonian.
- thisrod 10y ago
  The key word is an electron. The total angular momentum is conserved, but the way the electrons share it between them is not. As I recall, actual atomic Hamiltonians have terms like σ₁·σ₂. The reason involves some tricky correspondence between exchange and spin that I haven't thought about in a long time, and would have to look up. I agree with the broader point that this essay is odd and confusing.
- pdonis 10y ago
  > the Hamiltonian is invariant under rotations Yes, but that doesn't contradict what the article said. The article said, unpacking slightly, that the Hamiltonian operator does not commute with the "angular momentum of a single electron" operator. It only commutes with the "angular momentum of all the electrons taken together" operator.
GFK_of_xmaspast 10y ago
> Quantum mechanics can and does predict every element on the periodic table. Given that we can't confirm that, how do we actually know this?
- Xcelerate 10y ago
  Sorry. I should have just said "can". It can't be confirmed yet that it does for every element, but it would be very surprising if it didn't. A huge amount of computational power was used to calculate the gyromagnetic moment of the isolated electron to ten decimal places using QED, and it was found to exactly match those digits obtained experimentally. It's one of the greatest precision tests of QED ever performed. So there's not really any a priori reason to think that, were we given exponentially large computational resources, we couldn't compute any property of a compound or chemical reaction to as many decimal places as we desired (ignoring some details like the divergence of Feynman path integrals and the effects of quantum gravity, which are both expected to affect calculations at some far out point in their decimal expansion).
  - 21 10y ago
    It's quite a big leap from a single isolated electron to a system with hundreds of them (not to mention QCD interactions). While you are right that there is no a priori reason to dismiss that, it's not quite a bagged thing. Especially considering that we are missing some things for sure (neutrino mass, ...)
pdonis 10y ago
> Quantum mechanics can and does predict every element on the periodic table. Whether we can calculate what it predicts at the present time is a different matter entirely. If we can't calculate what it predicts, then we don't know that what it predicts matches what we actually observe. We might believe, with high confidence, that QM, once we can calculate what it predicts, will in fact predict what we observe; but until we've actually done it, we don't know for sure. [Edit: I see GFK_of_xmaspast already made this point.]

danbruc 10y ago

Some people argue there are fundamental limits on what a classical computer can discover about quantum chemical systems, but then others argue that quantum computing will surmount these barriers (and others argue that it won't). There is nothing a quantum Turing machine can compute that can not be computed by a classical Turing machine. However quantum Turing machines may be able to solve problems faster than classical Turing machines but, as far as I know, it is not known whether this is indeed the case. It is known that the speedup is at most exponential.

Xcelerate 10y ago
This is what I meant by "fundamental limits". I should have clarified that meant a practical computation time.

jpmattia 10y ago

> so he knows what he's talking about; I'm not so sure. For example: > The commutator of the Hamiltonian with the angular momentum of an electron does not vanish.8 Its eigenvalue is, therefore, not a constant of motion. In fact, the Hamiltonian is invariant under rotations (the electron does not care about which way the protons in the nucleus are rotated), which is why the angular momentum even exists. (From the fact that the Lagrangian is invariant under rotation: Noether's theorem tells you that there is a conserved quantity associated with that invariance. It turns out that rotational invariance gives rise to angular momentum.) I haven't checked his reference 8, but "Boston Studies Series in the Philosophy of Science" is not the first place I'd go to understand constants of the motion and how they relate to things that commute with the Hamiltonian.

thisrod 10y ago
The key word is an electron. The total angular momentum is conserved, but the way the electrons share it between them is not. As I recall, actual atomic Hamiltonians have terms like σ₁·σ₂. The reason involves some tricky correspondence between exchange and spin that I haven't thought about in a long time, and would have to look up. I agree with the broader point that this essay is odd and confusing.
pdonis 10y ago
> the Hamiltonian is invariant under rotations Yes, but that doesn't contradict what the article said. The article said, unpacking slightly, that the Hamiltonian operator does not commute with the "angular momentum of a single electron" operator. It only commutes with the "angular momentum of all the electrons taken together" operator.

GFK_of_xmaspast 10y ago

> Quantum mechanics can and does predict every element on the periodic table. Given that we can't confirm that, how do we actually know this?

Xcelerate 10y ago
Sorry. I should have just said "can". It can't be confirmed yet that it does for every element, but it would be very surprising if it didn't. A huge amount of computational power was used to calculate the gyromagnetic moment of the isolated electron to ten decimal places using QED, and it was found to exactly match those digits obtained experimentally. It's one of the greatest precision tests of QED ever performed. So there's not really any a priori reason to think that, were we given exponentially large computational resources, we couldn't compute any property of a compound or chemical reaction to as many decimal places as we desired (ignoring some details like the divergence of Feynman path integrals and the effects of quantum gravity, which are both expected to affect calculations at some far out point in their decimal expansion).
- 21 10y ago
  It's quite a big leap from a single isolated electron to a system with hundreds of them (not to mention QCD interactions). While you are right that there is no a priori reason to dismiss that, it's not quite a bagged thing. Especially considering that we are missing some things for sure (neutrino mass, ...)

pdonis 10y ago

> Quantum mechanics can and does predict every element on the periodic table. Whether we can calculate what it predicts at the present time is a different matter entirely. If we can't calculate what it predicts, then we don't know that what it predicts matches what we actually observe. We might believe, with high confidence, that QM, once we can calculate what it predicts, will in fact predict what we observe; but until we've actually done it, we don't know for sure. [Edit: I see GFK_of_xmaspast already made this point.]