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In quantum mechanics, quantum chemistry, and condensed matter physics courses, it seems like finding the ground state energy of a Hamiltonian is incredibly important. I can understand why having the ground state wave function (or behaviors of the wave function) could help, but I’m struggling to understand how the ground state energy helps us.

In Quantum Computing pitches by huge companies, I always hear something along the lines of “Quantum Chemistry simulations will allow for breakthroughs in the pharmaceutical industries,” but the quantum simulation algorithm in quantum computing they refer to is the Variational Quantum Eigensolver, which returns an estimate for the ground state energy. How does knowing this energy tell us anything about the material? I’m struggling to bridge this gap.

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    $\begingroup$ To understand the excited states, you need to understand the ground state. If you can't get the ground state correct, why would you believe any excited states? $\endgroup$
    – Jon Custer
    Jun 10 at 19:47
  • $\begingroup$ I can understand this argument for the states themselves, but does the ground state energy on it's own tell us anything? $\endgroup$
    – Jlee523
    Jun 10 at 22:11
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    $\begingroup$ DFT for narrow-gap semiconductors quite often results in a ground state with a negative band gap - that is, DFT says it is a metal. So, how do you trust any DFT results on the material? Well, you don't until you get a proper ground state. In other situations, what you 'know' the ground state to be might not be what it actually is, particularly for complex systems with long range interactions. $\endgroup$
    – Jon Custer
    Jun 10 at 23:13
  • $\begingroup$ So is the claim that knowing the GS energy tells us nothing, other than the fact that we can use the results to check against our other calculations? As in, knowing the true GS allows us to use and trust other computational methods? $\endgroup$
    – Jlee523
    Jun 10 at 23:26
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    $\begingroup$ Minimising the ground state energy WRT geometry gives us the equilibrium geometry, which is pretty important. Further Calculation of the ground state in a quantum calculation doesn't just give us the Energy, it gives us an (approximation to) the wavefunction as well, from which any ground state property can be calculated. Further perturbation theory of various forms applied to the ground state gives us a way into excited state properties, but you need the ground state first. $\endgroup$
    – Ian Bush
    Jun 25 at 8:50
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The ground state energy is used ubiquitously throughout various computational chemistry disciplines. I suppose a single ground state energy isn't particularly useful, but that's because all we can measure experimentally are relative energies.

  • We can probe the thermodynamic stability of a reaction (or set of reactions) by determining the energy difference between reactants and products. If you want to know if a drug will actually bind to a target protein/membrane site, you need to determine if the bound complex is actually stable relative to the separated form.

  • Differences in energy between reactants/products and transition states gives us insight about the kinetics of a process. If you want to know the timeframe for a drug to take effect or break down in the body, you need have some understanding of it's reaction kinetics.

  • A number of molecular properties (dipoles, polarizabilities, forces, etc) can be determined as derivatives of the ground state energy. These require either having an analytical formula for the ground state energy which can be differentiated or, for more complex methods where such a formula isn't available, using numerical/automatic differentiation, which requires evaluating the ground state energy (multiple times in the case of numerical derivatives). Molecular dynamics studies of diffusion of a drug in the blood or through a cell membrane require knowing the forces on the molecule.

  • The most common experimental observable for an excited state is the excitation energy, which to calculate requires both the energy of that state and the ground state. We don't really know the excited state energy experimentally, we only know it's energy relative to the ground state.

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    $\begingroup$ Speaking as a code developer one of the most important uses of the Energy is as the very first hurdle code has to pass before you actually believe the results - "if you can't get that right how can you get anything else right?" tends to be the view. $\endgroup$
    – Ian Bush
    Jun 26 at 15:43
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In addition to the excellent (thermodynamic) arguments listed by Tiberius, there is also a more technical reason why you want to know the energy. The variational principle implies that a trial wave function is a better approximation of the real wave function if it has a lower energy. Most methods in quantum chemistry take an expansion of eigen functions and try to find the combination of expansion coefficients that give the lowest overall energy in accordance with the variational principle. You therefore get the energy from the process of finding the best representation of your wave function.

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