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When looking back at my courses of quantum mechanics, I noticed that assumptions about the ground state of a quantum mechanical system where rather vague and unprecise. It is always assumed that a ground state exists and that it had a finite energy. So my questions are the following:

  1. Should every quantum mechanical system have a ground state? And how can we be sure of this?

  2. Should this ground-state have a finite energy?

  3. Or is an energy of $-\infty$ also allowed?

A neat proof for the ground state is given in "Introduction to Quantum Mechanics" by D. J. Griffiths for the simplest system (problem 2.2). There it is shown that if you have an energy-eigenstate $\psi(x)$ (working in the position space for simplicity) with energy $E$, so: $$\hat{H}\psi(x)=E\psi(x),$$ where we consider a simple non-relativistic point-particle, so $$\hat{H}=\hat{T}+\hat{V}.$$By applying this on this on the equation for the eigenstate $\psi(x)$ this yields:$$\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right]\psi(x)=E\psi(x),$$ or (after some simple rewriting) as:$$\frac{\partial^2\psi(x)}{\partial x^2}=\frac{2m}{\hbar^2}\left[V(x)-E\right]\psi(x).$$In the case that we would have an energy which is lower than the minimum of $V(x)$ (assuming that the potential has a minimum), the wave-function would be non-normizable since $\psi(x)$ and $\partial_{xx}\psi(x)$ whould have the same sign. This is because of the fact that $\psi(x)$ can only have a minimum if it's positive and a maximum if it's negative, which yields the non-normalizable character.

  1. So this still leaves me with the question on how they know it for sure for potentials of the Coulomb-kind?

  2. and how they do this in quantum field theory (and, by extension, classical physics)?

I know this question is in the same line of reasoning as (Why does a system try to minimize potential energy?) kindoff. But what I'm looking for is not "Why does everything tend to a minimal energy". But more "Why do we assume such a minimal energy-state exists?".

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The Hamiltonian of good physical systems is always assumed to be bounded from below, that is, there is a state with lowest energy, the ground-state. Because you can always shift all the energies (in non-relativistic QM), you could shift all energies by $-\infty$ in principle, though the spacing between the ground-state and the excited states would stay the same.

Not having a ground-state would imply some kind of instability in the system. For instance, non-interacting bosons in the grand-canonical ensemble at zero-temperature do not form a good system at positive chemical potential: you can put an infinite number of bosons in the state of zero momentum, which implies an infinite density and an infinite (negative) energy. This is not a sound state of matter.

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  • $\begingroup$ So more from the reasoning of statistical physics, where for example if T→0 the system should go to some kind of steady state ? Because If I think of a Coulomb-system, the potential can go to $-\infty$, doesn't that mean that the energy can keep on decreasing so that the ground-state would be undefined ? $\endgroup$ – Nick Jun 8 '14 at 0:39
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    $\begingroup$ Be careful that quantum mechanically, it is not because the potential goes (naively) to $-\infty$ that the groundstate energy goes to $-\infty$. That's exactly how QM solves the atom stability. The boundedness of the Hamiltonian is not the same that that of the potential. $\endgroup$ – Adam Jun 8 '14 at 6:40
  • $\begingroup$ so clasically we shouldn't always have a ground-state ? Yes indeed, I had to solve the hydrogen-atom in my second year and there the electron did not crash into the proton due to quantum mechanics. But I was wondering how we can be sure of this ground state? I added a proof in my question for simple systems. But I'm still unsure how we can say it's true in general. Of course a system without a ground state might be silly from a physical point of view. I was perhaps wondering more about the mathematical. Why can't you shift your energies in relativistic quantum mechanics ? $\endgroup$ – Nick Jun 8 '14 at 10:21
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    $\begingroup$ In special relativity, the energy is not independent, you have to think of it as part of a 4-vector, with the momentum. So you can't just shift the energy without changing the momentum. $\endgroup$ – Adam Jun 8 '14 at 16:20
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    $\begingroup$ @Nick: Well, no. I depends of the (physical) system. Having $N$ atoms in a closed box is not the same than having a fluctuating number of bosons due to a reservoir of particle. In the thermodynamic limit (if it is well defined! and it's not the case here in the GCE), both would give the same answer for standard problem, but that's not necessarily true. For example, long range interaction makes the different ensembles give different answer (for example, gravitational systems). $\endgroup$ – Adam Jun 9 '14 at 13:21
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I know it's been asked 2 years ago, but I want to answer to help anyone that finds this.

As far as I understand:

Let's assume $E=-\infty$ states are not allowed because then all particles would probably want to be in this system if they come into contact with it, as it's the global and local minimum of energy in the whole world. You could also get infinite energy out of it, and we know that's not ok according to thermodynamics. Thus it would be unstable and unphysical.

So we are assuming the minimum energy a particle can have is $E>-\infty$, and as such, there must be a lower limit to the energy. If a particle is bound in this system, it has quantized states. But if there is a lower limit to the energy we can have, and the energies are quantized, this means that there has to be one or more states (or going towards a continuum) with the lowest energy $E=E_{0}>-\infty$. So here is your minimum in energy.

Also, this problem is solved by Dirac's equation which does not allow negative energies (if you get negative energies, then you actually have an antiparticle with positive energy).

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  • It is observed that the quantum mechanical oscillator always have a ground state energy of *

h cross omega by 2

* and it is necessary we have the concept of

zero point energy

in the oscillator(quantum) and if the quantum systems becomes large then it is seen as the classical approximation of the oscillator.

you can even find the examples in the scattering theory shows a existence of the ground state energy.

the concept of the negative energy is actually solved by the Dirac for the interpretation of negative energy states.

If you are familiar with the angular momentum of qm we find that spin is not included and in the dirac equations the spin is readily comes while deriving the spin of a particle.

this is the advantage of the Dirac equation which is the linear equation and the schrodinger representation is one of the easiest method to understand the physical state of the state.

the Dirac equation fully solves the problem and it is in relativistic qm that the spin and the physical significance of the negative energy states are explained. pls study the relativistic quantum mechanics carefully. in fact the Dirac solved successfully to get the linear equation to complete the full form of the quantum mechanics.

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    $\begingroup$ I'm aware of the Dirac-problem and so on. I'm mostly wondering why we can assume that we have a ground state. For the case of the harmonic oscillator the proof is given above. But for example for other systems this is not so clear to me :(. $\endgroup$ – Nick Jun 8 '14 at 20:08
  • $\begingroup$ So it's more a kind of existence-reasoning/proof that I'm looking for. $\endgroup$ – Nick Jun 8 '14 at 20:18

protected by Qmechanic Nov 11 '16 at 10:30

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