# Why should energy have a lower bound but not an upper bound?

This question seems to be naive, but I really want some intuitive way of understanding the reason.

Consider the harmonic oscillator problem in quantum mechanics, there are the creator (operator) $a^\dagger$ and the annihilator (operator) $a$. The lower bound of energy exists due to the fact that there is a vacua state which is invariant under annihilator $\left|0\right>$:

$$a \left|0\right> = 0$$

(while there is no such state for the creator) therefore the lowest energy state would be $\left|0\right>$ - a lower bound.

Is there a more general explanation or principle that demands the existence of lower bound of energy?

Is there any case there is a upper-vacua state (or should I say: saturated state)?

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$$a \left|0\right> = 0$$ not $$a \left|0\right> = \left|0\right>.$$ See a related question physics.stackexchange.com/q/112807 –  user1906035 Jul 6 '14 at 9:24
yeah thanks for the correction! –  Mr.T Jul 6 '14 at 9:45
What do you mean by trivial explanations? Is experimental evidence trivial? Because as far as I know there is no experimental evidence of completely unstable systems with energy unbounded from below. Mathematically it is possible to define an unbounded from below self-adjoint operator, but then you should be able to attribute some meaning to it as an Hamiltonian. –  yuggib Jul 6 '14 at 10:01
If one considers experimental evidence is the origin of a fact, that fact should be made an axiom or a principle. –  Mr.T Jul 6 '14 at 10:04

I can think of two reasons why we need a lower bound, one statistical, one inuititive.

First, the intuitive: The annihiliation/creation operators represent adding/removing particles (or excitations, or whatever). The vacuum state as lowest energy state represents obviously the empty state from which no further particles can be removed. It is also clear (in the case of bosons) that I can just keep adding particles and nothing stops me from doing that, so an upper bound of energy would mean that there is a state $|\text{max}\rangle$ on which the creation operator acts as $a^\dagger|\text{max}\rangle = 0$. This is, from an intuitive point of view, obviously nonsense. Why should adding a particle destroy all the others? Where has all the energy gone? Demanding a general upper bound for energies is thus a bad idea. [I want to stress again that this is only heuristic. It is not meant to translate to a rigourous argument directly.]

Now, the statistical. We know that the time evolution drives systems towards the state with the lowest (free) energy. If the Hamiltonian is not bounded below, there is no global minimum of the energy, and thus there are only metastable states. But since there is no ground state, such a system would be able to drop to ever lower energy levels. In consequence, a system not bounded below would be able to radiate energy infinitely. That is obviously nonsense, no physical system can hold an infinite amount of accessible energy. So though the physics can describe such a system, its time evolution behaviour is not something we observe in the real world.

Both arguments do not prohibit that there are systems with upper bounds. The easiest example are pure spin systems: Take a number of particles with spin an put them in a (constant) magnetic field. Now the lowest energy state is when all spins are flipped one way, and the highest energy state is when all spins are flipped the other way. You cannot add further energy to this system, it is naturally bounded below and above.

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I think $a$ and $a^\dagger$ are pretty somewhat mathematically equivalent, after all one is adjoint of the other (before we interpret them as annihilator and creator). What is the main different that make them different? –  Mr.T Jul 6 '14 at 13:39
Adjointness is not equivalence. Their fundamental commutation relation is $[a,a^\dagger] = 1$ (disregarding annying deltas). From this alone it follows that if one is the creation operator, the other is the annihilation operator. If adjointness were somehow close to equivalence, then self-adjointness would not be such a powerful criterion for having a real spectrum. –  ACuriousMind Jul 6 '14 at 13:42
I didn't mean equivalent in that sense. I mean they are like i and -i, different, but they are square roots of -1. And we choose the signal i just by convention. –  Mr.T Jul 6 '14 at 13:53

1) Operator a is called anihilator and so on. 2)The answer depends on a point of view. Formally,one can consider a charged particle in a homogenous magnetic field and treat problem relativistically using Dirac equation. In this case the harmonic oscillator ladder of energy levels does not have lower bound of energy. If you agree that Dirac equation is a proper one, your question is meaningless, I think. 3) Is this a reliable website? Who is allowed to change other people questions? When I saw the question first time there was a mistake in naming operators. Next somebody changed the names so they are now proper and the information about who did that dissapeared!

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I just point out a possible explanation of the lower bound and I don't say that it is true. The lower bound of energy, as I see, is prevalent in physics: the spectral of Hydro atom, the spectral of harmonic oscillator,... If the existence of lower bound is not necessary, why does it exist everywhere? –  Mr.T Jul 6 '14 at 9:35
The calculation is just for describing. I'm not good at math enough to study all the complicated theory. I just want to have an idea. Everyone is able to edit, but the origins still exist if one wants to see. –  Mr.T Jul 6 '14 at 9:40