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Yes, a classical system would have a definite energy ##E##. And, in general this energy could be any number.Hazel Appraiasal said:H
Hmm, so let me get this straight
In QM, a state of a quantum system is denoted by the symbol "psi" or ##\psi## or ##|\psi \rangle##.
A quantum state can be in a state where they're at that point it just denoted by ##\psi = \psi_5## or ##\psi = \psi_{16}## for example (I don't know if the importance of the numbers at the bottom)
But when it is in a superposition of states it, for example, could be in
$$\psi = \psi_1 + \psi_2$$
$$\psi = 2\psi_1 - \psi_4 + 3\psi_8$$
And the numbers next to the psi symbols are the possibilities you will get that state just like in your example
##\psi = 2\psi_1 - \psi_4 + 3\psi_8## would be ##4, 1## and ##9##. In other words, on average out of ##14## measurements of energy in a system in that state, we would get ##E_1## four times, ##E_4## once and ##E_8## nine times.
Correct me if I am wrong but this is what I'm getting
Some quantum systems - like an atom - may only take certain energy values. There is a lowest possible energy ##E_1##, then a next energy level ##E_2## etc. These are called energy eigenstates.
That property is called discreteness. I.e. energy can only take certain quantised values.
Quantum systems may also be in a superposition of many energy eigenstates. They do not have a definite energy until measured. That's a second characterictic that is not found in classical systems.