It seems that the eigenvalue problem rules out the possibility of E=0?

[BHL 20]

TL;DR Summary

Since the eigenvalue problem can't distinguish between a non-existent wavefunction (and therefore a non-existent particle), and the energy being zero?

Since the eigenvalue problem can't distinguish between a non-existent wavefunction (and therefore a non-existent particle), and the energy being zero. This is the next thing that has started bothering me on my journey to understand quantum mechanics.

For example, in the algebraic derivation of the ground state of the harmonic oscillator, it is assumed that the ground state has that wavefunction which vanishes when acted upon by the lowering operator. In the video I linked below this step is taken around the 12.00 minute mark.

But writing that a_-ψ₀=0 guarantees that this ground state wavefunction will have E>0. Because if it had E=0, then it would itself have a zero on the right side of its eigenvalue equation. So it seems that because the eigenvalue problem can't tell apart a non-existent particle from a zero energy particle, any quantum system with discrete energy levels will have a ground state with E>0.

Maybe this is a stupid question, and maybe this is an intentional effect of the formalism. But I'd like to know if my reasoning here is correct. And if this is inherent to the formalism of quantum mechanics, what is the justification for it?

 

[Dale]

the eigenvalue problem can't distinguish between a non-existent wavefunction (and therefore a non-existent particle), and the energy being zero

Neither can I.

 

[Gaussian97]

Since the eigenvalue problem can't distinguish between a non-existent wavefunction (and therefore a non-existent particle), and the energy being zero. This is the next thing that has started bothering me on my journey to understand quantum mechanics.

A general eigenvalue problem is stated as finding a non-zero state fulfilling the equation $A\left|\psi\right>=\lambda \left|\psi\right>$
This means of course that the equation itself is not enough, you always need to assure that your state is not the zero one (the neutral element of the addition). I don't know what do you mean about energy zero, the zero state has no definite energy since is not a physical state being non-normalizable.

For example, in the algebraic derivation of the ground state of the harmonic oscillator, it is assumed that the ground state has that wavefunction which vanishes when acted upon by the lowering operator. In the video I linked below this step is taken around the 12.00 minute mark.

That the lowering operator vanishes when acting on the ground state is not an assumption, is something you can prove.

But writing that a_-ψ₀=0 guarantees that this ground state wavefunction will have E>0. Because if it had E=0, then it would itself have a zero on the right side of its eigenvalue equation. So it seems that because the eigenvalue problem can't tell apart a non-existent particle from a zero energy particle, any quantum system with discrete energy levels will have a ground state with E>0.

Not true, look at the Hamiltonian $H=a^\dagger a$, the ground state has energy 0, there's no problem at all.

 

[BHL 20]

A general eigenvalue problem is stated as finding a non-zero state fulfilling the equation $A\left|\psi\right>=\lambda \left|\psi\right>$
This means of course that the equation itself is not enough, you always need to assure that your state is not the zero one (the neutral element of the addition). I don't know what do you mean about energy zero, the zero state has no definite energy since is not a physical state being non-normalizable.

I'm just wondering why lecturers and textbooks make such a big deal at the end of a derivation when the ground state energy comes out to be zero. "In classical physics we can have a harmonic oscillator at rest with zero energy, but in quantum mechanics it must always have some energy." If it cannot be any other way, why is it presented as a surprising result?

That the lowering operator vanishes when acting on the ground state is not an assumption, is something you can prove.

In the derivation I linked, it is used as the assumption from which to derive the ground state eigenfunction for the system. Is it something you can prove without first having the expression for this eigenfunction?

Not true, look at the Hamiltonian $H=a^\dagger a$, the ground state has energy 0, there's no problem at all.

Sorry, I don't understand this example.

 

[Gaussian97]

I'm just wondering why lecturers and textbooks make such a big deal at the end of a derivation when the ground state energy comes out to be zero. "In classical physics we can have a harmonic oscillator at rest with zero energy, but in quantum mechanics it must always have some energy." If it cannot be any other way, why is it presented as a surprising result?

The surprising result is that the minimum energy for the quantum HO system is greater than the minimum energy for the classical HO. If we take this last one as the 0 by definition then yes, you will have that the energy for all the quantum states will be positive. Whether this causes you surprise or not is purely subjective, but I think that lots of times it is mentioned as a connection with Heisenberg's uncertainty principle.
Anyway, it has nothing to do with the energies being positive in general and nothing to do with the equation $a\left|0\right>=0$.

In the derivation I linked, it is used as the assumption from which to derive the ground state eigenfunction for the system. Is it something you can prove without first having the expression for this eigenfunction?

The proof is usually something like this:
We prove that the eigenvalues of our Hamiltonian are bounded from below.
We find an operator that maps eigenstates to other eigenstates with the property that the energy of the latter is strictly lower than the initial energy.
Therefore such an operator need necessarily to annihilate the ground state.

Sorry, I don't understand this example.

What you don't understand? Is a Hamiltonian with a spectrum $E=0,1,2,3,...$ which clearly shows that there's nothing forbidding a state with $E\leq 0$

 

[kith]

Maybe it helps to look at a simple example from linear algebra. What are the eigenvalues and eigenvectors of the following matrix?
$$\left( \begin{array}{rrr}
1 & 0 \\
0 & 0 \\
\end{array}\right)$$