On the "spectrum" of an operator in quantum mechanics Very simple question, I'm new to this. I'm reading Griffiths book on QM and have a question about the "spectrum" of an observable operator. Does the spectrum of an operator require specification of a particular system? Or is the spectrum of an operator just every possible eigenvalue that can be obtained by every possible eigenfunction of an operator?
 A: In general, the spectrum of an operator depends on the Hilbert space (= space of possible wavefunctions) on which it is defined. For example, the Hamiltonian for a free particle in one dimension $H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$ has eigenvalues that depend on the boundary conditions for the wavefunction $\psi$. In a box (= infinite potential well) of length $a$, the eigenvalues are $E_n = \frac{\hbar^2}{2m}\left(\frac{n\pi}{a}\right)^2$ which depend on $a$, even though the expression for the operator does not. For the same operator without any boundary conditions, any real number is an eigenvalue.
The Hamiltonian is a special operator because it defines the system. A "system" is nothing more than a choice of Hilbert space and a Hamiltonian defined on it. The spectrum of other operators, such as position $x$ and momentum $p = \frac{\hbar}{i} \frac{\partial}{\partial x}$, depends on the Hilbert space but not on the Hamiltonian.
A: Each operator has it's own spectrum. For the hamiltonian the spectrum is the set of allowed energy levels. This depends on masses and potentials.
A: The spectrum of a particular operator is the set of all possible eigen values. 
In regards to the other answer: since the Hamiltonian of a system has a different form for different systems, also the spectrum of that Hamiltonian will vary with different systems. So talking about the spectrum of a Hamiltonian is equivalent to “the set of all possible energies a system can have”. However, once you specify that it’s the Hamiltonian of a harmonic oscillator, this fixed the shape of $H$ and therefore its spectrum. 
