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Apolgies in advance if this is a confusing ramble and multitude of questions, I'm not quite sure how to articulate myself. I am currently reading up on quantum mechanics and seem to have confused myself when trying to understand the difference between a discrete spectrum and continuous spectrum.

I understand the basic starting example- points on a line lying at integer values with spacing $\epsilon$ vs when $\epsilon \to 0$. I can also see how the mathematics involved differs- summation vs integration (limit of this summation).

Where I believe my confusion begins is when dealing with real examples. Am I correct in assuming that there are no conditions on the possible values of position (eigenvalues can take any values) which is why when normalising a wave-function dependent on time and position, we integrate over all of space?

However angular momentum, spin, energy- are all quantised, so are only able to take discrete values. Does the time-independent Schrodinger equation only deal with discrete values for energy? Is this the difference between continuous and discrete- simply whether there are constraints placed on the eigenvalues and hence the wave function?

And how does this relate to the potential energy and its implications of bounded vs scattering states? Are bounded states analogous to discreteness?

If anybody could clear up any of the above I'd greatly appreciate it. I thought I had some grasp on the topic until I started thinking harder.

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Where I believe my confusion begins is when dealing with real examples. Am I correct in assuming that there are no conditions on the possible values of position (eigenvalues can take any values) which is why when normalising a wave-function dependent on time and position, we integrate over all of space?

Typically yes. Of course, you could consider restricted spaces (e.g. the particle in a box) in which case this is no longer true; you could also discretize the position variable to model e.g. a particle hopping on a lattice.

Does the time-independent Schrodinger equation only deal with discrete values for energy?

No. The free-particle Hamiltonian on the domain $L^2(\mathbb R)$ is simply $$\big(H_{free}\psi\big)(x) = - \frac{\hbar^2}{2m} \psi''(x)$$ and the spectrum of this Hamiltonian is $\sigma(H_{free})=[0,\infty)$. As another example, the spectrum of the Hamiltonian of the Hydrogen atom is $$\sigma(H_{Hydrogen}) = \left\{-\frac{13.6 \mathrm{ eV}}{n^2} \ | \ n = 1,2,3\ldots\right\}\cup [0,\infty)$$ which is comprised of both a discrete part and a continuous part.

Is this the difference between continuous and discrete- simply whether there are constraints placed on the eigenvalues and hence the wave function? And how does this relate to the potential energy and its implications of bounded vs scattering states? Are bounded states analogous to discreteness?

It's hard to give a one-size-fits-all answer to this. However, it is worth noting that an element of the discrete part of the spectrum is called an eigenvalue and corresponds to a (normalizable) eigenfunction; such an eigenfunction is necessarily spatially localized, and corresponds physically to a bound state. On the other hand, elements of the continuous part of the spectrum are called generalized eigenvalues and correspond to (non-normalizable) generalized eigenfunctions which are not spatially localized and correspond physically to scattering states.

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  • $\begingroup$ Thank you for responding! So the equation $H\psi_{n}(x)=E_{n}\psi_{n}$ which I had been referring to as the TISE equation is actually just the discrete spectrum case and the general form can be used for a continuous spectrum case such as the free particle. Is this also the case for potential steps and tunnelling? I understand the infinite potential well case is discrete due to the boundary conditions. Finally, are these cases where the connection with the potential energy comes from? As I have read, whether a state is bounded can be determined by the potential value at $\pm\infty$. $\endgroup$
    – rileygrey65
    Commented May 17, 2022 at 21:00
  • $\begingroup$ @rileygrey65 The equation $H\psi_n=E_n\psi_n$ is indeed the TISE. For a particle on a line, $\psi$ is a function of $x$ and $H$ is a differential operator usually of the form $H:=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)$, where $V$ is the potential. As a result, $H\psi=E\psi$ becomes a differential equation. It's possible that some solutions to this differential equation are square-integrable, in which case we call them eigenfunctions; if they are not, they may be generalized eigenfunctions which are useful tools but not strictly valid states in their own right. $\endgroup$
    – J. Murray
    Commented May 17, 2022 at 21:29
  • $\begingroup$ @rileygrey65 Generally, if $E>V(\pm \infty)$ then the corresponding wavefunction is not normalizable and it corresponds to what we'd call a scattering state. If $E<V(\pm \infty)$ then it corresponds to what we'd call a bound state. $\endgroup$
    – J. Murray
    Commented May 17, 2022 at 21:34

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