I Can the Time Independent Schroedinger Equation Be Used to Find Unbound States?

[TeethWhitener]

Even then, it is usually called a hydrogen atom only when the electron is bound to the nucleus, not when it is light years away from it.

Even the ground state of the hydrogen atom has nonzero probability of the electron being light-years away.

Moreover, the space spanned by all bound states (i.e., what you'd like to call ''bound'') is only a small (nondense) subspace of the Hilbert space. A (normalized) state of the hydrogen atom contains (under any reasonable distribution) with probability one a nontrivial contribution form the (unbound) scattering states.

I'm unfamiliar with these two statements (Edit: for the first one, I'm assuming you're referring to pure vs mixed states). Do you mind giving me some resources so that I can understand them better?

There are even lots of (normalized) states whose projection to the bound state sector is zero - against your intuition.

I'm thinking of free wavepackets, but are there others (which aren't superpositions of scattering states)?

Therefore it is better to get used to the established terminology rather than complain about the difference in meaning between intuition and the formal concept.

I would say it's better to understand why the established terminology was chosen rather than simply to get used to it. Which is what I'm trying to do.

 

[A. Neumaier]

I'm unfamiliar with these two statements (Edit: for the first one, I'm assuming you're referring to pure vs mixed states). Do you mind giving me some resources so that I can understand them better?

Everything refers to pure states.

One learns in functional analysis that for a selfadjoint operator $H$ with discrete and continuous spectrum, the Hilbert space is the direct sum of two orthogonal closed invariant subspaces, one spanned by the bound states, on which $H$ has a purely discrete spectrum, and its orthogonal complement, on which $H$ has a purely continuous spectrum. An arbitrary wave function is a sum of two wave functions from each subspace; if both are normalizable, the result is normalizable. A typical state has contributions from both subspaces,

Even the ground state of the hydrogen atom has nonzero probability of the electron being light-years away.

So your idea of calling it a bound state is even less convincing.

I'm thinking of free wavepackets, but are there others (which aren't superpositions of scattering states)?

There are lots of normalized superpositions of scattering states, which according to your proposal (bound = normalizable) should be called bound states, though they are not bound in any meaningful sense.

 

[TeethWhitener]

Everything refers to pure states.

Ok then I'm definitely not familiar with it.

So your idea of calling it a bound state is even less convincing.

Are you asserting that the ground state of the hydrogen atom isn't a bound state? But it's an eigenstate of the Hamiltonian! (without the phase factor)

There are lots of normalized superpositions of scattering states, which according to your proposal (bound = normalizable) should be called bound states, though they are not bound in any meaningful sense.

Yes, I agree with this. In post 14 above, I mentioned that they should be normalizable for all time (as $x,t\to\pm\infty$), which I think should exclude free wavepackets based on the presence of a factor of $e^{(\infty-\infty)}$ in the probability density (that is, the limit doesn't exist for free wavepackets). But I don't know what the formalism for that is. (Edit: this probably doesn't work: by evaluating the limit along the line as x=vt, you still get zero).

 

[DrClaude]

@TeethWhitener : If you can get your hands on Messiah's book, it is explained in Chapter 2, III-17.

Bound and unbound states are qualifications given to eigenstates. An eigenstate is bound if $\int | \psi (\mathbf{r}) |^2 \mathrm{d}\mathbf{r}$ converges, and thus the probability of finding the particle at infinity is 0.

 

[TeethWhitener]

@TeethWhitener : If you can get your hands on Messiah's book, it is explained in Chapter 2, III-17.

Bound and unbound states are qualifications given to eigenstates. An eigenstate is bound if $\int | \psi (\mathbf{r}) |^2 \mathrm{d}\mathbf{r}$ converges, and thus the probability of finding the particle at infinity is 0.

Thanks, I'll see what I can do. I don't have a problem with eigenstates being bound. My misgiving was about superpositions of bound eigenstates not being called bound states:

Moreover, the space spanned by all bound states (i.e., what you'd like to call ''bound'')

This was basically right. I was calling "bound" the space spanned by all bound states. Since this space is closed, all of the states therein will converge (and be "bound by the potential" in the spirit of post 28). My assertion that bound states are precisely the normalizable states (post 21) was wrong, because of the existence of normalizable wavepackets made up of superpositions of the continuum states. I'm still not sure whether they (wavepackets, that is) remain normalizable out to $t\to\infty$. The simple limit of the probability density goes to zero for all $\mathbf{x}$ as $t\to\infty$, meaning the state is non-normalizable at infinity, but this is not rigorous.

 

[A. Neumaier]

I'm still not sure whether they (wavepackets, that is) remain normalizable

The norm is preserved by the dynamics, so normalizable at one time implies it at all times.

I was calling "bound" the space spanned by all bound states.

But this is not an interesting space as once you perturb the Hamiltonian you will get contributions from the scattering part even in your perturbed bound states.

The notion of a bound state is useful only for eigenstates, and this is why it is traditionally applied only to this case.

 

[TeethWhitener]

But this is not an interesting space as once you perturb the Hamiltonian you will get contributions from the scattering part even in your perturbed bound states.

Isn't this true for all states (even eigenstates)? I don't see how this statement makes the notion of a bound state more useful for eigenstates than for superpositions of eigenstates.

EDIT: I'm content to let things lie here. I'll simply change my usage to match standard usage and stop complaining.

 

[A. Neumaier]

Isn't this true for all states (even eigenstates)? I don't see how this statement makes the notion of a bound state more useful for eigenstates than for superpositions of eigenstates.

The point is that there is virtually no use for the class of superpositions of bound states, hence there is no point in giving these a special name. on the other hand, the discrete eigenstates are very important for many reasons and hence deserve having a short name for them, to facilitate communication.

 

[A. Neumaier]

Isn't this the definition of stationary states, not bound states?

These two concepts just differ by a phase; see the explanation in post #24.

 

[George Jones]

Here is my impression of common usage for "bound state" in standard (non-rigourous) quantum mechanics texts.

The energy spectrum for a Hamiltonian can have two parts, a continuous part, and a discrete part.

Bound states are stationary states that correspond to energies in the discrete part of the spectrum, and are normalizable.

Scattering states are stationary states that correspond to energies in the continuous part of the spectrum, and are not normalizable (or are delta function normalizable).

The stationary states of a harmonic oscillator are all bound states. The finite square well has both scattering and bound stationary states.

 

[George Jones]

To back up my previous post, I have attached a couple of pages from the third edition of "Quantum Mechanics: Foundations and Applications" by Arno Bohm, which is an advanced, somewhat rigourous text (e.g., it treats rigged Hilbert spaces).

From the second of these pages:

But the spectrum $\left\{ E_\alpha \right\}$ of $H$ is the combination of a continuous spectrum ... and a discrete spectrum ... Physically the continuous spectrum corresponds to scattering states and the the discrete spectrum to bound states

I have several other standard quantum texts that say similar things.

 

[vanhees71]

I admit, I am not familiar with the definition of bound states as eigenstates and only eigenstates. If that is the case, then you're right: superpositions of eigenstates are not, in general, eigenstates. I always considered a state to be "bound" if it had a vanishing chance of escaping to infinity (in a more precise sense, this would mean a bound state is normalizable as $x,t\to\pm\infty$--this properly excludes free wavepackets). But I could very well be using an incorrect term.

Indeed the common terminology is that "bound states" are the normalizable eigenstates of the Hamiltonian of the system under consideration. Since they are normalizable (or, in the position representation, the corresponding wave function is square-integrable) they represent true pure states of the system and since they are eigenstates of the Hamiltonian the time-dependence is just a phase factor $\exp(-\mathrm{i} E t/\hbar)$, where $E$ is the eigenstate. The eigenvalues of the bound states are in the discrete part of the Hamiltonian's spectrum. There can also be "scattering states", which are generalized eigenstates (distributions) of the Hamiltonian with the spectral value in the continuous part of the spectrum. These you can "normalize only to a $\delta$ distribution". The corresponding wave functions behave like oscillating exponential functions for large distances, and thus describe scattering states of particles, i.e., far away from the scattering potential the particles behave nearly like free particles ("asymptotic free states").