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

[fog37]

Hello Forum,

Just checking my correct understanding of the following fundamental concepts:

Stationary states: these are states represented by wavefunctions $\Psi(x,y,z,t)$ whose probability density function $|\Psi(x,y,z,t)|^2 = |\Psi(x,y,z)|^2$, that is, the pdf is only a function of space and does not vary in time at different points in space.

Bound states: these are states described by wavefunctions $\Psi(x,y,z,t)$ that are nonzero only within a certain region of space and zero everywhere else.

• It is surely possible to have bound stationary states. But is it possible to have bound nonstationary states? I don't think so.
• What about unbound stationary states? I don't think so either since the unbound wavefunction would evolve.
• The time independent Schroedinger equations (TISE) is solved to find bound and stationary states. Is it possible to use the TISE and its solutions to find the wavefunction for an unbound state, either stationary or nonstationary? The TISE presumes that the wavefunction is separable and I wonder if separability is only a feature of stationary and bound states...

Thanks,
Fog37

 

[TeethWhitener]

Bound states: these are states described by wavefunctions Ψ(x,y,z,t)Ψ(x,y,z,t)\Psi(x,y,z,t) that are nonzero only within a certain region of space and zero everywhere else.

No. Think of the ground state of the hydrogen atom: it's a bound state, but its wavefunction is proportional to $e^{-r}$; that is, greater than zero everywhere.

 

[fog37]

ok, thanks.

As far as the bound state, Wikipedia defines a bound state as the state of a particle subject to a potential such that the particle has a tendency to remain localized in one or more regions of space.

 

[TeethWhitener]

I think a more useful heuristic definition is that a bound state is one where the particle can't escape to infinity. In other words, the energy expectation value is always less than the value of the potential energy in the infinite limit. So for a hydrogen atom, since $\lim_{r\to\infty}V(r) = 0$, the bound states are linear combinations of the stationary states with energy less than zero. For the harmonic oscillator, since $\lim_{r\to\infty}V(r) = +\infty$, all possible states are bound states.

bound nonstationary states

See my comment about the harmonic oscillator above. More to the point, google "coherent states."

unbound stationary states?

Using your definition of stationary as $\partial_t |\psi(x,t)|^2 = 0$, what about free particle plane waves ($\psi(x,t) = exp[i(kx-\omega t)]$)?

 

[TeethWhitener]

Is it possible to use the TISE and its solutions to find the wavefunction for an unbound state, either stationary or nonstationary?

Again, what about a free particle plane wave? Caveat: I don't really know much about life in an uncountably infinite Hilbert space, so there are probably some more stringent mathematical requirements placed on these free particle "eigenstates."

EDIT: As for separability, $e^{i(kx+\omega t)}=e^{ikx}e^{i\omega t}$.

 

[Khashishi]

Any superposition of bound states is a bound state. Most of these are not stationary states. For example, a hydrogen atom in a (|1s> + |2s>)/sqrt(2) state.

 

[A. Neumaier]

Any superposition of bound states is a bound state.

No, not according to the standard conventions. Bound states are defined as eigenstates of the Hamiltonian. Superpositions of eigenstates for different energies are not longer eigenstates.

 

[TeethWhitener]

No, not according to the standard conventions. Bound states are defined as eigenstates of the Hamiltonian. Superpositions of eigenstates for different energies are not longer eigenstates.

Is this true? I can't believe one would call a coherent state of a harmonic oscillator unbound.

Edit: this also implies that the standard picture of ammonia is not a bound state (the vibrational eigenstates of ammonia are linear combinations of that state and its umbrella-inverted partner).

 

[A. Neumaier]

this also implies that the standard picture of ammonia is not a bound state (the vibrational eigenstates of ammonia are linear combinations of that state and its umbrella-inverted partner).

This is special as the ground state is degenerate in case the Hamiltonian has a symmetry. Nonvanishing superposititions of eigenstates corresponding to the same eigenvalue (in this case the ground state energy) are again eigenstates.

 

[TeethWhitener]

This is special as the ground state is degenerate in case the Hamiltonian has a symmetry.

I'm pretty sure this isn't right. The wavepackets where the ammonia is localized with hydrogens pointing to the right (call it $\psi_R$) and left (call it $\psi_L$) are degenerate, but they aren't eigenfunctions. The ground state (eigenstate) is $\frac{1}{\sqrt{2}}(\psi_R+\psi_L)$, and the first excited state is $\frac{1}{\sqrt{2}}(\psi_R-\psi_L)$. Here's Feynman to explain:
http://www.feynmanlectures.caltech.edu/III_09.html

 

[A. Neumaier]

I'm pretty sure this isn't right. The wavepackets where the ammonia is localized with hydrogens pointing to the right (call it $\psi_R$) and left (call it $\psi_L$) are degenerate, but they aren't eigenfunctions. The ground state (eigenstate) is $\frac{1}{\sqrt{2}}(\psi_R+\psi_L)$, and the first excited state is $\frac{1}{\sqrt{2}}(\psi_R-\psi_L)$. Here's Feynman to explain:
http://www.feynmanlectures.caltech.edu/III_09.html

This is because the interaction with the electromagnetic field breaks the symmetry. In the absence of external forces, the Hamiltonian is invariant under parity, and both states are eigenstates ($A=0$ in Feynman's notation). An external field breaks the symmetry and causes level splitting. See https://en.wikipedia.org/wiki/Energy_level_splitting .

 

[TeethWhitener]

This is because the interaction with the electromagnetic field breaks the symmetry. In the absence of external forces, the Hamiltonian is invariant under parity, and both states are eigenstates ($A=0$ in Feynman's notation). An external field breaks the symmetry and causes level splitting. See https://en.wikipedia.org/wiki/Energy_level_splitting .

No, look closer. The electromagnetic field breaks the symmetry of the non-eigenstate $\psi_L$ and $\psi_R$ states (which are degenerate in the absence of an E field). The eigenstates of the Hamiltonian without an electric field are given by the linear combinations I mentioned above, with energies given by equations 9.8 and 9.9. Adding the electric field shifts the eigenstate energies as well as the non-eigenstate energies (see eqn. 9.30). Also take a look at figure 9.2 for a graphical depiction of this. The non-eigenstate $\psi_L$ and $\psi_R$ have value $E_0$ in the absence of an electric field. The eigenstates do not.

Edit: BTW: the $A$ term has nothing to do with the electric field. It's effectively a measure of the inversion energy barrier of the ammonia molecule.

 

[A. Neumaier]

Sorry, yes, you are right; the degeneracy is broken by the energy barrier, and $A=0$ would correspond to a zero barrier height. Thus the eingenstates (i.e., the bound states) are indeed superpositions of the chiral states.

However this does not prove your earlier, wrong claim that superpositions of bound states are bound states, since one gets the two bound states as superpositions of other states, not of bound states.

 

[TeethWhitener]

However this does not prove your earlier, wrong claim that superpositions of bound states are bound states, since one gets the two bound states as superpositions of other states, not of bound states.

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.

 

[A. Neumaier]

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.

This is the uncommon and somewhat vague definition given in Wikipedia. A bound state according to their definition is more conventionally called a normalizable state.

In the applications of quantum mechanics you'll usually find the notion of bound states applied to eigenstates only, in contrast to scattering states, which are unnormalizable eigenstates (in a rigged Hilbert space).

 

[PeterDonis]

Bound states are defined as eigenstates of the Hamiltonian.

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

AFAIK bound states are states with energy less than the potential energy.

 

[TeethWhitener]

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

AFAIK bound states are states with energy less than the potential energy.

Less than the potential energy at infinity (otherwise not even the HO ground state counts). But yes, that was my impression as well.

 

[TeethWhitener]

In the applications of quantum mechanics you'll usually find the notion of bound states applied to eigenstates only, in contrast to scattering states, which are unnormalizable eigenstates (in a rigged Hilbert space).

I always figured coherent HO states are bound (they don't escape to infinity). But again, I could be wrong.

 

[A. Neumaier]

AFAIK bound states are states with energy less than the potential energy.

This is the classical definition, with energy less that the energy of any trajectory that can escape to infinity. But there is no general precise quantum version of it, only for the special case of eigenstates.
I know of no precise definition of what it could mean for a quantum system to be bound when not in an eigenstate, unless one takes ''normalizable=bound''.
Note that there are quantum systems with bound (eigen)states of positive energy even though the potential vanishes at infinity! See https://physics.stackexchange.com/a/350394

 

[A. Neumaier]

I always figured coherent HO states are bound (they don't escape to infinity).

If you assert that coherent states are bound and superpositions of bound states are bound then, at least for the harmonic oscillator, every normalizable state is bound. So the qualification ''bound'' loses its meaning.

 

[TeethWhitener]

I know of no precise definition of what it could mean for a quantum system to be bound when not in an eigenstate, unless one takes ''normalizable=bound''.

This is what I would propose.

Note that there are quantum systems with bound (eigen)states of positive energy even though the potential vanishes at infinity! See https://physics.stackexchange.com/a/350394

This is very interesting. It reminds me once again of $f(x) = x^2 e^{-x^8\sin^2x}$, a square-integrable function with undefined limits at infinity.

If you assert that coherent states are bound and superpositions of bound states are bound then, at least for the harmonic oscillator, every normalizable state is bound.

Yes, I would assert that every possible HO state is bound.

So the qualification ''bound'' loses its meaning.

For the harmonic oscillator, yes. But for general potentials, no.

 

[A. Neumaier]

This is what I would propose.

But there is already a name ''normalizable'' for this notion, and this is universally used. There is no need to introduce a second name for it. especially since, strictly speaking, all states are normalizable (unnormalizable wave functions cannot be realized in the Hilbert space), so all states would be bound...

 

[TeethWhitener]

But there is already a name ''normalizable'' for this notion, and this is universally used. There is no need to introduce a second name for it.

This is semantics, not science. We have $\text{Normalizable states} \supset \text{Eigenstates}$ and you want to call "bound states" eigenstates, whereas I want to call "bound states" normalizable states. If this is standard usage, so be it, but

especially since, strictly speaking, all states are normalizable (unnormalizable wave functions cannot be realized in the Hilbert space), so all states would be bound...

But if all states are normalizable, there is already a name "states" for this notion, and this is universally used. There is no need to introduce a second name for it.

Maybe there are different usages in different areas of science?

 

[A. Neumaier]

But if all states are normalizable, there is already a name "states" for this notion, and this is universally used. There is no need to introduce a second name for it.

The common usage is unambiguous, with distinct names for distinct items:

• Wave functions can be normalized (of norm 1), normalizable (elements of the Hilbert space, representing after normalization states) or unnormalizable (e.g., plane waves, not in the Hilbert space but in a rigged extension of it, used in scattering theory).
• Pure states (often used without the qualification ''pure'') are described by normalized wave functions.
• Scattering states are solutions of the time-independent Schroedinger equation described by unnormalizable wave functions.
• Bound states are solutions of the time-independent Schroedinger equation described by normalized wave functions. They are the eigenstates of the Hamiltonian corresponding to the discrete spectrum.
• Stationary states are normalized time-independent solutions of the time-dependent Schroedinger equation, corresponding to bound states with an additional time-dependent phase factor.

If you prefer to deviate from the common usage, people will just have difficulties understanding you.

 

[TeethWhitener]

Bound states are solutions of the time-independent Schroedinger equation described by normalized wave functions. They are the eigenstates of the Hamiltonian corresponding to the discrete spectrum.

Just so we're clear, the normalized solutions of the time independent Schrodinger equation are a superset of the eigenstates of the Hamiltonian. Your usage of "bound" here means "eigenstate" specifically. Am I correct?

I understand the standard usage, and if I'm incorrect with my usage, so be it. I just don't like what it insinuates in everyday language. Like that an electron in a hydrogen atom with a wavefunction $\Psi=\frac{1}{\sqrt{2}}(\phi_{1s}+\phi_{2s})$ is not bound to the nucleus, even though it will never escape to infinity. Or that a particle in a superposition of states in an infinite well potential is not bound by the potential, even though it can't even wander beyond the walls of the box. Maybe it would be easier to swallow if I weren't a native English speaker and didn't associate "bound" with "restricted" and "unbound" with "free."

 

[A. Neumaier]

the normalized solutions of the time independent Schrodinger equation are a superset of the eigenstates of the Hamiltonian.

No. They are exactly the same things.

 

[TeethWhitener]

No. They are exactly the same things.

Yes, you're right. Sorry about that.

 

[A. Neumaier]

I just don't like what it insinuates in everyday language. Like that an electron in a hydrogen atom with a wavefunction $\Psi=\frac{1}{\sqrt{2}}(\phi_{1s}+\phi_{2s})$ is not bound to the nucleus, even though it will never escape to infinity.

The standard terminology is to say that the electron is bound to the nucleus whenever its wave function is normalizable (and hence remains normalizable all the time). But it is not in a bound state, since this has a more specific meaning.

In your example, the electron will have this wave function only at a particular time, which has to be specified in order for the statement to make sense. This is why one usually doesn't talk like that. While if it is in a stationary state, only the phase changes, and the state (namely the ray defined by it, or the rank 1 desity operator associated with it, or the positive linear functional defined by it) will be invariant. Then talking about its state without reference to the time makes sense, and this state is a bound state.

 

[TeethWhitener]

In your example, the electron will have this wave function only at a particular time, which has to be specified in order for the statement to make sense. This is why one usually doesn't talk like that. While if it is in a stationary state, only the phase changes, and the state (namely the ray defined by it, or the rank 1 desity operator associated with it, or the positive linear functional defined by it) will be invariant. Then talking about its state without reference to the time makes sense, and this state is a bound state.

Correct me if I’m wrong, but the Schrodinger equation can’t evolve a superposition of stationary states into an unnormalizable state, right? (Edit: I would be interested to see the behavior of a superposition of states with positive eigenvalue from your earlier example.) So a reference to time is unnecessary if you’re simply talking about whether or not a particle can escape being bound by a potential.

Again, I concede that my terminology was nonstandard. I just find it extraordinarily counterintuitive to say that a bound system is not in a bound state. It seems to imply that the system is not in a state, since one can’t assert that it’s in an unbound state. But we can only assert that if we assert that states are invariant in time. But I assumed any vector in the Hilbert space was a state. Maybe I’m wrong with that terminology too, but if I’m not, then we have a system that’s bound and in a state but not in a bound state.

 

[A. Neumaier]

we have a system that’s bound and in a state but not in a bound state.

For historical reasons, technical terminology is often slightly different in meaning from the corresponding intuitive notion. This also holds for terms like ''energy'' or ''space''.

Moreover, in reality the situation is more complicated. For example, a hydrogen atom has a continuous spectrum due to the motion of the center of mass. it becomes bound only when the center of mass motion is decoupled and the reduced system is considered in the rest frame. 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. But as a formal physical system the hyddrogen atom also has states for the latter. 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. There are even lots of (normalized) states whose projection to the bound state sector is zero - against your intuition.

Even a free particle is always in a normalized state though its spectrum is purely continuous though its mean position will be unbound for large times.

Thus your intuition cannot be made formally precise without contradiction other intuition. 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.

 

[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.