Can we physically realize unbound states?

[Haorong Wu]

TL;DR

Can we physical realize unbound states?

In principle, unbound states exist when $E \gt V$ at infinity. Are these unbound states just for calculation, or can we physical realize unbound states? After all, an unbound state means its possibility at infinity is not zero, and it seems impossible to realize such a state.

 

[Dale]

Sure. Such states are abundant. They are called ions.

 

[Keith_McClary]

Are you talking about Quantum Mechanics?

 

[Haorong Wu]

Sure. Such states are abundant. They are called ions.

It is just hard for me to imagine how would an electron appear at infinity. Maybe I misunderstand some concepts here?

 

[Haorong Wu]

Are you talking about Quantum Mechanics?

Yes. It is a problem in a exam. The answer it provides is that this kind of states can not be realized. But I feel skeptical about it.

My answer was that it could be physically realized, because I thought about free electrons. Also, I cannot search other answers in Chinese, and I do not know how to properly transfer the problem to english for searching in google.

 

[Keith_McClary]

Yes. It is a problem in a exam. The answer it provides is that this kind of states can not be realized. But I feel skeptical about it.

My answer was that it could be physically realized, because I thought about free electrons. Also, I cannot search other answers in Chinese, and I do not know how to properly transfer the problem to english for searching in google.

For $V=0$ everywhere, $e^{ikx}$ is a free state. If you integrate its probability density over a length $L$ it is $L$ times the amplitude squared, so if its total probability is normalised to $1$, the amplitude must be $\frac{1}{\sqrt(L)}$. As $L \to \infty$ the amplitude $\to 0$ . We say it is not normalisable. So it is not mathematically or physically realisable.

 

[Dale]

It is just hard for me to imagine how would an electron appear at infinity. Maybe I misunderstand some concepts here?

An unbound state is just an ion. There is no requirement that the electron appear at an infinite distance away from the nucleus. All that is needed is that the energy is higher than is needed for the electron to be at an infinite distance. Even if they happen to be near each other, if the energy is high enough they are unbound.

 

[PeroK]

It is just hard for me to imagine how would an electron appear at infinity. Maybe I misunderstand some concepts here?

An unbound state is one that is not bound. Any talk of "infinity" is just a simple way to say that the particle is lost to the core system: an ionised atom, a particle escaping from a potential well, a space probe escaping the solar system.

None of these things ever reaches "infinity", but they never come back. Hence are not bound solutions.

 

[Keith_McClary]

In principle, unbound states exist when $E \gt V$ at infinity.

I assumed you meant that $E$ is an energy corresponding to a (non-normalisable) free-particle solution of the Schrödinger equation. If not, what is $E$ exactly?

 

[Dale]

We say it is not normalisable. So it is not mathematically or physically realisable.

I looked into it a bit and you are correct that it is not normalizable. And yet ions clearly exist. So how is that reconciled? It seems like normalizability must not be required for physical realization.

 

[PeterDonis]

how is that reconciled?

Non-normalizable states are idealizations; I don't think anyone would claim that an actual free electron that came from ionizing an atom is exactly described by such a state. But using such a state in the math makes the math tractable, and makes reasonably accurate predictions for many scenarios, so clearly such an idealized state is a reasonably good approximation for many purposes, even though it has a property (non-normalizability) that means it can't be exactly realized physically.

 

[Keith_McClary]

I looked into it a bit and you are correct that it is not normalizable. And yet ions clearly exist. So how is that reconciled?

You can have a (continuous integral) superposition of positive energy solutions which is normalizable. In the case of $V=0$ everywhere this is just a Fourier transform.

 

[Dale]

I don't think anyone would claim that an actual free electron that came from ionizing an atom is exactly described by such a state. But using such a state in the math makes the math tractable, and makes reasonably accurate predictions for many scenarios,

So is there a known state which does accurately represent the quantum mechanical state of an ion, and if so how does it differ from the non normalizable state?

 

[PeterDonis]

is there a known state which does accurately represent the quantum mechanical state of an ion

I don't know. Possibly a superposition of the sort @Keith_McClary describes might.

 

[HomogenousCow]

I would think "realistic" ions are best described by thermalized mixed states, the actual pure states are probably some horribly complicated entangled wavepackets.

 

[vanhees71]

Summary:: Can we physical realize unbound states?

In principle, unbound states exist when $E \gt V$ at infinity. Are these unbound states just for calculation, or can we physical realize unbound states? After all, an unbound state means its possibility at infinity is not zero, and it seems impossible to realize such a state.

Eigenstates of a self-adjoint operator to an eigenvalue in the continuum of its spectrum are usually not normalizable, i.e., they do not belong to the Hilbert-space of square integrable functions, and only such wave function represent (pure) quantum states. The generalized eigenvectors for eigenvalues in the continuous spectrum are distributions. The math is somewhat involved, if you want to get it rigorous. For physicists, if they bother at all about mathematical rigorous formulations, the socalled "rigged Hilbert space formalism" is the most convenient one.

For the practitioner the non-rigorous treatment of standard textbooks are enought, though one should be aware that trouble can occur when one is not carefully remembering that there's some non-rigorous handwaving used.

The most simple example is a particle moving in one dimension and considering the momentum "eigenstates". The momentum operator is $\hat{p}=-\mathrm{i} \hbar \partial_x$. The eigenvalue problem thus reads
$$-\mathrm{i} \hbar \partial_x u_p(x)=p u_p(x).$$
The solution for any $p \in \mathbb{C}$ is obviously
$$u_p(x)=A_p \exp(\mathrm{i} p x/\hbar).$$
Now for any $p \notin \mathbb{R}$ no "scalar-product integral" of two such solutions makes sense, because if diverges for sure when integrated over the real axis. Only for $p \in \mathbb{R}$ it makes sense within the theory of distributions since then you get
$$\langle p_1|p_2 \rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p_1}^*(x) u_{p_2}(x)=A_{p_1}^* A_{p_2} \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i}(p_1-p_2)/\hbar]=A_{p_1}^* A_{p_2} 2 \pi \hbar \delta(p_1-p_2)=|A_{p_1}|^2 2 \pi \hbar \delta(p_1-p_2),$$
where $\delta$ is the Dirac-$\delta$ distribution. For $p_1=p_2$ the integral of course diverges, and thus a momentum eigenfunction cannot represent a proper state of the particle. It's convenient to "normalize" the eigenfunctions "to a $\delta$ distribution", i.e., you make
$$A_{p}=\frac{1}{\sqrt{2 \pi \hbar}}.$$

Nevertheless these generalized eigenstates are very important, because you can represent any proper wave function as a "generalized superposition" of them, because of the completeness relation,
$$\int_{\mathbb{R}} \mathrm{d} p u_{p}^*(x_1) u_p(x_2)=\frac{1}{2 \pi \hbar} \int_{\mathbb{R}} \mathrm{d} p \exp[\mathrm{i} p (x_2-x_1)/\hbar]=\delta(x_2-x_1).$$

In this case the entire spectrum of the momentum operator is continuous, and thus the "superposition" is not a sum but an integral:
$$\psi(x)=\langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d} p \tilde{\psi}(p) u_p(x).$$
The "momentum-space wave function" is uniquely defined because of the "normalization to a $\delta$ distribution"
$$\tilde{\psi}(p)=\int_{\mathbb{R}} \mathrm{d} x u_{p}^*(x) \psi(x).$$
If you write out the eigenfunctions you see that this is of course nothing else than the well-known Fourier transformation (with some conventions concerning the constants appearing in the Fourier integral and its inverse Fourier integral adapted to the needs of quantum mechanics).

 

[A. Neumaier]

So is there a known state which does accurately represent the quantum mechanical state of an ion, and if so how does it differ from the non normalizable state?

An $H^+$ ion is just a proton. In general, ions are just systems consisting of more or fewer electrns than the total nuclear charge. The unbound states of a hydrogen atom are states of the joint system of ion plus electron that in the absence of external forces separate in finite time. The unbound eigenstates are unnormalizable scattering states, and theiir integral describes normalized states realizable in the lab.