# What exactly is a bound state and why does it have negative energy?

Could you give me an idea of what bound states mean and what is their importance in quantum-mechanics problems with a potential (e.g. a potential described by a delta function)?

Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative?

I figured it out, mathematically (for instance in the case of a potential described by a Delta function), but what is the physical meaning?

If you have a copy of Griffiths, he has a nice discussion of this in the delta function potential section. In summary, if the energy is less than the potential at $-\infty$ and $+\infty$, then it is a bound state, and the spectrum will be discrete: $$\Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right).$$ Otherwise (if the energy is greater than the potential at $-\infty$ or $+\infty$), it is a scattering state, and the spectrum will be continuous: $$\Psi\left(x,t\right) = \int dk \ c\left(k\right) \Psi_k\left(x,t\right).$$ For a potential like the infinite square well or harmonic oscillator, the potential goes to $+\infty$ at $\pm \infty$, so there are only bound states.

For a free particle ($V=0$), the energy can never be less than the potential anywhere***, so there are only scattering states.

For the hydrogen atom, $V\left(r\right) = - a / r$ with $a > 0$, so there are bound states for $E < 0$ and scattering states for $E>0$.

Update

*** @Alex asked a couple questions in the comments about why $E>0$ for a free particle, so I thought I'd expand on this point.

If you rearrange the time independent Schrödinger equation as $$\psi''= \frac{2m}{\hbar^2} \left(V-E\right) \psi$$ you see that $\psi''$ and $\psi$ would have the same sign for all $x$ if $E < V_{min}$, and $\psi$ would not be normalizable (can't go to $0$ at $\pm\infty$).

But why do we discount the $E<V_{min}=0$ solutions for this reason, yet keep the $E>0$ solutions, $\psi = e^{ikx}$, when they too aren't normalizable?

The answer is to consider the normalization of the total wave function at $t=0$, using the fact that if a wave function is normalized at $t=0$, it will stay normalized for all time (see argument starting at equation 147 here):

$$\left<\Psi | \Psi\right> = \int dx \ \Psi^*\left(x,0\right) \Psi\left(x,0\right) = \int dk' \int dk \ c^*\left(k'\right) c\left(k\right) \left[\int dx \ \psi^*_{k'}\left(x\right) \psi_k\left(x\right)\right]$$

For $E>0$, $\psi_k\left(x\right) = e^{ikx}$ where $k^2 = 2 m E / \hbar^2$, and the $x$ integral in square brackets is $2\pi\delta\left(k-k'\right)$, so

$$\left<\Psi | \Psi\right> = 2\pi \int dk \ \left|c\left(k\right)\right|^2$$ which can equal $1$ for a suitable choice of $c\left(k\right)$.

For $E<0$, $\psi_k\left(x\right) = e^{kx}$ where $k^2 = - 2 m E / \hbar^2$, and the $x$ integral in square brackets diverges, so $\left<\Psi | \Psi\right>$ cannot equal $1$.

• Exactly what I wanted to know. I don't have a copy of Griffiths, but you explained the topic satisfactorily. Jan 5 '14 at 20:34
• @Alex With the time independent Schrödinger equation rearranged as $\psi'' = 2m/\hbar^2 \left(V-E\right) \psi$, $\psi''$ and $\psi$ would have the same sign for all $x$ if $E < V_{min}$, and $\psi$ would not be normalizable (can't go to $0$ at $\pm \infty$). Oct 12 '16 at 12:34
• @EricAngle I see that if we consider $E < 0$ then as you say the eigenfunctions are not normalizable but are you then implying that if we consider energy eigenstates of the free particle ($V=0$) with $E > 0$: $$\psi'' = -\frac{2m}{\hbar^2}E\psi,$$ then considering $E >0$ gives us normalizable eigenfunctions? But as I understand the energy eigenfunctions are not normalizable for a free particle since a free particle cannot exist in a stationary state. So we reject $E < 0$ since the solutions are not normalizable but then we end up with non normalizable solutions anyway? Thanks for your time.
– Alex
Nov 7 '16 at 14:57
• @Alex good question. See my update above. Nov 11 '16 at 14:18
• @EricAngle Very nice answer. I think there's a typo, it should be $k^2 = \frac{-2mE}{\hbar^2}$.
– user100411
Nov 16 '16 at 18:26

It means the same thing it means in classical mechanics: if it is energetically forbidden to separate to arbitrarily large distance they are "bound".

The Earth is gravitationally bound to the Sun and the Moon to the Earth. Electrons in a neutral atom are electomagnetically bound to the nucleus. A pea rolling around in the bottom of a bowl is bound.

By contrast the Voyager probes are (barely) unbound and will fly (slowly) off into the galaxy.

• But when I have only one particle and a "attractive contact potential" ($g\delta(x-a)$) we talk about states bounded to... the potential? The $\delta$ rapresent a sort of potential created by another particle? Jan 3 '14 at 22:56
• Thinking of those states as "bound to the potential" is usually enough to get a good picture of the physics of the particle. Nevertheless "external potentials" are always used as a model of the interaction with something else, when you don't really care about the dynamics of the "something else". A very good example of this is how you derive the energy levels for an atom as the bound states of the electron in the external potential generated by the nucleus.
– fqq
Jan 4 '14 at 20:22

Mathematically, bound states are states that decay sufficiently fast at infinity, so that the probability of finding the particle they describe in far away regions of space is negligible.

It has long been conjectured, based on physical intuition, it is the case for the meaningful quantum mechanical states, such as the eigenfunctions of the Hamiltonian (it is not expected that an atomic electron has a sensible probability of being at infinite distance from its nucleus).

This has been proved mathematically in the eighties, mainly by S.Agmon. Roughly speaking, the result is the following: eigenfunctions of the Schrödinger operator (i.e. corresponding to the discrete spectrum) are exponentially decaying in space. So if $\psi_n(x)$ are such eigenfunctions, $\lvert \psi_n(x)\rvert\leq A e^{-B\lvert x\rvert}$, for some positive constants $A,B$.

Barry Simon writes:

One of the more intriguing questions concerns the presence of discrete eigenvalues of positive energy (that is, square-integrable eigenfunctions with positive eigenvalues) . There is a highly non-rigorous but physically appealing argument which assures us that such positive energy “bound states” cannot exist. On the other hand, there is an ancient, explicit example due to von Neumann and Wigner which presents a fairly reasonable potential $$V$$, with $$V(r) \to 0$$ as $$r \to \infty$$ and which possesses an eigenfunction with $$E = 1$$.
The potential$$V(r)=\frac{-32 \sin r[g(r)^3 \cos r-3g(r)^2\sin^3r+g(r)\cos r+sin^3r]}{[1+g(r)^2]^2}$$ with $$g(r)=2r-\sin2r$$ has the eigenvalue +1 with eigenfunction $$u(r)=\frac{\sin r}{r(1+g(r)^2)}$$ On Positive Eigenvalues of One-Body Schrodinger Operators

I can't offer any intuition as to why these unappealing solutions should occur.

(Simon's paper is concerned with characterising potentials which do not have positive eigenvalues.)

D. B. Pearson offers a physical interpretation for a related phenomenon, "singular continuous spectrum":

... a potential made up of an infinite sequence of bumps, their separation increasing rapidly with distance. The potential is spherically symmetric (or one dimensional), bounded, and locally non-singular. It is even possible to have $$\lim_{r \to \infty} V(r) = 0$$, in which case the condition for singular continuous spectrum is (roughly) $$\sum_1^\infty g_n^2 = \infty$$, where $$g_n$$ is the height of the $$n$$'th bump. ... This is exactly the condition for the particle, after possibly a large number of transmissions and reflections at successive bumps, ultimately with probability $$1$$ to be reflected back to a neighbourhood of the origin.

• Very interesting. Quantum mechanics is endlessly full of surprises. Aug 5 '17 at 15:36
• Interesting. Btw, I guess he forgot to say 'in three dimensions'.
– lcv
Feb 26 '19 at 9:13
• Any chance of a plot of that potential, and its eigenfunction? It would make this answer even more valuable. May 9 '19 at 19:18
• @EmilioPisanty I found a plot here, but it is the behaviour at $\infty$ that causes the unintuitive behaviour. May 9 '19 at 20:14
• Ooofff. That's one ugly beast. May 9 '19 at 20:17

Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative?

First, we need to understand that Energy is relative, both in Quantum and in Newtonian mechanics. This means that different reference frames will determine different values for the energy of a given object at a given point in space and time.

In this case, we are allowing to have negative Energy values for delta potential energy by the following.

𝑉(𝑥)=−𝜆 𝛿(𝑥) (where 𝜆 is a positive number)

This means, in this reference frame we consider all Energy values with minimum 𝑉(𝑥) value that is negative infinity. Therefore, we are allowed to solve the Schrodinger Equation for all the negative and positive values of Energy.

In contrast, for the free particle problem's reference frame, we consider minimum 𝑉(𝑥) value is zero and all the Energy values should be positive for well-defined solutions.

However, the conservation of energy holds in any reference frame. This means that, in a single reference frame the total energy (KE + PE) will be constant over time.