9
$\begingroup$

How is a bound state defined in quantum mechanics for states which are not eigenstates of the Hamiltonian i.e. which do not have definite energies? Can a superposition state like $$\psi(x,t)=\frac{1}{\sqrt{2}}\phi_1(x,t)+\frac{1}{\sqrt{2}}\phi_2(x,t), $$ where $\phi_1$ and $\phi_2$ are energy eigenstates be a bound state? How to decide?

$\endgroup$
8
$\begingroup$

Bound states are usually understood to be square-integrable energy eigenstates; that is, wavefunctions $\psi(x)$ which satisfy $$ \int_{-\infty}^\infty|\psi(x)|^2\text dx<\infty \quad\text{and}\quad \hat H \psi=E\psi. $$

This is typically used in comparison to continuum states, which will (formally) obey the eigenvalue equation $\hat H\psi=E\psi$, but whose norm is infinite. Because their norm is infinite, these states do not lie inside the usual Hilbert space $\mathcal H$, typically taken to be $L_2(\mathbb R^3)$, which is why the eigenvalue equation is only formally true if taken naively - the states lie outside the domain of the operator. (Of course, it is possible to deal rigorously with continuum states, via a construct known as rigged Hilbert spaces, for which a good reference is this one.)

$\endgroup$
8
$\begingroup$

The bound state is defined such that the probability density average will be finite at some particular space region when time passes. While for unbounded states, as time passes, the probability density will tends to zero. See Landau Quantum Mechanics section 10.

This can be understand as this, if the state is bounded, i.e. it is exist only within some particular region, so the probability density should be finite in that region as time passes. On the contrary, when the state is a free motion, the wave package will spread out as time passes, thus the probability density at any point will tends to zero as time tends to infinity.

Edit: now I want to say that discrete eigenstates OR superpositions of these are bound states; while otherwise is an unbounded one.

Note that in this definition of bound state, the average energy $E<V(\pm\infty)$ always holds. However, $E<V(\pm\infty)$ cannot guarantee a state to be bounded. For example, a state (like this one) composed of both discrete spectra and continuum spectra, can have its average energy $E$ either larger or smaller than $V(\pm\infty)$. You might say it's an unbounded one, it depends.

The criterion $E<V(\pm\infty)$ guarantees a bounded state if and only if by $E$ you refer to the energy of an eigenstate.

$\endgroup$
  • $\begingroup$ What about 0.5 probability in finite region A and the other 0.5 smears to infinity? Seems contradicting with your 1st sentence def. $\endgroup$ – xiaohuamao Oct 5 '14 at 4:05
  • $\begingroup$ @huotuichang Yeah, I also though that yesterday. Then shall we say that the non-eigenstate bound state is the superposition of the eigenstates correspond to the discreet eigenvalues? While it is neither bound or unbound when it has infinite motion component? $\endgroup$ – an offer can't refuse Oct 5 '14 at 4:17
  • $\begingroup$ Why not using $E>V(\infty)$ as a criterion for unbounded states? $\endgroup$ – xiaohuamao Oct 5 '14 at 4:22
  • $\begingroup$ @huotuichang Using this criterion means that some state has infinite motion component is bounded while others may not be bounded. So I think it is ambiguous to talk bound or unbound when it is a mixture of discreet and continuous states. $\endgroup$ – an offer can't refuse Oct 5 '14 at 4:27
  • $\begingroup$ No, I reckon it univocal. I just found the finite A + infinity case aforementioned unrealistic since it only occurs when there stands an infinitely high wall, which makes the two regions totally unconnected de facto. So give me an example of mixture type if you can. $\endgroup$ – xiaohuamao Oct 5 '14 at 4:38
4
$\begingroup$

Since states which are not eigenstates of the Hamiltonian are also not eigenstates of the time evolution, it does not make sense to talk about "bound states" for these states, as they are continually changing into other states. For energy eigenstates, it makes sense to speak of "a bound state", since that state will stay the same forever unless acted upon.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.