1
$\begingroup$

Let's assume that we have delta potential well with $V = -\lambda\delta(x)$, where $\lambda >0$. Now if we solve Schrodinger equation, we get one eigenvalue $E_b=-\frac{m\lambda^2}{\hbar^2}$ with only one eigenfunction $\psi(x) = \sqrt{\frac{m\lambda}{\hbar^2}}\exp(-\frac{m\lambda}{\hbar^2}|x|)$. What does that even mean?

Having only one eigenfunction means no matter how many time we measure energy of the system, we would get $E_b$. So on average we will have $<E> = E_b$. It seems a bit problematic, since we can produce an electron beam with $E<0$ where $E$ can be any number like $E_p$. It's not restricted to only $E_b$. And that would mean conservation of energy would be violated. In other words we have $<E>\neq E_p$

Do note that in quantum mechanics, it doesn't matter if in the first few measurements we get an energy like $E_m$ where $E_m \neq E_p$, it's even natural. But on average we expect $<E> = E_p$. At least it seems the case for other stationary (constant potential w.r.t to time) systems. Or am I wrong, and we should throw conservation of energy, altogether?

After all it's possible to prove this point by Ehrenfest theorem. $$\frac{d}{dt}<A> = \frac{i}{h}<[A,H]>+<\frac{\partial A}{\partial t}>$$ We have $A=H$ here, so

$$\frac{d}{dt}<H> = 0$$

in other words, we do have conservation of energy, and it has nothing do to with uncertainty principle, which is something else entirely. If you don't like it, It's ok. Just assume that as $\Delta t \to \infty$, We don't have $<E> = E_p$ for this particular system.

One possible answer is only an electron with energy $E_b$ will be bounded to this system. That's, if we produce a beam with energy $E_b$ we would have a bound state, else we don't have a bound state at all. But it's not good, since it's possible to ask, What will be happened to a beam with $E<0$ and $E \neq E_b$ in this system? Not only it's not a bound state, It's not an unbounded state as well (just look at Schrodinger equation). How can I explain behavior of this kind of electrons with Schrodinger equation?

Edit for comments:

So in short I would like to ask what's meaning of $E_p<0$ where $E_p \neq E_b$. Is it a bound state?

If it's, then does that mean conservation of energy will be violated here? (As I stated in second paragraph).

If it is not a bound state, then what is it?! It can not be a unbound state since $E_p<0$. It is not a bound state, or unbound state?!

$\endgroup$
12
  • 2
    $\begingroup$ You seem to realise that there do exist unbound states, so I'm not sure what the problem you're having is. $\endgroup$ – jacob1729 Jun 1 '20 at 16:09
  • $\begingroup$ @jacob1729 1) Accepting violation of conservation of energy is my problem. Or 2) Having a system which can not be described by Schrodinger equation. In the case of $E<V$ and $E_p\neq E_b$ system is NOT in the unbound states! just look at Schrodinger equation. This particular case has to be in bound states, but that would mean violation of conservation of energy. $\endgroup$ – Paradoxy Jun 1 '20 at 16:12
  • $\begingroup$ As defined in your post, $V$ is not a number but a function of $x$ so I'm not sure what you mean by $E\lt V$. $\endgroup$ – jacob1729 Jun 1 '20 at 16:17
  • $\begingroup$ @jacob1729, I will change it to $E<0$ so it becomes like its Wikipedia page. V is indeed a distribution, and we work with its amplitude $\lambda$. But I got your point. $\endgroup$ – Paradoxy Jun 1 '20 at 16:21
  • 1
    $\begingroup$ It's only possible to have negative energy if there's some feature in the potential that lets you somehow "bend" the wavefunction so that the exponential tails could get to zero at infinity. But if you produce an electron beam somewhere far from the potential well, you can't ever make it propagate away: even if your emitter is in a potential well, the wavefunction of the electron it emitted will exponentially decay outwards, and is unlikely to get into your faraway delta-well. In any case, you'll then have to include the potential well from where the electron is emitted into description. $\endgroup$ – Ruslan Jun 1 '20 at 16:32
1
$\begingroup$

we can produce an electron beam with $E<0$ where $E$ can be any number like $E_p$.

We can't. All the states with $E<0$ are evanescent waves. They can only decay exponentially towards e.g. $x\to+\infty$, but then they'll grow exponentially towards $x\to-\infty$.

The potential well bends the wavefunction, so that exponential growth can be altered ($\exp(x)$ becomes a mix of $\exp(-x)$ and $\exp(x)$ after the well) and, when $E=E_b$, be turned into exponential decay.

So in short I would like to ask what's meaning of $E_p<0$ where $E_p \neq E_b$. Is it a bound state?

These are not stationary states. Actually, they are not quantum states at all. All solutions of the Schrödinger's equation with such values of $E$ are exponentially divergent at least on one side. This makes these "wavefunctions" not only non-square-integrable: none of these functions will even solve the boundary value problem if you set homogeneous Dirichlet or Neumann boundary conditions at finite points $x=\pm a$.

If it is not a bound state, then what is it?! It can not be a unbound state since $E_p<0$. It is not a bound state, or unbound state?!

A question to you: what are the solutions of the particle-in-a-box problem where $E$ is not one of the eigenvalues $E_n$? Exactly the same: they are not solutions of the boundary value problem.

if we solve Schrodinger equation, we get one eigenvalue $E_b=-\frac{m\lambda^2}{\hbar^2}$ with only one eigenfunction $\psi(x) = \sqrt{\frac{m\lambda}{\hbar^2}}\exp(-\frac{m\lambda}{\hbar^2}|x|)$. What does that even mean?

This means that the only state when a particle doesn't escape to infinity is $E=E_b$. All the other states correspond to infinite motion. A particle with $E\ge0$ can emit a photon and transition to this $E=E_b$ state. Conversely, a particle in $E=E_b$ state can absorb a photon and transition to a state with $E\ge0$.

$\endgroup$
3
  • $\begingroup$ Thanks, If you provide mathematical reasoning for "All solutions of the Schrödinger's equation with such values of E are exponentially divergent at least on one side" I will accept your answer. $\endgroup$ – Paradoxy Jun 2 '20 at 15:40
  • $\begingroup$ @Paradoxy well the reasoning is simple: there are only two solutions of a second-order ODE. This gives us two arbitrary constants that we can set to satisfy the boundary conditions. One of these constants can be chosen such that the BC at $x\to-\infty$ is satisfied. The other, if we fix this constraint, will only change overall normalization. The only other parameter left is $E$. This parameter is an eigenvalue of the Hamiltonian whenever the solution can satisfy all the boundary conditions. $\endgroup$ – Ruslan Jun 2 '20 at 15:52
  • $\begingroup$ In particular, as you set $E$ to an eigenvalue (e.g. $E_b$), and then start increasing it, the solution (that we've fixed to satisfy BC at left infinity) will get its zero from $x\to+\infty$ closer and closer into the oscillatory domain (that is only a single point in the delta potential), and eventually a second zero (or a finite value, for scattering states) comes to the point $x=+\infty$. This is your second point of Hamiltonian's spectrum. Further reading: Courant nodal domain theorem, and shooting method. $\endgroup$ – Ruslan Jun 2 '20 at 15:56

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.