What is the meaning of Schrodinger equation solution for bound state of delta potential well?

Question

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?!

Answer 1

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 $E\ge0$.