Finite Well Problem: Can Energy Be Equal to V0?

[Coelum]

TL;DR

Do bound solutions exist for E=V0 where V0 is the depth of the well?

Dear PFers,
I have a question about the finite potential well problem. Let's assume the well is centered in 0 and the potential is V₀ outside the well and 0 inside. For more details see Wikipedia: https://en.wikipedia.org/wiki/Finite_potential_well.

Now, the question is: can the energy of the particle in a bound state be equal to V₀?
I expect it cannot, as:

1. the wave-function outside the box should satisfy the free particle Schrödinger equation with null energy (essentially, the second derivative should be 0)
2. the solution of the equation described above is a linear function
3. a linear function cannot be normalized if the domain is unbound
4. the domain outside the well is unbound (in fact, we have two, separate, unbound regions on both sides).

Is my reasoning correct?

 

[PeroK]

TL;DR Summary: Do bound solutions exist for E=V0 where V0 is the depth of the well?

Dear PFers,
I have a question about the finite potential well problem. Let's assume the well is centered in 0 and the potential is V₀ outside the well and 0 inside. For more details see Wikipedia: https://en.wikipedia.org/wiki/Finite_potential_well.

Now, the question is: can the energy of the particle in a bound state be equal to V₀?

An energy of $E < V_0$ represents a bound state; and $E > V_0$ represents a scattering state. Is your question what happens if we look for $E = V_0$.

The short answer is that it must be something of an unstable state, where technically it cannot escape to infinity, but not is it in a stable bound state.

There is a finite set of bound states (with $E < V_0$), but the highest bound energy is always less than $V_0$. I guess you could play about with the solution to see what parameters allow you to get closest to $E = V_0$.

I expect it cannot, as:

1. the wave-function outside the box should satisfy the free particle Schrödinger equation with null energy (essentially, the second derivative should be 0)
2. the solution of the equation described above is a linear function

I think you are confusing the energy level (which is just a single eigenvalue) with a constant wavefunction. For $E = V_0$, the wavefunction would be zero outside the well.

1. a linear function cannot be normalized if the domain is unbound
2. the domain outside the well is unbound (in fact, we have two, separate, unbound regions on both sides).

Is my reasoning correct?

I don't think so.

 

[Coelum]

Thanks PeroK for your prompt reply.

I think you are confusing the energy level (which is just a single eigenvalue) with a constant wavefunction. For E=V0, the wavefunction would be zero outside the well.

From your answer, I guess I was not clear. Let me add a few details.

If $E=V_0$ the Schrödinger equation outside the well becomes
$$ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} = 0 $$
with solution a linear function ($A,B\in \mathcal R$)
$$ \Psi(x)=Ax+B. $$
Inside the well the S.E. is
$$ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} = E\Psi $$
$$ \frac{d^2\Psi}{dx^2} = -\frac{2mE}{\hbar^2}\Psi = -K^2\Psi $$
with solution ($C,D\in \mathcal R$)
$$ \Psi(x)=C\sin(Kx)+D\cos(Kx). $$
The normalization condition for a single particle imposes $A=B=0$ on both sides of the well. The boundary conditions for the continuity of $\Psi$ and its first derivative at the borders of the well lead to the null solution as the only solution. That's what I meant in my previous statements.

If you still believe my reasoning is wrong, please help me by pointing out the error in the derivation above (including computing the boundary conditions, whose boring details I shamelessly skipped).

 

[PeroK]

If $E=V_0$ the Schrödinger equation outside the well becomes

I'm not sure about setting $E = V_0$ at this stage. It should result in no solutions,

$$ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} = 0 $$
with solution a linear function ($A,B\in \mathcal R$)
$$ \Psi(x)=Ax+B. $$
Inside the well the S.E. is
$$ -\frac{\hbar^2}{2m}\frac{d^2\Psi}{dx^2} = E\Psi $$
$$ \frac{d^2\Psi}{dx^2} = -\frac{2mE}{\hbar^2}\Psi = -K^2\Psi $$
with solution ($C,D\in \mathcal R$)
$$ \Psi(x)=C\sin(Kx)+D\cos(Kx). $$
The normalization condition for a single particle imposes $A=B=0$ on both sides of the well. The boundary conditions for the continuity of $\Psi$ and its first derivative at the borders of the well lead to the null solution as the only solution. That's what I meant in my previous statements.

That's not the approach I understood in your OP. If you follow through with the mathematics here, then you end up with no solutions (for numerical reasons) with $E = V_0$.

If you still believe my reasoning is wrong, please help me by pointing out the error in the derivation above (including computing the boundary conditions, whose boring details I shamelessly skipped).

There's nothing wrong if you try to solve the equation fully where $E = V_0$ and show there are no solutions.

 

[Coelum]

I'm not sure about setting $E = V_0$ at this stage. It should result in no solutions,

That's not the approach I understood in your OP. If you follow through with the mathematics here, then you end up with no solutions (for numerical reasons) with $E = V_0$.

There's nothing wrong if you try to solve the equation fully where $E = V_0$ and show there are no solutions.

Great, I think we agree now. Thanks again.

 

[DrClaude]

From what I know for scattering, where the problem is slightly different as the spatial coordinate is then radial, $r \in [0,\infty)$, I expect the following.

The case $E=V_0$ would correspond to a scattering (unbounded) state, increasing linearly outside the well, corresponding to an infinite wavelength. That seems to be what you found. Inside the well, the wave function will be oscillating (the amount of oscillation increasing with the number of bound states).

The normalization condition for a single particle imposes A=B=0 on both sides of the well.

I think this is where it goes wrong. Scattering (continuum) wave functions are not square integrable, but there are normalization convention one can use. Of course, these states cannot correspond to actual physical states, but are useful for calculating things (just like plane waves).

 

[Coelum]

From what I know for scattering, where the problem is slightly different as the spatial coordinate is then radial, $r \in [0,\infty)$, I expect the following.

The case $E=V_0$ would correspond to a scattering (unbounded) state, increasing linearly outside the well, corresponding to an infinite wavelength. That seems to be what you found. Inside the well, the wave function will be oscillating (the amount of oscillation increasing with the number of bound states).

I think this is where it goes wrong. Scattering (continuum) wave functions are not square integrable, but there are normalization convention one can use. Of course, these states cannot correspond to actual physical states, but are useful for calculating things (just like plane waves).

I agree. In fact, I just worked out a non-normalizable solution. I started asking myself: can the energy of a particle in a free state be equal to V₀?

In order to simplify the treatment, I assume we have a stationary flow of particles. Then, since we do not require normalization, a solution to the stationary Schrödinger equation exists. It is a real constant outside the well and a sum of sine and cosine inside (details depend on the selection of the origin). More interestingly, we get the following condition: L*sqrt(2mV_0)/h must be an integer number. Otherwise, the wave-function is zero.

My question is about the interpretation of the solution. For E>V_0, the solution of the problem is a sum of two complex exponentials representing flows in opposite direction of the x axis. Can the same interpretation be applied to our case? Furthermore, it looks like this solution has no classical equivalent. Is that correct?

 

[PeroK]

The case $E=V_0$ would correspond to a scattering (unbounded) state, increasing linearly outside the well, corresponding to an infinite wavelength.

Are you sure this isn't a degenerate solution? Usually the non-normalisable wave-functions are combined in a wave-packet, with (I assume) the $E$ as the most probable energy. Here, $E = V_0$ is on the lower limit of possible scattering energies, so it ought to have a probability density of zero in any wave-packet.

 

[DrClaude]

More interestingly, we get the following condition: L*sqrt(2mV_0)/h must be an integer number. Otherwise, the wave-function is zero.

I think you are requiring the potential wave to a have quasi-bound state at $E = V_0$, which is not necessary. Solutions exist for any value of $L$ and $V_0$.

My question is about the interpretation of the solution. For E>V_0, the solution of the problem is a sum of two complex exponentials representing flows in opposite direction of the x axis. Can the same interpretation be applied to our case? Furthermore, it looks like this solution has no classical equivalent. Is that correct?

It a scattering state corresponding to zero momentum, so particles not moving/scattering. What this means exactly I can't say.

 

[DrClaude]

Are you sure this isn't a degenerate solution? Usually the non-normalisable wave-functions are combined in a wave-packet, with (I assume) the $E$ as the most probable energy. Here, $E = V_0$ is on the lower limit of possible scattering energies, so it ought to have a probability density of zero in any wave-packet.

As I said above, I don't know how to physically interpret zero-energy scattering. It won't appear in any realistic condition, but it is still part of the set of solutions to the problem, and it used for instance to calculate scattering lengths, as shown in this figure from the Sakurai textbook:

 

[Coelum]

Thanks to both PeroK and DrClaude for the very interesting discussion. Now I understand much better what I found.

 

[Vanadium 50]

As I said above, I don't know how to physically interpret zero-energy scattering

I don't think anybody does.

Note that Δp = 0 so Δx must be infinite. I don't know how one would prepare such a state, but it would be hard to describe it as "bound".

 

[Coelum]

I don't think anybody does.

Note that Δp = 0 so Δx must be infinite. I don't know how one would prepare such a state, but it would be hard to describe it as "bound".

Yes, it looks to me like a mix of bound and unbound states. On one hand, the classical corresponding state is bound. On the other hand, it is not normalizable, so it should represent a stream of particles. Also, there is a well-defined condition for it to exist. Not all wells allow such a state. Overall, really strange.