Can a Specific Width Parameter Ensure a Bound State in a 1D Quantum System?

[CAF123]

Homework Statement

Consider a one-dimensional quantum system described by the potential: $$V(x) = -V_o + \frac{1}{2}mw^2x^2\,\,,V_o > 0\,\,\text{for}\,\, |x| < b\,\,\text{and}\,\,0\,\,\text{otherwise}$$
Show that the state described by: $$\psi_{-}(x) = R_{-} \exp\left[-\frac{x^2}{2x_o^2}\right]\,\,\,\text{for}\,\,|x| < b$$
can be an eigenstate of the Hamiltonian describing a bound state (i.e. a state with
negative energy) only for a specific value of ##x_o##.

Homework Equations

##\hat{H}\psi_{-}(x) = E\psi_{-}(x)## if ##\psi_{-}## is an eigenstate of the Hamiltonian.

The Attempt at a Solution

I have sketched the potential and it looks parabolic in the region ##x \in (-b,b)## A bound state is a state of neg energy => E < 0 and so the path of some incoming particle intersects the potential plot in two places. For ##x \in (-b,b)## it makes sense that the wave function is a decaying exponential. Because ##V_o## is fixed, it also makes sense that the decaying exp is not proportional to m or w, since I think this means there is only one parabola that satisfies the boundary conditions. If ##\psi_{-}## is to be an eigenstate, it must satisfy the eigenvalue problem ##\hat{H}\psi_{-} = E \psi_{-}##. I tried subbing ##\psi_{-}## in, and this gave me an expression for E in terms of x. I am not really sure what else to try, any hints would be great.
Many thanks.

 

[mfb]

??

I have sketched the potential and it looks parabolic in the region ##x \in (-b,b)##

Then you should try other values of x_0, too.

A bound state is a state of neg energy => E < 0

Okay

and so the path of some incoming particle intersects the potential plot in two places.

Which path, which incoming particle?

For ##x \in (-b,b)## it makes sense that the wave function is a decaying exponential.

Why?
(it does not, and the wave function is not a decaying exponential)

Because ##V_o## is fixed, it also makes sense that the decaying exp is not proportional to m or w, since I think this means there is only one parabola that satisfies the boundary conditions.

Where is the relation between the statements here?

If ##\psi_{-}## is to be an eigenstate, it must satisfy the eigenvalue problem ##\hat{H}\psi_{-} = E \psi_{-}##.

Good.

I tried subbing ##\psi_{-}## in, and this gave me an expression for E in terms of x.

How does this expression look like?
How does it have to look like for an eigenstate?

 

[CAF123]

??
Then you should try other values of x_0, too.

I am not sure what you mean - if x is not in (-b,b) then V is zero. If x is in (-b,b) then the potential is parabolic with a minimum value V_o

Which path, which incoming particle?

For the purposes of the question, the particle has -V_o< E < 0. ##\psi_{-}## must be associated to a quantum particle which has negative energy. I am only trying to get an idea of what is going on, I don't know if it helps me solve the question or not.

Why?
(it does not, and the wave function is not a decaying exponential)

For the particle to be in (-b,b) the particle must have penetrated one side of the potential barrier. Is ##\psi_{-}## not a decaying exponential?

How does this expression look like?
How does it have to look like for an eigenstate?

Is this the right approach? Subbing in ##\psi_{-}##into the TISE: ##-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi_{-} + V(x)\psi_{-} = E\psi_{-}## gives the following; $$\psi_{-} \left[\frac{\hbar^2}{2m} \frac{1}{x_o^2} - \frac{\hbar^2}{2m} \frac{x^2}{x_o^6} + \frac{1}{2}mw^2x^2 - V_0\right] = E\psi_{-}. $$ For ##\psi_{-}## to be an eigenstate, the bracketed part must equal E. But why does this only give one ##x_o##?

Thanks for reply.

 

[mfb]

I am not sure what you mean - if x is not in (-b,b) then V is zero. If x is in (-b,b) then the potential is parabolic with a minimum value V_o

x_0 means x₀, that is not x.
Did you plot the function for different values of x₀? Try ##x_0=\frac{b}{5}##, for example. Do you see a deviation from a parabola?

For the purposes of the question, the particle has -V_o< E < 0. ##\psi_{-}## must be associated to a quantum particle which has negative energy.

The energy scale is arbitrary, there is no special meaning of "negative energy" - apart from the fact that this is the energy range where you are supposed to look for solutions.

For the particle to be in (-b,b) the particle must have penetrated one side of the potential barrier. Is ##\psi_{-}## not a decaying exponential?

It is not. It would be a decaying exponential outside of (-b,b) if (!) the potential there would be constant - it is not.

Is this the right approach?

It is.

For ##\psi_{-}## to be an eigenstate, the bracketed part must equal E.

The important point here: it must be E independent of x. How can the bracket be independent of x?

 

[CAF123]

x_0 means x₀, that is not x.
Did you plot the function for different values of x₀? Try ##x_0=\frac{b}{5}##, for example. Do you see a deviation from a parabola?

I think I see your point - at x =-b,b, you could have a spike. However, a parabolic form is still okay I think.

It is not. It would be a decaying exponential outside of (-b,b) if (!) the potential there would be constant - it is not.

Okay I see now that it is obviously not a decaying exponential, a Gaussian rather. Outside the region [-b,b], the potential is a constant (=0), so should I not expect a decaying solution? When I solve the Schrodinger eqn this is what I get?

The important point here: it must be E independent of x. How can the bracket be independent of x?

So that means the coefficient of ##x^2## in the bracketed part must be zero. I.e $$\frac{1}{2} mw^2 - \frac{\hbar^2}{2m} \frac{1}{x_0^4}= 0 \Rightarrow x_0^4 = \frac{\hbar^2}{m^2 w^2}$$ and so ##x_o = \pm \frac{\hbar^{1/2}}{w^{1/2}m^{1/2}}##.

The next part of the question is about finding the wave function for all x and the probabilities of finding the particle in or outside the well. I think it makes more sense to consider the particle already inside the well - I think if you consider instead an incoming particle then in this set up, it would have negative kinetic energy which is not physical. For x < -b and x > b, the Schrodinger eqn is $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi_{-} = E \psi_{-}\Rightarrow \frac{d^2}{dx^2} \psi_{-} = -\frac{2mE}{\hbar^2}\psi_{-}$$For a bound state, E < 0 and so ##-2mE/\hbar^2## will be positive. Hence ##\psi_{-} = A\exp(\sqrt{2mE}/\hbar x)## for x < -b and ##\psi_{-} = D\exp(-\sqrt{2mE}/\hbar x)## for x > b. (Reject the diverging exp terms) By the continuity of the wave function I can determine A and D in terms of ##R_{-}##. Does this seem ok?

 

[mfb]

I think I see your point - at x =-b,b, you could have a spike. However, a parabolic form is still okay I think.

I would not call this a spike, but it can be discontinuous. And I misinterpreted your potential, see below:

Okay I see now that it is obviously not a decaying exponential, a Gaussian rather.

Inside, okay.

Outside the region [-b,b], the potential is a constant (=0), so should I not expect a decaying solution? When I solve the Schrodinger eqn this is what I get?

Oh sorry, I misread the potential. Okay, outside you get a decaying exponential, I agree.

So that means the coefficient of ##x^2## in the bracketed part must be zero. I.e $$\frac{1}{2} mw^2 - \frac{\hbar^2}{2m} \frac{1}{x_0^4}= 0 \Rightarrow x_0^4 = \frac{\hbar^2}{m^2 w^2}$$ and so ##x_o = \pm \frac{\hbar^{1/2}}{w^{1/2}m^{1/2}}##.

Good. Both x₀ give the same wave function.

The next part of the question is about finding the wave function for all x and the probabilities of finding the particle in or outside the well. I think it makes more sense to consider the particle already inside the well - I think if you consider instead an incoming particle then in this set up, it would have negative kinetic energy which is not physical.

There is no incoming particle.

For x < -b and x > b, the Schrodinger eqn is $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi_{-} = E \psi_{-}\Rightarrow \frac{d^2}{dx^2} \psi_{-} = -\frac{2mE}{\hbar^2}\psi_{-}$$For a bound state, E < 0 and so ##-2mE/\hbar^2## will be positive. Hence ##\psi_{-} = A\exp(\sqrt{2mE}/\hbar x)## for x < -b and ##\psi_{-} = D\exp(-\sqrt{2mE}/\hbar x)## for x > b. (Reject the diverging exp terms) By the continuity of the wave function I can determine A and D in terms of ##R_{-}##. Does this seem ok?

The way you use the signs looks wrong. If E<0, your square root is imaginary, and you don't get a decay.

 

[CAF123]

The way you use the signs looks wrong. If E<0, your square root is imaginary, and you don't get a decay.

I am not really sure how I would correct this. The negative that appears there in the first place (in the h/2m term) is already there in the Schrodinger eqn. So I will definitely have a negative on the RHS. Then I said that since E is negative in our case, then this makes the RHS positive. So when I find the roots of the characteristic polynomial I have real solutions.

 

[mfb]

The minus sign should stay in the square root, as (-2mE) is positive and will give real roots, while (2mE) is/does not. Just check the individual steps, it has to be there.