Bound state of a square well, no allowed bound state mean?

[Outrageous]

Homework Statement

Show in the graph ,there will be no allowed bound states with odd-parity if the well depth is less than ${V_min}$
Find ${V_min}$ in terms of k and a.where a is the half of the well width.
What does no allowed bound state mean?

Homework Equations

$cotz=-pa/z$ where p^2 =$/frac^{2m(V-E)}{/hbar}$
z =ka ,where k=2mE/hbar

The Attempt at a Solution

Actually I just want to know when do we have no allowed bound state? What kind of condition I have to fulfill? ( eg.i think no allowed bound state, E≥V so that the particle will not be tracked, correct ? But how to show on graph.
The interception point can be used to find the E or V if one of the is given. no allowed bound state mean no interception point?

Thanks

 

[vela]

Homework Statement

Show in the graph there will be no allowed bound states with odd-parity if the well depth is less than $V_{min}$. Find $V_{min}$ in terms of k and a, where a is the half of the well width. What does no allowed bound state mean?

Homework Equations

$\cot z =-\frac{pa}{z}$ where $p^2=\frac{2m(V-E)}{\hbar}$ and $z=ka$ where $k=\frac{2mE}{\hbar}$

The Attempt at a Solution

Actually I just want to know when do we have no allowed bound state? What kind of condition I have to fulfill? (e.g. I think no allowed bound state, E≥V so that the particle will not be tracked, correct? But how to show on graph.)

I'm not sure what you mean by the particle not being tracked. If E≥V, then the particle is unbound. That's not the same as saying there is no bound state.

The interception point can be used to find the E or V if one of the is given. No allowed bound state mean no interception point?

Thanks

Right. The allowed energies are those for which the condition $\cot z = -\frac{pa}{z}$ is satisfied. (I'm trusting that you supplied the correct condition here.) As you vary V, the number of places where the two curves intersect will change. The question is asking you to identify the minimum value of V such that there is no bound state of odd parity. Note that there could still be a bound state of even parity.

 

[Outrageous]

Thanks for replying. So my answer is (-pa/z)< (pi /2), by keeping the energy constant and the solve for the minimum V.
One more to ask, actually what does it mean no bound state ? I wonder what is the physical meaning of no bound state? No wave function?

 

[hilbert2]

A square well always has at least one bound state, no matter how shallow it is. The lowest energy state always has even parity, so there need not exist a bound state with odd parity (if there's only one bound state). If there's no bound state, it means that the energy eigenfunctions form a continuous spectrum and can't be made real-valued with any choice of the normalization constant.

 

[Outrageous]

A square well always has at least one bound state, no matter how shallow it is. The lowest energy state always has even parity, so there need not exist a bound state with odd parity (if there's only one bound state). If there's no bound state, it means that the energy eigenfunctions form a continuous spectrum and can't be made real-valued with any choice of the normalization constant.

Thanks. Bound state also refer to the wavefunction $\psi_n$.
Just to confirm: my solution for the question is (pa/z)< (pi/2) ,solve for minimum V

 

[Outrageous]

But less than pi/2 doesn't mean it must have one interception point with the even parity, then ...?

 

[vela]

Your condition doesn't look correct to me. It should be independent of E.

Try looking over the section where the wave functions are found in http://en.wikipedia.org/wiki/Finite_potential_well.

 

[Outrageous]

Your condition doesn't look correct to me. It should be independent of E.

Try looking over the section where the wave functions are found in http://en.wikipedia.org/wiki/Finite_potential_well.

I am sorry, but I am still confused.
No need to keep E constant, because the interception point I want is $${\u_o}$$ (from Wikipedia ) . But I still wonder how to find out the minimum voltage? $$ {u_ o} < {\frac{\pi}{2}}$$ then solve for minimum V.