Solving Potential V with Schrodinger Equation

[KFC]

Assume the potential in question is

<br /> V = \left\{<br /> \begin{matrix}<br /> \infty, \qquad x&lt;0 \\<br /> -V_0, \qquad 0\leq x \leq a \\<br /> 0, \qquad x&gt;a<br /> \end{matrix}<br /> \right.<br />

where V_0 is positive.

if we need to find the bound state, we consider the energy is less than the potential. But the potential withn [0, a] is negative, is that mean the energy will be more negative (i.e. |E| &gt; V_0) ?

Generally, the Schrodinger equation will be written of the following form

<br /> \frac{d^2\psi}{dx^2} + k^2\psi = 0<br />

where

k = \sqrt{\frac{2m}{\hbar^2}(E-V)}. The general solution is of the form

\psi = A\exp(ikx) + B\exp(-ikx)

For x<0 or x>a region, the wavefunction must decay because the potential is larger than E, k=\sqrt{\frac{2m}{\hbar^2}(E-V)} = i \sqrt{\frac{2m}{\hbar^2}(V-E)} = i\kappa, the solutions in those region become

\psi = A\exp(\kappa x) + B\exp(-\kappa x)

But within [0, a], if we want to find bound state, energy must be negative and |E|&gt;V_0, so

k = \sqrt{\frac{2m}{\hbar^2}(-|E|-(-V_0))} = \sqrt{\frac{2m}{\hbar^2}(V_0-|E|)}

but this also lead to k be imaginary number, that is, the solution within [0, a] is decaying again? But I think the solution in that region should be oscillating. Where am I get wrong?

 

[ice109]

first of all your "general solution" isn't a general solution. it's the solution in the well and you produce the correct solution for x>a by luck.

energy should be less than zero but greater than -v_0.

for x<0 the wavefunction doesn't exist. write out the entire time independent shrodinger eqn like this:

\frac{-h^2}{2m}\frac{d^2 \psi(x)}{dx^2}+V(x) \psi (x) = E \psi(x)

for each region and don't skip steps and you should end up with

k=\pm\sqrt{\frac{2m}{\hbar^2}(-E+V_0)}=\pm\sqrt{-\frac{2m}{\hbar^2}(V_0 - E)}

inside the halfwell which produces standing waves in the well. note E is negative by assumption.

 

[KFC]

Thanks for reply.

first of all your "general solution" isn't a general solution. it's the solution in the well and you produce the correct solution for x>a by luck.

I don't understand why not. In that solution form, k is varied from region to region. In rightmost region, k is imaginary which decaying solution. So why can I say in all regions they have same form of solution but with different k?

energy should be less than zero but greater than -v_0.

This is my doubt. I wonder what's the requirement of bound state? Is E<V or |E|<|V| ?
If the potential V is positive, it will trap the particle has energy less than the potential, it is quite trivial. But how to understanding the situation when the potential is negative?

for x<0 the wavefunction doesn't exist. write out the entire time independent shrodinger eqn like this:

\frac{d^2 \psi(x)}{dx^2}+V(x) \psi (x) = E \psi(x)

for each region and don't skip steps and you should end up with

k=\pm\sqrt{-\frac{2m}{\hbar^2}(E+V_0)}

inside the halfwell which produces standing waves in the well.

 

[ice109]

Thanks for reply.


I don't understand why not. In that solution form, k is varied from region to region. In rightmost region, k is imaginary which decaying solution. So why can I say in all regions they have same form of solution but with different k?

but you have it wrong. in right most region k is real and therefore the solution decays, and in the well k is imaginary and hence oscillates. have you taken a differential equations class?

This is my doubt. I wonder what's the requirement of bound state? Is E<V or |E|<|V| ?
If the potential V is positive, it will trap the particle has energy less than the potential, it is quite trivial. But how to understanding the situation when the potential is negative?

the requirement for a bound state is |E|<|V|. you can understand the situation when the potential is negative by realizing that there's no such thing as absolute potentials, only relative potentials and therefore the whole situation is the same if we shift the potential up everywhere by V_0. then for 0&lt;x&lt;a ~~ V(x)=0, for x&gt;a ~~ V(x) = V_0 and the for x&lt;0 ~~ V(x) = \infty

 

[KFC]

Thanks for explanation :)

but you have it wrong. in right most region k is real and therefore the solution decays, and in the well k is imaginary and hence oscillates. have you taken a differential equations class?

the requirement for a bound state is |E|<|V|. you can understand the situation when the potential is negative by realizing that there's no such thing as absolute potentials, only relative potentials and therefore the whole situation is the same if we shift the potential up everywhere by V_0. then for 0&lt;x&lt;a ~~ V(x)=0, for x&gt;a ~~ V(x) = V_0 and the for x&lt;0 ~~ V(x) = \infty

 

[ice109]

no prob. just out of curiosity what is your native language?