Delta function potential; Schrodinger Equation

[unscientific]

Homework Statement

Consider the TISE for a particle of mass m moving along the x-axis and interacting an attractive delta function potential at origin:
Part(a): What is the difference between a bound state particle and a free particle?

Part(b): Show ##\psi _{(x)} = exp (-|k|x)## is a solution of TISE, given conditions:

Part(c): Explain why symmetric solution is of form:

Part(d): Use boundary conditions to show the following:

Part(e): Investigate this solution graphically

Part(f): Find the significance of this result

Homework Equations

The Attempt at a Solution

Part(a)

A bound particle as energy E<0 while a continuum eigenfunction gives energy E>0.

Part (b)

TISE is given by:

[tex]\hat {H}\psi = E\psi[/tex]

Using first boundary condition provided we get

[tex] 2k = \frac{\lambda}{a}[/tex]

Substituting ##\psi = e^{-k|x|}## into TISE and using what we have just found:

[tex]-\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 - \frac{\hbar ^2 \lambda}{2ma}\delta_{(x)} = E[/tex]

I'm not sure why the delta function goes to zero here?

[tex]E = -\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 [/tex]

Part (c)

For |x|>a, V = 0:
[tex]-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi}{\partial x^2} = E\psi[/tex]

Solving this equation gives an exponential ##\psi = exp (-k|x|)##.

For -a<x<a, the particle is bounded and symmetric about origin so ##\psi \propto exp(kx) + exp(-kx)## which is ##\propto cosh(kx) = A cosh(kx)##.

Part (d)

Boundary Conditions:

At x = a, ##\psi_{|x|<a} = \psi_{|x|>a}##.
Thus we get:

[tex] e^{-ka} = A cosh (ka)[/tex]

Second boundary condition:
At x = a, ##\frac{\partial \psi}{\partial x}_{left} = - \frac{\partial \psi}{\partial x}_{right}##

The delta function gives opposite gradients about a.

[tex]e^{-ka} = A sinh (ka) [/tex]

Using boundary condition earlier,

[tex] tanh (ka) = 1[/tex]
[tex] tanh(ka) = 2 - 1 [/tex]

Can I simply substitute in ##2k = \frac{\lambda}{a}## here?

[tex] tanh(ka) = \frac{2ka}{ka} - 1 [/tex]
[tex] tanh (ka) = \frac{\lambda}{y} - 1 [/tex]

Part (e)

We can see that ##y_0 < \lambda##.

Since ##tanh(y_0) < 1##, thus ##y_0 > \frac{\lambda}{2}##.

Since for x≠a both delta functions are positive, it implies that the second term corresponding to the two delta functions ##\delta_{(x-a)}## and ##\delta_{(x+a)}## in the schrodinger equation are >0. Thus the total energy subtract that is less than ##-\frac{\hbar ^2\lambda^2}{8ma^2}##.

Part(f)

Two delta function corresponds to electron being attracted to the two protons in nucleus (which is more massive so can treat as stationary). This is a more stable (favourable) configuration, so

[tex]H + H^+ \rightarrow H_2^+[/tex]

 

[vela]

Homework Statement

Consider the TISE for a particle of mass m moving along the x-axis and interacting an attractive delta function potential at origin:
Part(a): What is the difference between a bound state particle and a free particle?

Part(b): Show ##\psi _{(x)} = exp (-|k|x)## is a solution of TISE, given conditions:

Part(c): Explain why symmetric solution is of form:

Part(d): Use boundary conditions to show the following:

Part(e): Investigate this solution graphically

Part(f): Find the significance of this result

Homework Equations

The Attempt at a Solution

Part(a)

A bound particle as energy E<0 while a continuum eigenfunction gives energy E>0.

Part (b)

TISE is given by:

[tex]\hat {H}\psi = E\psi[/tex]

Using first boundary condition provided we get

[tex] 2k = \frac{\lambda}{a}[/tex]

Substituting ##\psi = e^{-k|x|}## into TISE and using what we have just found:

[tex]-\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 - \frac{\hbar ^2 \lambda}{2ma}\delta_{(x)} = E[/tex]

I'm not sure why the delta function goes to zero here?

[tex]E = -\frac{\hbar ^2}{2m}(\frac{\lambda}{2a})^2 [/tex]

You're verifying the solution for x>0 and x<0. At x=0, there's a cusp, so ##\psi''## isn't defined, so the Schrodinger equation doesn't really apply.

Part (c)

For |x|>a, V = 0:
[tex]-\frac{\hbar^2}{2m}\frac{\partial ^2 \psi}{\partial x^2} = E\psi[/tex]

Solving this equation gives an exponential ##\psi = exp (-k|x|)##.

For -a<x<a, the particle is bounded and symmetric about origin so ##\psi \propto exp(kx) + exp(-kx)## which is ##\propto cosh(kx) = A cosh(kx)##.

Part (d)

Boundary Conditions:

At x = a, ##\psi_{|x|<a} = \psi_{|x|>a}##.
Thus we get:

[tex] e^{-ka} = A cosh (ka)[/tex]

Second boundary condition:
At x = a, ##\frac{\partial \psi}{\partial x}_{left} = - \frac{\partial \psi}{\partial x}_{right}##

The delta function gives opposite gradients about a.

[tex]e^{-ka} = A sinh (ka) [/tex]

This isn't correct. Integrate the Schrodinger equation between ##x=a-\varepsilon## and ##x=a+\varepsilon## and take the limit as ##\varepsilon \to 0^+##. This will yield the proper boundary condition on the derivatives.

Using boundary condition earlier,

[tex] tanh (ka) = 1[/tex]
[tex] tanh(ka) = 2 - 1 [/tex]

Can I simply substitute in ##2k = \frac{\lambda}{a}## here?

[tex] tanh(ka) = \frac{2ka}{ka} - 1 [/tex]
[tex] tanh (ka) = \frac{\lambda}{y} - 1 [/tex]

Part (e)

We can see that ##y_0 < \lambda##.

Since ##tanh(y_0) < 1##, thus ##y_0 > \frac{\lambda}{2}##.

Since for x≠a both delta functions are positive, it implies that the second term corresponding to the two delta functions ##\delta_{(x-a)}## and ##\delta_{(x+a)}## in the schrodinger equation are >0. Thus the total energy subtract that is less than ##-\frac{\hbar ^2\lambda^2}{8ma^2}##.

Part(f)

Two delta function corresponds to electron being attracted to the two protons in nucleus (which is more massive so can treat as stationary). This is a more stable (favourable) configuration, so

[tex]H + H^+ \rightarrow H_2^+[/tex]

 

[unscientific]

You're verifying the solution for x>0 and x<0. At x=0, there's a cusp, so ##\psi''## isn't defined, so the Schrodinger equation doesn't really apply.

So the equation##exp(-k|x|)## here only applies when x≠0. We can't find out what the solution is when x = 0? Is it because due to the nature of the delta function it is a sum of infinite series at x = 0?

This isn't correct. Integrate the Schrodinger equation between ##x=a-\varepsilon## and ##x=a+\varepsilon## and take the limit as ##\varepsilon \to 0^+##. This will yield the proper boundary condition on the derivatives.

But graphically you can see that it is symmetric about 0. You can also see that at x-ε the gradient is opposite that of at x+ε.

 

[vela]

So the equation##exp(-k|x|)## here only applies when x≠0. We can't find out what the solution is when x = 0? Is it because due to the nature of the delta function it is a sum of infinite series at x = 0?

At x=0, the lefthand side of the Schrodinger equation isn't defined, so it doesn't really tell you anything about ##\psi##. The continuity requirement, however, allows you to fill in the hole at x=0.

But graphically you can see that it is symmetric about 0. You can also see that at x-ε the gradient is opposite that of at x+ε.

It may look that way in the picture when you have a single delta function, but this time you have two. You no longer have the symmetry you had before. You need to justify the boundary condition mathematically.

 

[paranormal]

This result shows that the delta function potential is able to represent the interaction between the electron and protons in the nucleus, leading to the formation of a stable molecule. It also shows that the symmetric solution, with its characteristic exponential behavior, is able to accurately describe the behavior of the particle in this potential.