Quantum potential problem -- Particle confined in 1 dimension

[gabz220]

Homework Statement

A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential: V(x) = infinity when x<0, V(x) = -e^2/4*pi*ε*x when x≥0
For the region x ≥ 0 , by substituting in the Schrödinger equation, show that the wave function
u(x) = C*x*exp(-αx) can be a satisfactory solution of the Schrodinger equation so long as the constant α is suitably chosen. Determine the unique expression for α in terms of m , e and other fundamental constants. Note that C is a normalisation constant.then Show that the energy of the particle represented by u (x) is given by
E = -m*e^4/ 2(4*pi*ε*ħ)^2
This has me really stumped, can't show that the energy is equal to the equation above.

Homework Equations

Schrodinger equation

The Attempt at a Solution

attempt:
d^2(u(x))/dx^2 = ((E - V(x))2m/ħ^2)u(x) rearranging the Schrodinger equation
and
d^2(u(x))/dx^2 = (α^2 - 2α/x)C*x*exp(-αx)
so
((E - V(x))2m/ħ^2) = (α^2 - 2α/x) equating coefficients
this is where i get stuck. Any help apreciated

 

[PeroK]

Homework Statement

A particle with mass m and electric charge e is confined to move in one dimension along the x -axis. It experiences the following potential: V(x) = infinity when x<0, V(x) = -e^2/4*pi*ε*x when x≥0
For the region x ≥ 0 , by substituting in the Schrödinger equation, show that the wave function
u(x) = C*x*exp(-αx) can be a satisfactory solution of the Schrodinger equation so long as the constant α is suitably chosen. Determine the unique expression for α in terms of m , e and other fundamental constants. Note that C is a normalisation constant.then Show that the energy of the particle represented by u (x) is given by
E = -m*e^2/ 2(4*pi*ε*ħ)^2
This has me really stumped, can't show that the energy is equal to the equation above.

Homework Equations

Schrodinger equation

The Attempt at a Solution

attempt:
d^2(u(x))/dx^2 = ((E - V(x))2m/ħ^2)u(x) rearranging the Schrodinger equation
and
d^2(u(x))/dx^2 = (xα^2 - 2α)C*x*exp(-αx) (***)
so
((E - V(x))2m/ħ^2) = (xα^2 - 2α) equating coefficients
this is where i get stuck. Any help apreciated

(***) You might want to check your differentiation.

 

[gabz220]

I have, i can't see the mistake; i even wolfram alpha'd it and it gave me the same awnser

 

[PeroK]

I have, i can't see the mistake; i even wolfram alpha'd it and it gave me the same awnser

There must be one term without a factor of x. $f(x) = xg(x) \ \Rightarrow \ f''(x) = 2g'(x) + xg''(x)$

 

[gabz220]

There must be one term without a factor of x. $f(x) = xg(x) \ \Rightarrow \ f''(x) = 2g'(x) + xg''(x)$

Your right, sorry, i copied it down wrong. I have been working with the correct derivative though and i still can't do it.
question now corrected

 

[PeroK]

Your right, sorry, i copied it down wrong. I have been working with the correct derivative though and i still can't do it.
question now corrected

Hint: when are two polynomials equal?

 

[PeroK]

You may also have mistyped the answer. I get $E = \frac{-me^4}{2(4\pi \epsilon \hbar)^2}$

 

[gabz220]

You may also have mistyped the answer. I get $E = \frac{-me^4}{2(4\pi \epsilon \hbar)^2}$

Yeah that was the answer.
thanks for the help

 

[BvU]

Hello gabz,

Interesting exercise.
In the future, Could you use brackets in appropriate places when posting ? For example, I assume your potential is $$
V(x) = -{e^2\over 4 \pi \epsilon_0 \; x} \ \rm ?
$$

 

[gabz220]

Hello gabz,

Interesting exercise.
In the future, Could you use brackets in appropriate places when posting ? For example, I assume your potential is $$
V(x) = -{e^2\over 4 \pi \epsilon_0 \; x} \ \rm ?
$$

yeah, sorry about that. Made quite a few mistakes, ill be more careful next time