- #1

Dammes

- 15

- 0

in the infinite well with small potential shown in the attachment.

I calculated the total energy by using the time independent Schrodinger equation and adding the correction energy to the equation of the slope k=(Vo/L)x.

E=h^2/8mL^2 +∫ ψkψ dx

ψ=√(2/L) sin(∏/L x)

when integrating ∫ ψkψ dx between 0 and L

I got Zero, ∫ ψkψ dx=0

∴ total energy=h^2/8mL^2 +0

so what i don't understand is when adding a small potential it doesn't affect the total energy of the system? that is what it shows when i integrated it.

I calculated the total energy by using the time independent Schrodinger equation and adding the correction energy to the equation of the slope k=(Vo/L)x.

E=h^2/8mL^2 +∫ ψkψ dx

ψ=√(2/L) sin(∏/L x)

when integrating ∫ ψkψ dx between 0 and L

I got Zero, ∫ ψkψ dx=0

∴ total energy=h^2/8mL^2 +0

so what i don't understand is when adding a small potential it doesn't affect the total energy of the system? that is what it shows when i integrated it.