- #1
Dammes
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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.
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