- #1

Kaguro

- 221

- 57

Moved from a technical forum

The problem is:

Solve the time independent Schrodinger Equation for infinite square well centered at origin. Show that the energy is same as in the original case(well between x=0 and x=L). Also show that the solution to the this case can be obtained by setting x to x-L/2 in ##\psi## in the original one.

Now, in the original problem, E = ## \frac{n^2\pi^2ħ^2}{2mL^2}##.

Here, I got my answer as 4 times that value:

From the Schrodinger Equation, put V=0 in (-L/2,L/2):

##\frac{d^2\psi}{dx^2}## + ##\frac{2mE}{ħ^2}\psi## = 0

So,

##\psi## = Acos(kx) + Bsin(kx) where ##k^2 = \frac{2mE}{ħ^2}##

Now the boundary conditions require ##\psi## be 0 at x=-L/2 and x=+L/2

Putting these and solving them I get:

0 = Acos(kL/2) - Bsin(kL/2) and 0 = Acos(kL/2) + Bsin(kL/2)

From them:

tan(kL/2) = A/B and also A/B = -tan(kL/2)

which means: 2tan(kL/2) = 0

and so kL/2 = n##\pi##

From the definition of k, I have:

E = ##\frac{2n^2\pi^2ħ^2}{mL^2}##

What mistake did I make?

Solve the time independent Schrodinger Equation for infinite square well centered at origin. Show that the energy is same as in the original case(well between x=0 and x=L). Also show that the solution to the this case can be obtained by setting x to x-L/2 in ##\psi## in the original one.

Now, in the original problem, E = ## \frac{n^2\pi^2ħ^2}{2mL^2}##.

Here, I got my answer as 4 times that value:

From the Schrodinger Equation, put V=0 in (-L/2,L/2):

##\frac{d^2\psi}{dx^2}## + ##\frac{2mE}{ħ^2}\psi## = 0

So,

##\psi## = Acos(kx) + Bsin(kx) where ##k^2 = \frac{2mE}{ħ^2}##

Now the boundary conditions require ##\psi## be 0 at x=-L/2 and x=+L/2

Putting these and solving them I get:

0 = Acos(kL/2) - Bsin(kL/2) and 0 = Acos(kL/2) + Bsin(kL/2)

From them:

tan(kL/2) = A/B and also A/B = -tan(kL/2)

which means: 2tan(kL/2) = 0

and so kL/2 = n##\pi##

From the definition of k, I have:

E = ##\frac{2n^2\pi^2ħ^2}{mL^2}##

What mistake did I make?