- #26

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After giving the problem some rest when I looked at it again, I found that I was desperately trying to get a single function describe everything. Now that I've understood that it's not possible, I can easily present a piecewise defined function.

To summarise:

Write the TISE. Set V=0.

Find ψ= Acos (kx) + Bsin (kx)

Take help from boundary conditions to get:

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

This means Either A=0 or B=0.

Set A=0:

k = 2nπ/L

Set B=0:

k = (2n-1)π/L

So in general k = nπ/L

get energy from there: ##E_n= \frac{n^2\pi^2ħ^2}{2mL^2}##

Normalise to get ##A^2+B^2=\frac{2}{L}##

Then write the boundary condition again:

0 = ##Acos(n\pi/2) + Bsin(n\pi/2)##

If n is odd:

0 = B*1 or B=0

Then ##\psi = \sqrt{ \frac{2}{L}}cos(n\pi x/L)##

Similarly,

If n is even:

Then ##\psi = \sqrt{ \frac{2}{L}}sin(n\pi x/L)##

So:

##\psi (x) =\begin{cases} \sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ \sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases} ##

Is all good?

To summarise:

Write the TISE. Set V=0.

Find ψ= Acos (kx) + Bsin (kx)

Take help from boundary conditions to get:

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

This means Either A=0 or B=0.

Set A=0:

k = 2nπ/L

Set B=0:

k = (2n-1)π/L

So in general k = nπ/L

get energy from there: ##E_n= \frac{n^2\pi^2ħ^2}{2mL^2}##

Normalise to get ##A^2+B^2=\frac{2}{L}##

Then write the boundary condition again:

0 = ##Acos(n\pi/2) + Bsin(n\pi/2)##

If n is odd:

0 = B*1 or B=0

Then ##\psi = \sqrt{ \frac{2}{L}}cos(n\pi x/L)##

Similarly,

If n is even:

Then ##\psi = \sqrt{ \frac{2}{L}}sin(n\pi x/L)##

So:

##\psi (x) =\begin{cases} \sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ \sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases} ##

Is all good?

Last edited: