A 1D Quantum Well - Different width "naming" gives different result ?

Solmyros
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Boundary conditions give different result depending on whether the quantum well is defined from 0 to L or from -L/2 to +L/2.
Greetings everyone,

Exactly as the title says. I am reaching to something strange and I do not know what I am missing. It must be something obvious...

case 1: -L/2 to L/2

After taking the Schrodinger equation and considering potential equal to zero inside we reach at this: $$Ψ=A×e^{ikz}+B×e^{- ikz}$$
case 1.png

Boundary conditions
Outside the well the potential is assumed infinite, so the wavefunction is equal to zero:
  • $$\Psi = 0, when \left|{z}\right|>\frac{L}{2}$$
  • $$Ψ(−\frac{L}{2})=0 \ (1)$$
  • $$Ψ(+\frac{L}{2})=0 \ (2)$$
From 1. we have : $$B=−A×e^{−ikL} \ (3)$$

Substituting (3) in 2. we get: $$k=\frac{n \times \pi}{L}, where \ n=0,1,2,... (4)$$

Therefore from (3) we get: $$B=−A \times e^{−inπ}$$

And finally, if $$n=2m, where \ m=0,1,2,...$$ we get from the original Schrodinger Equation: $$\Psi_{even}(z) = A \times ( e^{i k z} - e^{- i k z)} = 2 i A \sin{(kz)} \ (5)$$

case 2: 0 to L

After taking the Schrodinger equation and considering potential equal to zero inside we reach at this: $$Ψ=A'×e^{ikz}+B'×e^{- ikz}$$

case 2.png


Boundary conditions
Outside the well the potential is assumed infinite, so the wavefunction is equal to zero:
  • $$\Psi = 0, \ when \ z \le 0 \ or \ z \ge L$$
  • $$Ψ(0)=0 \ (6)$$
  • $$Ψ(L)=0 \ (7)$$
From 1. we have : $$B'=−A'$$ (8)
So, we get from the original Schrodinger Equation: $$\Psi(z) = A' \times ( e^{i k z} - e^{- i k z)} = A' \sin{kz} \ (9)$$

and

$$Ψ(L)=0 = A' \sin{kL} \ which \ gives \ k=\frac{n \times \pi}{L} \ (10), where \ n=0,1,2,... (\ like \ before)$$

But, now if $$n=2m, where \ m=0,1,2,...$$ we get $$\Psi_{even}(z) = A' \sin({\frac {2m \pi }{L}z}) \ (11)$$

Normalization will eventually give the same amplitude, BUT, on the first case the wavefunction is imaginary. This added 'i' is what buffles me.

Thank you in advance for your help.
 

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