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

In summary, the conversation discusses two different cases of solving the Schrodinger equation for a potential equal to zero inside a well. In both cases, the boundary conditions are that the wavefunction is equal to zero outside the well. In the first case, the resulting wavefunction is imaginary due to an added 'i' factor, while in the second case it is real. However, both wavefunctions represent the same eigenfunction and can be normalized to have equal amplitudes.
  • #1
Solmyros
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TL;DR Summary
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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  • #2
The basis eigenfunctions have an arbitrary complex factor of modulus ##1##. You can take this to be ##1## or ##i## or any other unit complex number.
 
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Likes Solmyros and DrClaude
  • #3
Wave functions with differing coefficients, real or complex of any magnitude, represent the same wave function. Normalization is required if observational probabilities are desired. Normalization makes them equal.
 
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