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

Solmyros
Messages
21
Reaction score
0
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.
 

Attachments

  • case 1.png
    case 1.png
    6.6 KB · Views: 174
Physics news on Phys.org
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.
 
  • Like
  • Informative
Likes Solmyros and DrClaude
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.
 
I am not sure if this falls under classical physics or quantum physics or somewhere else (so feel free to put it in the right section), but is there any micro state of the universe one can think of which if evolved under the current laws of nature, inevitably results in outcomes such as a table levitating? That example is just a random one I decided to choose but I'm really asking about any event that would seem like a "miracle" to the ordinary person (i.e. any event that doesn't seem to...
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Back
Top