One-dimensional Schrodinger Equation

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one-dimensional Schrodinger's Equation

Hi !
I wonder how to solve one-dimensional Schrodinger's Equation :

[tex] \frac{d^2 \psi (x)}{dx^2}\ = \ -(\frac{2 \pi}{\lambda})^2 \ * \ \psi (x) [/tex]

I've to obtain [tex]\psi (x) [/tex] , when
[tex]-(\frac{2 \pi}{\lambda})^2 [/tex] is known
Can you solve it as an example, when [tex]-(\frac{2 \pi}{\lambda})^2 = 2 [/tex] ? Please...
 
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Answers and Replies

  • #2
Based on my rather basic knowledge, making
[tex]
- \left( \frac{2 \pi}{ \lambda} \right)^2 = 2
[/tex]
would make the psi function an exponential function rather than having the wave characteristics you want.The solution would be:
[tex]
\psi(x) = e^{ \sqrt{2} x } .
[/tex]
But you want a periodic wave function. By taking
[tex]
\frac{2 \pi}{\lambda} = \sqrt{2}
[/tex]
(i.e. without the minus sign), the solution would be a wave equation and complex:
[tex]
\psi(x) = e^{ \sqrt{2} i x } .
[/tex]
 
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  • #3
Tom Mattson
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Megus said:
Hi !
I wonder how to solve one-dimensional Schrodinger's Equation :

[tex] \frac{d^2 \psi (x)}{dx^2}\ = \ -(\frac{2 \pi}{\lambda})^2 \ * \ \psi (x) [/tex]

I've to obtain [tex]\psi (x) [/tex] , when
[tex]-(\frac{2 \pi}{\lambda})^2 [/tex] is known
Can you solve it as an example, when [tex]-(\frac{2 \pi}{\lambda})^2 = 2 [/tex] ? Please...
You shouldn't commit to a specific value of -(π/λ)2. Just define it to be a literal constant, say, -k2. Then your differential equation is:

ψ"+k2ψ=0

Surely you have the solution of that one somewhere!
 
  • #4
Tom Mattson
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speeding electron said:
Based on my rather basic knowledge, making
[tex]
- \left( \frac{2 \pi}{ \lambda} \right)^2 = 2
[/tex]
would make the psi function an exponential function rather than having the wave characteristics you want.The solution would be:
[tex]
\psi(x) = e^{ \sqrt{2} x } .
[/tex]
No, an exponential function arises when the first derivative is negatively proportional to the wavefunction itself. The problem here involves the second derivative.

But you want a periodic wave function. By taking
The solution to his equation is indeed sinusoidal, but let's let him work it out. :wink:
 
  • #5
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In fact [tex]$\frac{{2\pi }}
{\lambda } = k$
[/tex] when k is the wavenumber, so the k^2 substution can be made even without the knowledge of how it helps solve the DE.
 
  • #6
Tom Mattson
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heardie said:
In fact [tex]$\frac{{2\pi }}
{\lambda } = k$
[/tex] when k is the wavenumber, so the k^2 substution can be made even without the knowledge of how it helps solve the DE.
Yes, my choice in symbols was not accidental. :wink:
 
  • #7
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Is it the correct answer ?:
[tex]\psi (x) = e^{ikx + \varphi_R + i \varphi_I} = e^{\varphi_R} \cdot e^{i(kx + \varphi_I)} = A(\cos (kx + \varphi_I) + i \sin (kx + \varphi_I))[/tex]
 
  • #8
Tom Mattson
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Megus said:
Is it the correct answer ?:
[tex]\psi (x) = e^{ikx + \varphi_R + i \varphi_I} = e^{\varphi_R} \cdot e^{i(kx + \varphi_I)} = A(\cos (kx + \varphi_I) + i \sin (kx + \varphi_I))[/tex]
Yes, though you might find it more convenient to replace the constants A and φI with A and B as follows:

ψ(x)=Acos(kx)+Bsin(kx),

and applying the boundary conditions from there.
 
  • #9
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Ok - thanks :biggrin:
 

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