Nonreal solutions to the TISE

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The discussion focuses on expressing non-real solutions to the Schrödinger equation, specifically \(\psi(x)\), as a linear combination of real solutions. The equation under consideration is \(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V\psi(x) = E\psi(x)\). Participants confirm that non-real solutions \(\psi\) and \(\psi^*\) can be represented as \((\psi + \psi^*)\) and \(i(\psi - \psi^*)\), which are real solutions. The suggested approach involves substituting \(\psi(x) = \psi_{1}(x) + i\psi_{2}(x)\) and separating the real and imaginary components to solve the equation.

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G01
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1. I am try to show that a non-real solution to the shrodinger eq., [tex]\psi(x)[/tex] can be expressed as a linear combination of real solutions.
2. Here's what I know:
Shrodinger Eq. [tex]-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} +V\psi(x) = E\psi(x)[/tex]

[tex]\psi[/tex] [tex]\psi*[/tex] are non-real solutions

[tex](\psi + \psi*)[/tex] and[tex]i(\psi -\psi*)[/tex] are real solutions.


3. Well, to start, since I am given two real solutions, I am assuming [tex]\psi[/tex] can be expressed as a linear combination of the two real solutions above. I tried this two different ways:

First I tried plugging the real solutions into the shrodinger eq. and seeing where it got me. I was able to split the equation into two separate equations involving only the original psi and psi* but I don't think this helps me.

I also tried setting psi = A +Bi and seeing if I could combine the real solutions in a way so they equal psi. Also, this hasn't been of much help so far. If I am somewhat on track but am confused, please point me in the right direction. Or am I completely off track? Thank you for your insight.
 
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Take [itex]\psi (x)= \psi_{1}(x)+i\psi_{2}(x) \ , \psi_{1,2}(x)\in \mathbb{R}[/itex]. Plug the psi in the SE and and separate the real and imaginary part. That's all.
 
Thanks Dexter.
 

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