G01
Jan26-07, 06:54 PM
1. I am try to show that a non-real solution to the shrodinger eq., \psi(x) can be expressed as a linear combination of real solutions.
2. Here's what I know:
Shrodinger Eq. -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} +V\psi(x) = E\psi(x)
\psi \psi* are non-real solutions
(\psi + \psi*) andi(\psi -\psi*) are real solutions.
3. Well, to start, since I am given two real solutions, I am assuming \psi 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.
2. Here's what I know:
Shrodinger Eq. -\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} +V\psi(x) = E\psi(x)
\psi \psi* are non-real solutions
(\psi + \psi*) andi(\psi -\psi*) are real solutions.
3. Well, to start, since I am given two real solutions, I am assuming \psi 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.