Re-arranging the Schrodinger equation

  • Thread starter vorcil
  • Start date
  • #1
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I just have a small question,

In my book it says that the schrodinger equation,

[tex]
i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
[/tex]

rearranged is,

[tex]
\frac{\partial\Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial\Psi ^2 psi}{\partial x^2} - \frac{i}{\hbar}V\Psi
[/tex]

how does the complex number, move over, and in the numerator? instead of the denominatior?

I can see how [tex] A\hbar = B\hbar ^2 becomes A = B \hbar [/tex]

but I don't understand how

[tex] A i = B + V\Psi becomes A = iB - i V\hbar [/tex]

could someone please explain to me the mathematical rules behind rearranging complex numbers in equations,

or give me some links that explain it, (in simple terms) please :P
 

Answers and Replies

  • #2
Matterwave
Science Advisor
Gold Member
3,965
327
[tex]\frac{1}{i}=-i[/tex]

This is one of the properties of imaginary numbers. I don't exactly recall a proof for this...hopefully someone else can answer your question in more detail.
 
  • #3
393
0
Do you know what I would search for If i wanted to understand how to manipulate i?

I tried the wikipedia log of complex numbers, but it is jungle of crap that is too hard to understand
tl;dr
 
  • #4
283
0
[tex]\frac{1}{i}=-i[/tex]

This is one of the properties of imaginary numbers. I don't exactly recall a proof for this...hopefully someone else can answer your question in more detail.
Like this?
[tex]\frac{1}{i} = \frac{1}{i}\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i[/tex]
 
  • #5
393
0
Like this?
[tex]\frac{1}{i} = \frac{1}{i}\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i[/tex]

JaWiB

how do i do this

[tex]
A i = B + V\Psi to A = iB - i V\Psi
[/tex]
 
  • #6
283
0
I don't think what you have is correct.
[tex]
i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
[/tex]
If you multiply both sides by [tex]i/\hbar[/tex], you get
[tex]
-\frac{\partial\Psi}{\partial t} = i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} + \frac{i}{\hbar}V\Psi
[/tex]
or
[tex]
\frac{\partial\Psi}{\partial t} = -i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi
[/tex]
 
  • #7
393
0
mmmm well that's closer then to what I had,

thank you i'll ask my tutors tomorrow
 
  • #8
Cyosis
Homework Helper
1,495
0
You have the Schrodinger equation wrong. It should be:

[tex]
i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
[/tex]
 
  • #9
393
0
You have the Schrodinger equation wrong. It should be:

[tex]
i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi
[/tex]

so Jawib's way does work!??
 
  • #10
393
0
I need to take some math papers...
I'm not as good as the rest of the physics majors at mathematics,

I've only done 1 math paper and 5 physics ones

I didn't even think to multiply both sides by i/h, like Jawib said :(

study study study
 
  • #11
459
0
I need to take some math papers...
I'm not as good as the rest of the physics majors at mathematics,

I've only done 1 math paper and 5 physics ones

I didn't even think to multiply both sides by i/h, like Jawib said :(

study study study

If you had kept at it, you would have got it eventually. You gave up. If you give up a lot that is a problem. Half-hearted studying won't get you anywhere.
 

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