Re-arranging the Schrodinger equation

AI Thread Summary
The discussion revolves around the rearrangement of the Schrödinger equation and the manipulation of complex numbers within it. A participant seeks clarification on how to correctly move complex numbers in mathematical equations, particularly in the context of the Schrödinger equation. Another participant explains that multiplying both sides by i/ħ can help achieve the desired form. There is also a correction regarding the original equation, emphasizing the importance of accuracy in mathematical expressions. The conversation highlights the need for persistence in studying complex mathematical concepts.
vorcil
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I just have a small question,

In my book it says that the schrodinger equation,

<br /> i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi<br />

rearranged is,

<br /> \frac{\partial\Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial\Psi ^2 psi}{\partial x^2} - \frac{i}{\hbar}V\Psi<br />

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

I can see how A\hbar = B\hbar ^2 becomes A = B \hbar

but I don't understand how

A i = B + V\Psi becomes A = iB - i V\hbar

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
 
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\frac{1}{i}=-i

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.
 
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
 
Matterwave said:
\frac{1}{i}=-i

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?
\frac{1}{i} = \frac{1}{i}\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i
 
JaWiB said:
Like this?
\frac{1}{i} = \frac{1}{i}\frac{i}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i

JaWiB

how do i do this

<br /> A i = B + V\Psi to A = iB - i V\Psi<br />
 
I don't think what you have is correct.
<br /> i\hbar\frac{\partial\Psi}{\partial t} = \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi<br />
If you multiply both sides by i/\hbar, you get
<br /> -\frac{\partial\Psi}{\partial t} = i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} + \frac{i}{\hbar}V\Psi<br />
or
<br /> \frac{\partial\Psi}{\partial t} = -i\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2} - \frac{i}{\hbar}V\Psi<br />
 
mmmm well that's closer then to what I had,

thank you i'll ask my tutors tomorrow
 
You have the Schrodinger equation wrong. It should be:

<br /> i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi<br />
 
Cyosis said:
You have the Schrodinger equation wrong. It should be:

<br /> i\hbar\frac{\partial\Psi}{\partial t} =- \frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi<br />

so Jawib's way does work!??
 
  • #10
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
vorcil said:
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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