1. ## simple substitution problem

2. Originally Posted by transgalactic

3. my link works kust press it
maybe you have a popup blocker

4. i used this property

$\frac{\partial}{\partial t}[ X(x)T(t) ] = X(x)[\frac{\partial}{\partial t}T(t)]$ because X(x) won't vary with t

i applied it and still didnt get the needed expresion

http://i50.tinypic.com/67uyc7.jpg

5. your RHS is wrong,.. it should be $- T(t) \frac{h^2}{2m}\frac{\partial^2}{\partial x^2}X(x)$

rearrange it and you'll get the result.

$i h\frac{\partial \Psi}{\partial t}= \frac{h^2}{2m}\frac{\partial^2\Psi}{\partial x^2}$
If you let $\Psi= X(x)T(t)$, you have $\frac{\partial\Psi}{\partial t}= X(x)\frac{dT}{dt}$ and $\frac{dX}{dx}T(t)$ (since X and T are functions of a single variable you should not have " $\partial$") so your equation becomes
$i h X(x)\frac{dT}{dt}= \frac{h^2}{2m}T\frac{d^2X}{dx^2}$
which is what you have. Now divide both sides by XT:
$\frac{i h}{T}\frac{dT}{dt}= \frac{h^2}{2m X}\frac{d^2X}{dx^2}$
the equation you want to get to.

The importance of this, by the way, is the fact that the right side now depends only of x while the left side depends only on t. In order to be equal for all x and t, each must be a constant! That allows you to separate the partial differential equation into two ordinary differential equations,
$\frac{ih}{T}\frac{dT}{dt}= \lambda$
and
$\frac{h^2}{2m X}\frac{d^2X}{dx^2}= \lambda$
which are the same as
$ih \frac{dT}{dt}= \lambda T$
and
$\frac{h^2}{2m}\frac{d^2X}{dx^2}= \lambda X$.

Of course, you have to determine possible values for $\lambda$ and you usually do that by looking at boundary conditions on X. Notice that $\lambda$ is an "eigenvalue" for both differential operators.

7. wow super
THANKS