simple substitution problem

**transgalactic** · December 8th 2009, 02:19 PM

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**Prove It** · December 8th 2009, 03:07 PM

Originally Posted by transgalactic

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Your link does not work...

**transgalactic** · December 8th 2009, 09:09 PM

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

**transgalactic** · December 9th 2009, 02:40 AM

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

**dedust** · December 9th 2009, 02:54 AM

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.

**HallsofIvy** · December 9th 2009, 05:10 AM

Your equation is originally,
$i h\frac{\partial \Psi}{\partial t}= \frac{h^2}{2m}\frac{\partial^2\Psi}{\partial x^2}$
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.

**transgalactic** · December 9th 2009, 07:37 AM

wow super
THANKS

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