Time-independent SE linear combination solution help

[gfd43tg]

Hello,

I am trying to derive the TISE, but I am having many questions, and the textbook (Griffiths) does not give any adequate explanation and I have minimal access to my professor. My goal is to find $\Psi (x,t)$. The book says the solution is

$$ \Psi (x,t) = \sum_{n=0}^{\infty} c_{n} \psi_{n}(x) exp(\frac {-iE_{n}t}{\hbar}) $$

So I start with the general SE (I am just taking this as a fact since the first page of Griffiths puts this equation and says its right)

$$ i \hbar \frac {\partial \Psi}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {\partial^{2} \Psi}{\partial x^{2}} + V \Psi $$

I can separate $\Psi (x,t) = \psi (x) f(t)$ and substitute

$$ i \hbar \frac {\partial [\psi (x) f(t)]}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {\partial^{2} [\psi (x) f(t)]}{\partial x^{2}} + V [\psi (x) f(t)]$$

Then divide by $\psi (x) f(t)$ in order to have the potential, $V$, stand alone

$$ i \hbar \frac {1}{f} \frac {\partial [f(t)]}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {1}{\psi (x)} \frac {\partial^{2} [\psi (x)]}{\partial x^{2}} + V $$

Now comes the first question. Griffiths says (without explanation)

$$ i \hbar \frac {1}{f(t)} \frac {\partial [f(t)]}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {1}{\psi (x)} \frac {\partial^{2} [\psi (x)]}{\partial x^{2}} + V = E $$

Now, what exactly is $E$? Is this "equality" actually just a definition, such that $E$ is just defined this way? Also, does $E$ stand for "energy"?

Anyways, they go on to find $f(t)$, which with trivial integration is exponential, and $f(t) = exp( \frac {-iE}{\hbar} t)$

So now I'm at this point, and I am not seeing how I will find $\Psi (x,t)$. I do know $\Psi (x,t) = \psi (x) exp( \frac {-iE}{\hbar} t)$ My guess is to try and substitute back into the SE

$$ i \hbar \frac {\partial [\psi (x) exp( \frac {-iE}{\hbar} t)]}{\partial t} = - \frac {\hbar^{2}}{2m} \frac {\partial^{2} [\psi (x) exp( \frac {-iE}{\hbar} t)]}{\partial x^{2}} + V [\psi (x) exp( \frac {-iE}{\hbar} t)] $$

Okay, well where the heck do I get this linear combination? This just got ugly, and the book doesn't show the steps.

 

[sk1105]

The equality in your first question comes about because you have two functions, each dependent on a different variable, which are equal. The only way to get this is if both functions are equal to something that depends on neither variable i.e. a constant. The choice of calling this constant $E$ is made a posteriori and does indeed refer to energy.

I'm afraid I can't remember how to answer your second question, which is a shame since I only learned it last year out of the same textbook!

 

[gfd43tg]

Anyone know how to make that critical step to go from the differential equation to a solution that is an infinite series?

 

[soarce]

The solution constructed as an infinite series is derived from the linearity of the Schroedinger equation. If you found two solutions then any linear combination of the two solutions is a solution of the Schroedinger equation.