# Time-independent SE linear combination solution help

1. Feb 26, 2015

### Maylis

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.

Last edited: Feb 26, 2015
2. Feb 26, 2015

### 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!

3. Feb 28, 2015

### Maylis

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

4. Mar 1, 2015

### 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.