# Thread: Intro to PDE's, Schrodinger's Equation

1. ## Intro to PDE's, Schrodinger's Equation

(a) Using the separation of variables technique, and letting the separation constant be denoted E (which turns out to be the total energy of the particle), show that the resulting differential equation that is independent of time—the so-called time-independent Schroedinger equation (TISE)—has the (1-D) form

$\displaystyle -\frac{\hbar}{2m}\frac{d^2\psi}{dx^2} + U(x)\psi(x) = E\psi{x}$

where we have set $\displaystyle \Psi(x,t) = \psi(x) T(t)$.

This I have done this already... And I've narrowed down to these two differential equations:

The one above

$\displaystyle i\hbar \frac{dT}{dt} = ET(t)$

(b) Also, show that the total wavefunction in this case has the form

$\displaystyle \Psi(x,t) = \psi(x) e^{-i\frac{E}{\hbar}t}$

Here's what I did:

$\displaystyle i\hbar \frac{dT}{dt} = ET(t)$

$\displaystyle \frac{dT}{T} = \frac{E}{i\hbar}dt$

$\displaystyle ln|T| = \frac{Et}{i\hbar} + C$

$\displaystyle T = e^{\frac{Et}{i\hbar} + C}$

$\displaystyle T = Ae^{\frac{Et}{i\hbar}}$

This looks only similar to the actual solution for T(t)... What am I doing wrong?

2. Originally Posted by Aryth
(a) Using the separation of variables technique, and letting the separation constant be denoted E (which turns out to be the total energy of the particle), show that the resulting differential equation that is independent of time—the so-called time-independent Schroedinger equation (TISE)—has the (1-D) form

$\displaystyle -\frac{\hbar}{2m}\frac{d^2\psi}{dx^2} + U(x)\psi(x) = E\psi{x}$

where we have set $\displaystyle \Psi(x,t) = \psi(x) T(t)$.

This I have done this already... And I've narrowed down to these two differential equations:

The one above

$\displaystyle i\hbar \frac{dT}{dt} = ET(t)$

(b) Also, show that the total wavefunction in this case has the form

$\displaystyle \Psi(x,t) = \psi(x) e^{-i\frac{E}{\hbar}t}$

Here's what I did:

$\displaystyle i\hbar \frac{dT}{dt} = ET(t)$

$\displaystyle \frac{dT}{T} = \frac{E}{i\hbar}dt$

$\displaystyle ln|T| = \frac{Et}{i\hbar} + C$

$\displaystyle T = e^{\frac{Et}{i\hbar} + C}$

$\displaystyle T = Ae^{\frac{Et}{i\hbar}}$

This looks only similar to the actual solution for T(t)... What am I doing wrong?
Note $\displaystyle \frac{1}{i} = - i$ and absorb $\displaystyle A$ into $\displaystyle \psi(x)$.