0
$\begingroup$

So starting from the time dependent schrodinger equation I perform separation of variables and obtain a time and spatial part. The spatial part is in effect the time independent schrodinger equation.

Since we are dealing with a free particle I can take the time independent equation, set V = 0 and solve.

I can do this successfully to obtain :

$Ae^{+i\sqrt{{2mE}/{\hbar^{2}}}x}+Be^{-i\sqrt{{2mE}/{\hbar^{2}}}x}$

My lecturer has a small section titled :

Solving for the Free Schrodinger Equation

$$V=0$$

$$\frac{\hbar^{2}}{2m}\frac{\partial^2\psi}{\partial x^2}+E\psi=0$$

$$E=\frac{p^2}{2m}$$

$$\psi=Ce^{-{iEt}/{\hbar}+{ipr}/{\hbar}}$$

This is the solution to the free TISE and TDSE.

He seems to be doing the same thing as me initially but he's obtained a different result ?

Also, the first section of his answer :$e^{-{iEt}/{\hbar}}$ is the solution to the time part of the equation (described above).

| cite | improve this question | |
$\endgroup$
1
$\begingroup$

The lecturer's answer assumes $p$ can be positive or negative, and your answer assumes $p=\sqrt{2mE}$ is positive. The factor of $\exp{(-iEt/\hbar)}$ is just the time-dependent part of the separated solution.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So are both valid answers ? Also, when I solve his equation with E = p^2/2m I obtain the solution : Ae^(+ip/hbar)+Be^(-ip/hbar). Why has conveniently ignored one half of this solution ? $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 17:49
  • 1
    $\begingroup$ I couldn't tell you why the lecturer chose to write it a certain way, but the solution works for $-p$ just as well as for $+p$. This just corresponds to a particle moving right or moving left. The lecturer's way says there are two solutions, and your way is a superposition of those two. Both are valid. $\endgroup$ – George G Apr 3 '14 at 18:19
  • $\begingroup$ So as a final solution if I were to write : $exp(-iEt/\hbar)$*$[A'e^{+ip/\hbar}+B'e^{-ip/\hbar}]$ Would this be also correct ? $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 18:25
  • $\begingroup$ I know I'm really drawing out the issue but I just want to be sure. $\endgroup$ – Ari Ben Canaan Apr 3 '14 at 18:26
  • 1
    $\begingroup$ The particle here is moving in both the +x and -x directions, but it's a valid solution to the free particle hamiltonian. $\endgroup$ – George G Apr 3 '14 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.