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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).

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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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