More babby QM

[Anonymous No.4437049]

Hey /sci/,

You guys really helped me yesterday, so I come again for (hopefully) one more problem that I'm fucking clueless on. I'm supposed to show by direct substitution/differentiation that the Green's function <span class="math">G(\mathbf{r}) = \frac{i}{8 \pi^{2} r} \left [ \left ( i \pi e^{ikr} \right ) - \left ( - i \pi e^{ikr} \right ) \right ] = - \frac{e^{ikr}}{4 \pi r}[/spoiler] satifies <span class="math">(\nabla^{2} + k^{2})G(\mathbf{r}) = \delta^{3}(\mathbf{r})[/spoiler]. This bitch is reminiscent of electrodynamics so I don't know why my prof pulled this one on me. I don't even know where to start as I haven't even seen a regular proof for this. Any suggestions?

[Anonymous No.4437079]

Bump

[Anonymous No.4437099]

Please /sci/ I know you guys can do this.

I'm running out of guidos

[Anonymous No.4437118]

Another bump.

[Anonymous No.4437156]

Please? :(

[Anonymous No.4437160]

Don't know how to do this, bro

[Anonymous No.4437166]

/Sci/ is high school students and undergraduates.

[Anonymous No.4437179]

>>4437166
But this is undergraduate :(

[Anonymous No.4437182]

>>4437166
and by undergraduates we mean jobless basement dwellers with less than 60k points on khan academy

[theta !vacuaB.oEw No.4437236]

<div class="math">G = - \frac{e^{ikr}}{4 \pi r} \Rightarrow \nabla G = - \frac{1}{4 \pi} \left ( \frac{1}{r} \nabla e^{ikr} + e^{ikr} \nabla \frac{1}{r} \right ) \Rightarrow</div>
<div class="math">\nabla^{2} G = \nabla \cdot (\nabla G) = - \frac{1}{4 \pi} \left [ 2 \left ( \nabla \frac{1}{r} \right ) \cdot (\nabla e^{ikr}) + \frac{1}{r} \nabla^{2} (e^{ikr}) \nabla^{2} \left ( \frac{1}{r} \right ) \right ]</div>
we know
<div class="math">\nabla \frac{1}{r} = - \frac{1}{r^{2}} \hat{r};~ \nabla(e^{ikr}) = ike^{ikr} \hat{r}; ~ \nabla^{2} e^{ikr} = ik \nabla \cdot (e^{ikr} \hat{r}) = ik \frac{1}{r^{2}} \frac{d}{dr} (r^{2} e^{ikr})</div>
<div class="math">\Rightarrow \nabla^{2} e^{ikr} = \frac{ik}{r^{2}} (2re^{ikr} + ikr^{2} e^{ikr}) = ike^{ikr} \left ( \frac{2}{r} + ik \right ); ~ \nabla^{2} \left ( \frac{1}{r} \right ) = -4 \pi \delta^{3}(\mathbf(r))</div>
so
<div class="math">\nabla^{2} G = - \frac{1}{4 \pi} \left [ 2 \left ( - \frac{1}{r^{2}} \hat{r} \right ) \cdot (ike^{ikr} \hat{r}) + \frac{1}{r} ike^{ikr} \left ( \frac{2}{r} + ik \right ) - 4 \pi e^{ikr} \delta^{3} (\mathbf{r}) \right ]</div>
<div class="math">\nabla^{2} G = \delta^{3} (\mathbf{r}) - \frac{1}{4 \pi} e^{ikr} \left [ - \frac{2ik}{r^{2}} + \frac{2ik}{r^{2}} - \frac{k^{2}}{r} \right ] = \delta^{3} (\mathbf{r}) + k^{2} \frac{e^{ikr}}{4 \pi r} = \delta^{3} (\mathbf{r}) - k^{2} G</div>
<div class="math">(\nabla^{2} + k^{2})G = \delta^{3} (\mathbf{r})</div>
qed

[Anonymous No.4437302]

>>4437236
Wow, thank you! How did you get from <span class="math">\nabla^{2} G = - \frac{1}{4 \pi} \left [ 2 \left ( - \frac{1}{r^{2}} \hat{r} \right ) \cdot (ike^{ikr} \hat{r}) + \frac{1}{r} ike^{ikr} \left ( \frac{2}{r} + ik \right ) - 4 \pi e^{ikr} \delta^{3} (\mathbf{r}) \right ][/spoiler] to <span class="math">\nabla^{2} G = \delta^{3} (\mathbf{r}) - \frac{1}{4 \pi} e^{ikr} \left [ - \frac{2ik}{r^{2}} + \frac{2ik}{r^{2}} - \frac{k^{2}}{r} \right ] = \delta^{3} (\mathbf{r}) + k^{2} \frac{e^{ikr}}{4 \pi r} = \delta^{3} (\mathbf{r}) - k^{2} G[/spoiler] though?

[theta !vacuaB.oEw No.4437318]

>>4437302
simple simplification -
<div class="math">e^{ikr} \delta^{3} (\mathbf{r}) = \delta^{3} (\mathbf{r})</div>