- #1
beyondlight
- 65
- 0
Homework Statement
Show that this solution:
[tex]\psi(x,y,z)=Ae^{i\mathbf{k}\cdot\mathbf{r}}[/tex]
is a solution for this equation:
[tex]\frac{\hbar^{2}}{2m}\Delta \psi(x,y,z)=E\,\psi(x,y,z),[/tex]
How is K related to E?
How does the corresponding timedependent function look like?
The Attempt at a Solution
My second derivative is:
[tex] \psi''(x,y,z)=i^{2}k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r} = -k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r}=-k_x^{2} \cdot \psi[/tex]
[tex]\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) \cdot \psi = E \cdot \psi[/tex]
[tex]\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) = E[/tex]
I suppose it is enough to say that
[tex]\frac{\hbar^{2}}{2m} (k_x^{2} + k_y^{2} + k_z^{2})[/tex] is energy? So it automatically satisfies the S.E.? What is the unit of [tex]\psi[/tex]?
But the relation between k and E should then be:
[tex]\frac{\hbar^{2}}{2m} \cdot (k_x^{2} + k_y^{2} + k_z^{2}) = \frac{\hbar^{2}}{2m} \cdot K = E[/tex]
[tex]\frac{E}{K}= \frac{\hbar^{2}}{2m}[/tex]
But the last task is a bit more tricky...I will just separate the time-depentent S.E.
[tex]\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}[/tex]
[tex]\frac{\hbar^{2}}{2m}\Delta = i\hbar \frac{d\psi}{dt} \cdot \frac{1}{\psi}[/tex]
Can someone help me from here?