Is This Wavefunction a Valid Solution for the Schrödinger Equation?

[beyondlight]

Homework Statement

Show that this solution:

$$\psi(x,y,z)=Ae^{i\mathbf{k}\cdot\mathbf{r}}$$

is a solution for this equation:

$$\frac{\hbar^{2}}{2m}\Delta \psi(x,y,z)=E\,\psi(x,y,z),$$

How is K related to E?

How does the corresponding timedependent function look like?

The Attempt at a Solution

My second derivative is:

$$\psi''(x,y,z)=i^{2}k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r} = -k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r}=-k_x^{2} \cdot \psi$$

$$\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) \cdot \psi = E \cdot \psi$$

$$\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) = E$$


I suppose it is enough to say that

$$\frac{\hbar^{2}}{2m} (k_x^{2} + k_y^{2} + k_z^{2})$$ is energy? So it automatically satisfies the S.E.? What is the unit of $$\psi$$?

But the relation between k and E should then be:

$$\frac{\hbar^{2}}{2m} \cdot (k_x^{2} + k_y^{2} + k_z^{2}) = \frac{\hbar^{2}}{2m} \cdot K = E$$

$$\frac{E}{K}= \frac{\hbar^{2}}{2m}$$


But the last task is a bit more tricky...I will just separate the time-depentent S.E.

$$\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}$$

$$\frac{\hbar^{2}}{2m}\Delta = i\hbar \frac{d\psi}{dt} \cdot \frac{1}{\psi}$$

Can someone help me from here?

 

[Redbelly98]

Welcome to Physics Forums.

Homework Statement

Show that this solution:

$$\psi(x,y,z)=Ae^{i\mathbf{k}\cdot\mathbf{r}}$$

is a solution for this equation:

$$\frac{\hbar^{2}}{2m}\Delta \psi(x,y,z)=E\,\psi(x,y,z),$$

How is K related to E?

How does the corresponding timedependent function look like?

The Attempt at a Solution

My second derivative is:

$$\psi''(x,y,z)=i^{2}k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r} = -k_x^{2}A \cdot e^{ik_x*r} \cdot e^{ik_y*r} \cdot e^{ik_z*r}=-k_x^{2} \cdot \psi$$

Small correction: it really should be k_xx, k_yy, etc. here. And it looks like you are using ψ'' to mean ∂²ψ/∂x². But you have the right idea.

$$\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) \cdot \psi = E \cdot \psi$$

$$\frac{\hbar^{2}}{2m}(k_x^{2} + k_y^{2} + k_z^{2}) = E$$


I suppose it is enough to say that

$$\frac{\hbar^{2}}{2m} (k_x^{2} + k_y^{2} + k_z^{2})$$ is energy? So it automatically satisfies the S.E.? What is the unit of $$\psi$$?

But the relation between k and E should then be:

$$\frac{\hbar^{2}}{2m} \cdot (k_x^{2} + k_y^{2} + k_z^{2}) = \frac{\hbar^{2}}{2m} \cdot K = E$$

$$\frac{E}{K}= \frac{\hbar^{2}}{2m}$$

Close, but not quite. You are saying that

$k_x^2 + k_y^2 + k_z^2 = k^1$ ?​

I think you need to modify the k¹ term, slightly, to have a true statement here.

But the last task is a bit more tricky...I will just separate the time-depentent S.E.

$$\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}$$

$$\frac{\hbar^{2}}{2m}\Delta = i\hbar \frac{d\psi}{dt} \cdot \frac{1}{\psi}$$

Can someone help me from here?

Let's back up 1 step to

$$\frac{\hbar^{2}}{2m}\Delta \psi = i\hbar \frac{d\psi}{dt}$$

Since ψ is a function of (x,y,z) only, we really need another symbol for the time-dependent part. It's common to assume the full wavefunction is the product of a spatial function (that's ψ(x,y,z) ) and a time-dependent function (call it f(t)). So you want to put ψ(x,y,z)·f(t) into the time-dependent Schrodinger Equation, and work out what f(t) is.

Hope that helps. By the way, \partial will give you the partial derivative symbol in LaTeX.