1. ## Schrödinger equation

Show that a solution for S.E. for a free particle in three dimensions can be written as:

ħ^2/2m(d^2/dx^2 + d^2/dy^2 + d^2/dz^2)Ψ(x,y,z) = EΨ(x,y,z)

Ψ(x,y,z)= A*exp(iK*r)

How is K related to E?

How does the corresponding timedependent function look like?

(Hint: Suppose Ψ(x,y,z)=Ψ(x)*Ψ(y)*Ψ(z) and separate the function )

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I just derivated Ψ(x,y,z) two times and inserted it into the equation, but i dont think its enough....

2. ## Re: Schrödinger equation

Just to clarify: you're asked to show that the function

$\displaystyle \psi(x,y,z)=Ae^{i\mathbf{k}\cdot\mathbf{r}}$ satisfies the 3D Schrödinger equation

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

and you're asked to find the relationship between $\displaystyle E$ and $\displaystyle \mathbf{k},$ right?

3. ## Re: Schrödinger equation

You got it chief.

4. ## Re: Schrödinger equation

Originally Posted by techmath
Show that a solution for S.E. for a free particle in three dimensions can be written as:

ħ^2/2m(d^2/dx^2 + d^2/dy^2 + d^2/dz^2)Ψ(x,y,z) = EΨ(x,y,z)

Ψ(x,y,z)= A*exp(iK*r)

How is K related to E?

How does the corresponding timedependent function look like?

(Hint: Suppose Ψ(x,y,z)=Ψ(x)*Ψ(y)*Ψ(z) and separate the function )

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I just derivated Ψ(x,y,z) two times and inserted it into the equation, but i dont think its enough....
I doubt there is even one textbook on Quantum Mechanics that does NOT answer this question. Have you refered to any?

5. ## Re: Schrödinger equation

This is from a booklet with exercies for introductory physics, for my university course. There is no answers on these types of tasks because it includes a rather long mathematical verification. The list of answers only includes numerical values.

6. ## Re: Schrödinger equation

Originally Posted by techmath
Show that a solution for S.E. for a free particle in three dimensions can be written as:

ħ^2/2m(d^2/dx^2 + d^2/dy^2 + d^2/dz^2)Ψ(x,y,z) = EΨ(x,y,z)

Ψ(x,y,z)= A*exp(iK*r)

How is K related to E?

How does the corresponding timedependent function look like?

(Hint: Suppose Ψ(x,y,z)=Ψ(x)*Ψ(y)*Ψ(z) and separate the function )

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I just derivated Ψ(x,y,z) two times and inserted it into the equation, but i dont think its enough....
First...Your Schrodinger equation needs a minus sign.
-ħ^2/2m(d^2/dx^2 + d^2/dy^2 + d^2/dz^2)Ψ(x,y,z) = EΨ(x,y,z)

$\displaystyle \Psi = A \cdot e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$

Now take your derivative with respect to x:
$\displaystyle \Psi ' = A \cdot k_x \cdot \cdot e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$

Now do it again:
$\displaystyle \Psi '' = A \cdot k_x^2 e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$

Now do this for y and z.

Give that a try and let us know how you are doing with it. And yes it looks horrible but most of it cancels out in the end.

If you need to actually solve this, then separate the equation as your hint suggests. You will get equations of the form
$\displaystyle -\frac{\bar /2m}X''(x) = E X(x)$

Solve this for X(x) and do the other coordinate (Y(y) and Z(z)) functions.

-Dan

7. ## Re: Schrödinger equation

Originally Posted by topsquark
First...Your Schrodinger equation needs a minus sign.
-ħ^2/2m(d^2/dx^2 + d^2/dy^2 + d^2/dz^2)Ψ(x,y,z) = EΨ(x,y,z)

$\displaystyle \Psi = A \cdot e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$
I think your i's got poked out.

Now take your derivative with respect to x:
$\displaystyle \Psi ' = A \cdot k_x \cdot \cdot e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$

Now do it again:
$\displaystyle \Psi '' = A \cdot k_x^2 e^{k_x x} \cdot e^{k_y y} \cdot e^{k_z z}$

Now do this for y and z.

Give that a try and let us know how you are doing with it. And yes it looks horrible but most of it cancels out in the end.

If you need to actually solve this, then separate the equation as your hint suggests. You will get equations of the form
$\displaystyle -\frac{\bar /2m}X''(x) = E X(x)$

Solve this for X(x) and do the other coordinate (Y(y) and Z(z)) functions.

-Dan

8. ## Re: Schrödinger equation

My second derivative is:

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

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

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

I suppose it is enough to say that

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

But the relation between k and E should then be:

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

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

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

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

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

Can someone help me from here?