# Planar wave solution to zero potential Schrödinger equation

1. Feb 8, 2016

### Schwarzschild90

1. The problem statement, all variables and given/known data

2. Relevant equations
\begin{align}
\begin{split}
\psi(x, t) = e^{(ikx- i \omega t)}
\\
V(x) = 0
\end{split}
\end{align}

3. The attempt at a solution
For a free particle, the Schrödinger equation can be put in the form of $\psi(x, t) = e^{(ikx- i \omega t)}$. With constant potentials for all x, $\forall x : V = 0$, or put equally succinctly: $V(x) = 0$ (the function of which is time independent for all x in the domain)

\begin{align}
\begin{split}
i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi
\end{split}
\end{align}

Make the substitution $\psi(x, t) = e^{i(kx-\omega t)}$

\begin{align}
\begin{split}
i \hbar \frac{\partial}{\partial t} e^{i(kx- \omega t)} = \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} e^{i(kx- \omega t)}
\end{split}
\end{align}

Now, take the partial derivative with respect to time of the left hand side of the equation
\begin{align}
\begin{split}
i \hbar \frac{\partial}{\partial t} e^{i(kx- \omega t)} = -i^2 \hbar \omega e^{(ikx-\omega t)}
\end{split}
\end{align}

And the second order partial derivative with respect to x, of the right hand side of the equation

\begin{align}
\begin{split}
\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} e^{i(kx- \omega t)} = \frac{\hbar^2}{2m} (i^2k^2) e^{i(kx- \omega t)}
\end{split}
\end{align}

7. Feb 8, 2016

### BvU

Bingo !

-- kudos for using the \split goody. Didn't know it. But it seems to confuse the \align thingy, perhaps because of the label rhs appearing twice ...​

8. Feb 8, 2016

### Schwarzschild90

I see no reason not to cancel the following term

\begin{align}
\begin{split}
e^{i(kx- \omega t)}
\end{split}
\end{align}

but I've been wrong before :P

9. Feb 8, 2016

### BvU

(6) has to be true for all t and all x, independently.
What is happening here is separation of the variables x and t.

Oh, and $e^{iy} \ne 0$ always, so why not ?

10. Feb 8, 2016

### Schwarzschild90

So for a free particle (in a box) we have

\begin{align}
\begin{split}
k^2 = \frac{p^2}{\hbar^2}
\end{split}
\end{align}

But the wavenumber

\begin{align}
\begin{split}
k = \frac{\pi}{\lambda} \\
\left( \frac{\pi}{\lambda} \right)^2 = \frac{p^2}{\hbar^2} \to \left( \frac{\pi}{\lambda} \right) = \frac{p}{\hbar}
\end{split}
\end{align}

11. Feb 8, 2016

### Staff: Mentor

Work with the equations ou already have. Don't introduce p here (which is an operator, since you are working in position space).

12. Feb 8, 2016

### BvU

And wasn't $k = {2\pi\over \lambda }$ ?
(But I agree with Claudius)

13. Feb 8, 2016

### Schwarzschild90

So I can use the following identity to separate variables

\begin{align}
\begin{split}
e^{iy} = \cos(y)+ i \sin(y)
\end{split}
\end{align}

And use the fact that

\begin{align}
\begin{split}
e^{i(kx- \omega t)} = e^{ikx} \times e^{- i \omega t}
\end{split}
\end{align}

14. Feb 8, 2016

### Schwarzschild90

Since e^(iy) \neq 0 always, I could go ahead and cancel the term on both sides of the equation?

15. Feb 8, 2016

### BvU

All in favor !

16. Feb 8, 2016

### Schwarzschild90

Therefore k and omega need to be related in the following way

\begin{align}
\begin{split}
\frac{\hbar}{2m} k^2 = \omega(k)
\end{split}
\end{align}

17. Feb 8, 2016

### BvU

Isn't it nice ? And if we now take a sneak preview: with $E = \hbar\omega\ \ \& \ \ \vec p = \hbar \vec k\ \$ it looks like $E = {p^2\over 2 m}$ !

18. Feb 8, 2016

### Schwarzschild90

It's a ridiculously satisfying result

Should I let the angular velocity depend on k?

19. Feb 8, 2016

### BvU

20. Feb 8, 2016

### Schwarzschild90

Thanks

#### Attached Files:

• ###### Schrödinger b.PNG
File size:
4.5 KB
Views:
149
Last edited: Feb 8, 2016