Planar wave solution to zero potential Schrödinger equation

[Schwarzschild90]

Homework Statement

Homework Equations

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

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}

Moderator's note: post edited to fix the LaTeX. Use double # for inline equations, not $.

 

[DrClaude]

You should get rid of the $i^2$. Otherwise, do you have a question?

 

[BvU]

And (2) $\Rightarrow$ (4) + (5) = 0 gives you ?

 

[Schwarzschild90]

@BvU: A homogenous partial differential equation

@DrClaude: I imagined that I would simplify a little later

 

[BvU]

@BvU: A homogenous partial differential equation

@DrClaude: I imagined that I would simplify a little later

$$ a = b\\c = d $$ gives a pde $a=c$, I agree .

 

[Schwarzschild90]

\begin{align}
\begin{split}
\frac{\hbar^2}{2m} (k^2) e^{i(kx- \omega t)} - \hbar \omega e^{(ikx-\omega t)} = 0
\\
\frac{\hbar}{2m} (k^2) e^{i(kx- \omega t)} = \omega e^{(ikx-\omega t)}
\end{split}
\end{align}

And cancelling an $\hbar$ right away

 

[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 ...​

 

[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

 

[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 ?

 

[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}

 

[DrClaude]

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}

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

 

[BvU]

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

 

[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}

 

[Schwarzschild90]

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

 

[BvU]

All in favor !

 

[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}

 

[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 $E = {p^2\over 2 m}$ !

 

[Schwarzschild90]

It's a ridiculously satisfying result

Should I let the angular velocity depend on k?

 

[BvU]

This is about a free particle !
Check out Broglie wavelength

 

[Schwarzschild90]

Thanks

 

[Schwarzschild90]

Which function do you think the author of exercise b asks me to use?

 

[BvU]

The one in (1).
You were already embarking on dissecting it in

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}

(and you remember that coefficients are complex numbers in QM)

 

[Schwarzschild90]

That yields the following

\begin{align}
\begin{split}
e^{i(kx-\omega t)} = \cos{(i(kx-\omega t))} + i \sin{(i(kx-\omega t))} \\
Closest: \ cosh(k x-t \omega)-sinh(k x-t \omega)
\end{split}
\end{align}

 

[BvU]

Re:

It's a ridiculously satisfying result
Should I let the angular velocity depend on k?

This $\omega$ is a totally different angular velocity.
I have become quite enthousiastic about these Feynman lectures (the link came from Simon Bridge) -- Richard Feynman talks about photons, but later on you understand that this is true for all and everything (except radioactivity and gravity ).

Re:

That yields the following

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

I seem to remember something like $e^{ix} = \cos x + i\sin x$

 

[Schwarzschild90]

The linear combination resulting is then

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

I'm very fond of watching Feynman's lectures as a way to cover more ground and to develop insight in physics

 

[BvU]

The linear combination resulting is then
\begin{align}
\begin{split}
e^{i(kx-\omega t)} = \cos{(kx-\omega t)} + i \sin{(kx-\omega t)}
\end{split}
\end{align}

So, as answer to b), you write $\ \Psi_s = \ \$ ?? and $\ \Psi_c = \ \$ ??

 

[Schwarzschild90]

\begin{align}
\begin{split}
\Psi_c = \cos{(kx-\omega t)} \\
\Psi_s = i \sin{(kx-\omega t)}
\end{split}
\end{align}

I can't see how I get rid of the complex coefficient

 

[BvU]

There is no reason to want to get rid of the complex coefficient (they are just fine, see #22). Wave functions have complex values.

However, you have another problem on your hand: From Euler you have an expression for $\cos x$ and it doesn't just contain $e^{ix}$ but also $e^{-ix}$. In other words: you need to check that $\ \Psi = e^{-i(kx-\omega t)}\$ also satisfies the Schroedinger equation (1.1) and then determine $C_1$ and $C_2$ in $\ \Psi_s = \ \ C_1 e^{i(kx-\omega t)} + C_2 e^{-i(kx-\omega t)}$ .

Idem $\Psi_c$

Note that (7) can't be the answer: they are part of the problem statement for (b) !

 

[Schwarzschild90]

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

 

[BvU]

Are we running into trouble here ?

 

[Schwarzschild90]

I think the solution might just be antisymmetric: Equal in magnitude but opposite in sign.

 

[BvU]

That's for $\Psi_s$. $\Psi_c$ is symmetric.

 

[Schwarzschild90]

Omega being angular frequency then just means that a negative sign will have the wave traveling in the opposite direction

 

[BvU]

No ! $\omega$ is a frequency.
And I thought about $+\omega t$ too, but that gives you standing waves. $\Psi_s$ and $\Psi_c$ both 'travel to the right'.

 

[Schwarzschild90]

Hbar is not negative, neither can mass or the wavenumber k^2 be negative.