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.

 

[BvU]

However, if $k$ satisfies the SE, then so does $-k$

I am struggling along with you (temporarily not very bright or something ). Didactically not good for you. Let's ask an expert, e.g. @Orodruin

 

[Schwarzschild90]

http://physics.stackexchange.com/qu...ve-is-negative-is-the-wavelength-negative-too

So it's because the wave is traveling in the opposite x-direction

 

[BvU]

That's the vector aspect. From the $k^2$ condition you get $\pm k$ in one dimension.
I'm still wondering what they mean in (b) with 'wave functions of the previous form' where I only see one: $$\Psi = \exp[i(kx -\omega t)] $$
(or should that be $|i(kx-\omega t)|$ ?? )

 

[Schwarzschild90]

Knowing them, then either way of doing it is correct.

How do I continue from here?

 

[BvU]

"Knowing them, then either way of doing it is correct" ?

Have we seen that $\Psi = e^ { -i(kx -\omega t) }$ satisfies the schroedinger equation ?
If so, then it's back to Euler

(have to run for work now... )

 

[Schwarzschild90]

Yes, since we've found the dispersion relation.

\begin{align}
\begin{split}
\Psi = C_1 e^{i(kx-\omega t)} + C_2 e^{-i(kx-\omega t)} \to \\
e^{i(kx-\omega t)} = C_1 e^{i(kx-\omega t)} + C_2 e^{-i(kx-\omega t)}
\end{split}
\end{align}

 

[Orodruin]

The functions quoted are both traveling to the right (if $k$ is positive). They are linear combinations of this form

Yes, since we've found the dispersion relation.
\begin{align}
\begin{split}
\Psi = C_1 e^{i(kx-\omega t)} + C_2 e^{-i(kx-\omega t)} \to \\
e^{i(kx-\omega t)} = C_1 e^{i(kx-\omega t)} + C_2 e^{-i(kx-\omega t)}
\end{split}
\end{align}

All you need to do is to identify the coefficients. Of course, negative $k$ will also give a solution to the SE, but that is besides the point - you have been given two particular wave functions and need to express them in terms of the functions you have.

 

[BvU]

Hi again,

We're still in trouble; I asked on the advisor lounge forum and got very good counsel from @vanhees71 . I'll ask him if I can quote him (or better: that he quotes himself in this thread).
The judgment so far is that part (a) is just fine, but in part (b) the author goes astray (for the reason you found: the functions are not solutions to the Schroedinger equation). Hence the question: what book does this exercise come from ?

 

[vanhees71]

You are overlooking that these wave functions do not make the slightest sense, because they do not solve the time-dependent Schrödinger equation. Is there a fully formulated question somewhere in this thread or a reference to the textbook where this question appears?

What makes sense is to ask to investigate the degeneracy of the Hamiltonian's eigenstates for $E \neq 0$. Indeed there are two solutions for each energy eigenvalue $E>0$. I set $\hbar=1$. For the beginners you should keep $\hbar$
$$\hat{H} u_E(x)=E u_E(x) \; \Rightarrow \; -\frac{1}{2m} u_E''(x)=E u_E(x) \;\Rightarrow u_E(x)=\exp(\pm \mathrm{i} k x).$$
The modes of the time-dependent Schrödinger equation have always the time dependence $\exp(-\mathrm{i} \omega_k t)$, $\omega_k=k^2/(2m)$.

To distinguish the twice degenerate energy eigenvalues you need an additional observable. Here you can use parity, i.e., the operator
$$\hat{P} \psi(x)=\psi(-x)=\pm \psi(x).$$
Since $\hat{P}$ commutes with $\hat{H}=\hat{p}^2/(2m)$, you can distinguish the degenerate eigenvectors by additionally determining the parity. This you get by the even and odd linear combinations
$$u_{E,P=1}(x)=\cos(k x), \quad u_{E,P-1}=\sin(k x), \quad k = \sqrt{2mE}.$$
The orresponding eigenmodes of the time-dependent Schrödinger equation are
$$u_{E,P=\pm 1}(t,x)=\exp(-\mathrm{i} E t) u_{E,P=\pm 1}(x), $$
i.e., both have the same time-dependent exponential factor, not $\sin(\omega t-k x)$.

 

[Oxvillian]

Sorry for barging into this thread!

I agree that the author of the problem has made a booboo in part (b)!

Wrt the parity thing, it's probably worth noting that the momentum eigenstates aren't parity eigenstates - that is, they're not states of definite parity. Indeed, if you do a parity transformation on a right-going wave, you get a left-going wave, and vice versa.

To distinguish between the momentum eigenstates, just look at the momentum!

I think what the author is getting at is that you can build sin's and cos's (states of definite parity) by adding together momentum eigenstates.

 

[vanhees71]

Yes, and the correct time-dependent states are then $\cos(k x) \exp[-\mathrm{i} E(k) t]$ and $\sin(k x) \exp[-\mathrm{i} E(k) t]$. These are the energy eigenstates in the Heisenberg picture of time evolution, by the way.

 

[BvU]

It seems we've lost OP a little. How are things? What did teacher have to say and from which book was this exercise ?

 

[Oxvillian]

Yes, and the correct time-dependent states are then $\cos(k x) \exp[-\mathrm{i} E(k) t]$ and $\sin(k x) \exp[-\mathrm{i} E(k) t]$. These are the energy eigenstates in the Heisenberg picture of time evolution, by the way.

In the Heisenberg picture, the time dependence is shifted from the states to the operators, so there's no \exp(-iEt/\hbar) factor attached to the energy eigenstates. Otherwise things are much the same. But I think we're getting sufficiently off-topic to thoroughly confuse the OP!

Perhaps it's worth writing down what part (b) of the problem probably should have said:

(b) Write \Psi_s = e^{-iwt}\sin(kx) and \Psi_c = e^{-iwt}\cos(kx) as linear combinations of wavefunctions of the previous form.

 

[vanhees71]

As the operators that represent operators carry the full time dependence in the Heisenberg picture their eigenvectors also carry that time dependendence.

In the Schrödinger picture these operators are time-independent and thus also their eigenvectors, but that's really not to the topic of this thread.

 

[Oxvillian]

Agreed - but what I'm trying to say is slightly different - if in the Heisenberg picture the wavefunction of the system starts out as sin(kx), then it will always remain sin(kx). Its inner products with eigenvectors of some given observable will however be time dependent, because as you rightly say, those eigenvectors inherit time dependence from the operator representing the observable.