Planar wave solution to zero potential Schrödinger equation

[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.

 

[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.

Notice also that the Heisenberg eigenstates you're talking about evolve the other way: in the above \exp(-iEt) should be replaced by \exp(iEt).

 

[vanhees71]

Well, it depends what are you looking on. What I've written down is the wave function, representing the state of a system. It's in fact independent of the picture used to calculate it. It's easy to see in the bra-ket formalism:

Heisenberg picture
-----------------------

The state vector is time-independent:
$$|\psi(t) \rangle_{\text{H}}=|\psi(t_0) \rangle_{\text{H}},$$
while the eigenstates (e.g., the position eigenstates) are time dependent) according to
$$|x,t \rangle_{\text{H}}=\exp[\mathrm{i} \hat{H} (t-t_0)] |x,t_0 \rangle_{\text{H}}.$$
Thus the wave function (which is picture independent!) is given by
$$\psi(t,x)=_{\text{H}} \langle x,t|\psi,t \rangle_{\text{H}}=_{\text{H}} \langle x,t_0| \exp[-\mathrm{i} \hat{H} (t-t_0)|\psi,t_0 \rangle_{\text{H}}.$$
If the state vector is an energy eigenstate, you have
$$u_E(t,x)=\exp[-\mathrm{i}(t-t_0) t] u_E(t_0,x).$$

Schrödinger picture
------------------------

The state ket carries the full time dependence
$$|\psi,t \rangle_{\text{S}}=\exp[-\mathrm{i}(t-t_0) \hat{H}]|\psi,t_0 \rangle_{\text{S}},$$
and the eigenkets of the observable operators are time-independent
$$|x,t \rangle_{\text{S}}=|x,t_0 \rangle_{\text{S}}.$$
Again you find the same time dependence of the wave function (which is picture independent)
$$\psi(t,x)=_{\text{S}} \langle x,t|\psi,t \rangle_{\text{S}}=_{\text{S}} \langle x,t_0|\exp[-\mathrm{i}(t-t') \hat{H}] \psi,t_0 \rangle_{\text{S}}.$$
If the particle is in an energy eigenstate at $t_0$, you find again the same result as when using the Heisenberg picture, as it must be
$$u_E(t,x)=\exp[-\mathrm{i} E (t-t_0)] u_{E}(t_0,x).$$
Of course, this tells us that the state doesn't change in this case, because the time evolution boils down to a simple phase factor, and the physical meaning is just the $|\psi(t,x)|^2$, which is the probability distribution for finding the particle at $x$ when looking for it at time $t$. For the energy eigenstate as the state vector, the phase factor cancels, and thus the energy eigenstates are precisely the stationary states of the particle.

 

[Oxvillian]

Sure - if you define the wavefunction in this nice picture-independent way, then it goes as e^{-i\omega t}. That is, it evolves "forwards" in time. And indeed in this post you said the correct thing about the Heisenberg picture position eigenstates, which evolve "backwards" in time:

the eigenstates (e.g., the position eigenstates) are time dependent) according to
$$|x,t \rangle_{\text{H}}=\exp[\mathrm{i} \hat{H} (t-t_0)] |x,t_0 \rangle_{\text{H}}.$$

But now you should just apply the same logic to the Heisenberg picture energy eigenstates- they likewise evolve "backwards" in time. Instead, you confused these states with what you've called the wavefunction of the system:

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.

That's the only part where I'd disagree with you.

[with apologies to the OP for this thread hijack!]

 

[vanhees71]

In this strange problem they talked about wave functions. Maybe it was a bit misleading to bring in different pictures. The solutions of the Schrödinger equation give wave functions, and energy eigenwavefunctions always have the time dependence $\exp(-\mathrm{i} E t)$.

 

[Oxvillian]

Indeed, the whole active/passive transformation business can always be relied upon to generate confusion.

 

[BvU]

So, Schwarzschild, you have seen what can happen on PF, even with an innocent looking exercise. Happens quite bit here. My estimate is you are being introduced to QM and a lot of the posts may be going over your head. How are you ? Would you care to answer #47 ?

 

[Schwarzschild90]

#47: I don't know which book the exercise is from. I think the authors invented the exercise themselves.

 

[BvU]

Hehe, authors but not a book isn't much to go by. If not a book, what is it and where did it come from ? It's not like we are going to jump on them or anything ! Just curiosity and perhaps a gentle nudge in the form of a polite question .
However, if you have good reason to be somewhat evasive that will be respected.

Personally,I really liked the thread because it made me re-think things I thought I knew at least a little about -- and it turned out I didn't get much further than severe doubt about the answer...

"In the end, we're all beginners.."​

 

[Schwarzschild90]

It's nothing of that sort, hehe. I'm not instructed to hide course material

We use the book "introduction to quantum mechanics 2nd edition" for the course.

 

[BvU]

Most widely known by that title is Griffiths, but I don't think it has a "1. Historical introduction"