Two Ways of Calculating the Solution to the Infinite Square Well?

[JohnH]

After watching this video: which explains why the wavefunction in an infinite square well is flattened, I tried running the calculation in both, what seems, the more more traditional way of using sin and by the method of, what seems to be, adding the wavefunction and its complex conjugate. Though both results integrated to one and had an arc spanning from one side to the other, the sin version has more curvature at the bottom and a peak of 2.5 while the other had not as much curvature and a peak of two. What am I missing here?

I've attached a picture of my mathematica notebook.

 

[BvU]

Can't read what you are plotting (too many brackets for my brain), but one is a sine and the other is a sine squared

$\$

 

[JohnH]

If a=Pi/4 is the length of the infinite square well and k=4, then the first graph (and integral) is Psi+Psi* and plotted over a range of a. The second graph and integral is (2/a)*Sin^2(kx) also plotted over a. Sorry for all the artefacts. I hadn't cleaned that up since manipulating it.

To be clear, what I'm wondering is if Psi+Psi* is supposed to equate to the more traditional solution to the infinite square well which is Abs[Psi sub x]^2=(2/a)*Sin^2((n*Pi*x)/a) and if so why I'm not getting that. And if not, why does the video seem to be suggesting that (starting at about 1:59).

 

[vanhees71]

I don't understand your calculation. Obviously you want to solve the energy-eigenvalue problem for
$$V(x)=\begin{cases} 0 & \text{for} \quad x \in [0,L] \\ \infty & \text{otherwise}. \end{cases}.$$
That means you want to solve
$$-\frac{\hbar^2}{2m} \psi_E''(x)=E \psi_E(x), \quad \psi_E(0)=\psi_E(L)=0.$$
For any real $E$ the solution of the differential equation obviously is
$$\psi_E(x)=A \sin(k x) + B \cos(k x), \quad k=\sqrt{2 m E}/\hbar.$$
It's clear that $E \geq 0$. To fulfill the boundary conditions, note that $\psi_E(0)=B=0$. And then you must have $k$ such that $\sin(k L)=0$, i.e.
$$k=n \pi/L, \quad n \in \mathbb{N}=\{1,2,3,\ldots \}.$$
The eigenfunctions are unique up to a factor, which is choosen such that
$$\int_0^L \mathrm{d} x |\psi_E(x)|^2=1 \; \Rightarrow \; A=\sqrt{2/L}.$$
Thus the unique energy eigenfunctions are
$$\psi_E(x)=\sqrt{\frac{2}{L}} \sin(n \pi x/L).$$

 

[JohnH]

Okay, yes, that's what I was trying to say. Does all that equate to a second way of getting this solution which involves adding Psi and Psi* as seems to be implied in the linked video at time stamp 1:59~ and if not why does the video seem to be suggesting that?

 

[vanhees71]

As I said, I don't understand what it is about this $\psi$ and $\psi^*$ thing. Up to an indetermined phase factor (which one usually sets tacitly to 1, as done above) the normalized energy eigenfunctions are unique and real. The latter is, because the potential is symmetric under reflection at the point $L/2$.

 

[JohnH]

So is the video wrong or am I misinterpreting it?

 

[vanhees71]

The video isn't even wrong ;-). SCNR.

 

[JohnH]

So how would you put what the video is saying into mathematical terms?

 

[vanhees71]

That's the introductory part of a quantum mechanics textbook. For the traditional wave-mechanics approach, I'd recommend Messiah, vol. 1.

 

[JohnH]

Alright. I'll look for it. It seems like I'm wrong to add the complex conjugate to Psi which I always knew was a bit figurative at best but anyway thanks for the replies.

 

[Haborix]

The crux of the problem in the video is that they get to the correct answer by wrong means. "A particle of known momentum moving to the right," the mathematical equivalent being a complex exponential , is simply not an allowed state in the infinite square well. Such a state doesn't satisfy the boundary condition that the wave function be zero at the walls. Now it so happens that real trigonometric functions can be made to satisfy the boundary conditions. They also can be written as sums and differences of complex exponentials, which is why the video happens to get the right answer.

Finally, what you call in post #3 the traditional solution is the traditional solution for the probability density, not the wavefunction (probability density is the norm squared of the wavefunction). Your plots don't compare like with like.

 

[Haborix]

I'll offer one piece of unsolicited advice. If you find some source which starts making claims about physical attributes of single particles in quantum mechanics, then warning bells should start going off in your head.

 

[vanhees71]

The infinite square well is NOT a simple example. It only appears to be one, because it's easy to calculate the energy eigenstates and eigenvalues, but if you think a bit further, you'll see that there is no momentum observable for this system, i.e., there is no self-adjoint operator that generates translations.

 

[JohnH]

I was wondering if there was some problem because upon further research I found
ψ = sin kx = [exp(ikx) – exp(-ikx)]/2i
and I was wondering how "nature" arrived at dividing it by 2i. But perhaps the analogy in the video is just incorrect, as you're saying Haborix.

 

[Haborix]

As vanhees said, this is a deceptively simple example. If you ignored the infinite potential walls for a moment, it would look like a flat potential where the normal basis states are the complex exponentials. But the walls enforce conditions on the wavefunction at the boundaries and cannot be ignored. When you start learning quantum mechanics this example basically let's you practice using the machinery of quantum mechanics. And it is good to come back to it once you've got experience with quantum mechanics to understand the momentum self-adjointness problem vanhees describes above.

If you know integration by parts and the definition for self-adjoint operators, you might try proving the momentum operator is self-adjoint. You'll eventually get to a point where you stop and go "huh?"

 

[vanhees71]

The definition of the sine function is just as it is because it is useful in this way in trigonometry. That's why
$$\sin z=\frac{1}{2} [\exp(\mathrm{i} z)-\exp(-\mathrm{i} z)], \quad z \in \mathbb{C}.$$
It's a definition of the sine function in terms of the (complex) exponential function.

Now I'm still puzzled by, what the question concerning the quantum mechanics problem(s) really is.

Let's first take the case of a free particle moving along a line. Then you can represent the Hilbert space in terms of the square Lebesgue-integrable position wave functions $\psi(x)$, $x \in \mathbb{R}$, and the position and momentum operators are both self-adjoint being densely defined on the said Hilbert space $\mathrm{L}^2(\mathbb{R})$:
$$\hat{x} \psi(x)=x \psi(x), \quad \hat{p} \psi(x) = -\mathrm{i} \hbar \partial_x \psi(x).$$
Now you can ask for general eigenfunctions of various operators. For $\hat{x}$ and $\hat{p}$ these are
$$u_{x'}(x)=\delta(x-x'), \quad u_{p}(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar),$$
where the norm is chosen such that these generalized functions are "normalized to a $\delta$ distribution". For the position eigenfunction it's obvious. For the momentum eigenfunction you see it as follows
$$\langle p|p' \rangle=\int_{\mathbb{R}} \mathrm{d} x u_{p}^*(x) u_{p'}(x) = \frac{1}{2 \pi \hbar} \int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i}(p'-p) x/\hbar=\frac{1}{2 \pi \hbar} 2 \pi \delta[(p-p')/\hbar] = \delta(p-p').$$
Since the Hamiltonian is
$$\hat{H}=\frac{1}{2m} \hat{p}^2,$$
any momentum eigenfunction is also an energy eigenfunction,
$$\hat{H} u_p(x)=\frac{p^2}{2m} u_p(x).$$
Except the ground state with $E_p=p^/(2m)=0$ all the energy eigenstates are degenerate, because for any $p \neq 0$ also for $-p$ you get the same energy eigenvalue $E_p$. That's, why any linear combination of the two momentum eigenfunctions with eigenvalues $\pm p$ are energy eigenstates to the same eigenvalue $E_p$.

You can lift this degeneracy by diagonalizing other operators simultaneously. In this case the degeneracy is obviously, because the particle moving to the right or the left with the same magnitude of momentum have the same energy. So to make the energy eigenvectors unique, you can ask for the eigenfunctions, which are at simultaneously eigenfunctions of the parity operator, which represents space reflections,
$$\hat{P}\psi(x)=\psi(-x),$$
and the eigenvalues are $P \in \{1,-1\}$ (even and odd functions under space reflections). The two energy eigenstates for a given $E_p=p^2/(2m)$ obviously are
$$u_E^{(+)}=\frac{1}{\sqrt{2}}[u_p+u_{-p}], \quad u_E^{-}=\frac{1}{\sqrt{2}} [u_p-u_{-p}].$$
For the infinite-square well, there is no momentum operator, because if there were one it should also be given by the same operator $-\mathrm{i} \hbar \partial_x$, because by definition $\hat{p}$ should generate space translations. For the infinite square well, however, the corresponding eigenfunctions belong not to the Hilbert space, because they do not fulfill the necessary boundary conditions $\psi(0)=\psi(L)=0$.

The result is that there are energy eigenstates, and they are nondegenerate, i.e., to each energy eigenvalue there's only one linearly independent eigenfunction, which thus are unique up to a factor (the factor doesn't play a role in quantum mechanics anyway, because all you need is to calculate the probability distributions, which are given by normalized wave functions modulus squared).

 

[Haborix]

The OP is trying to come to grips with an explanation given in the video. Your answer is nice, but written, I think, for a more advanced audience. This thread was marked "I", not "A", so I don't think talking about Lebesgue-integrable functions and densely defined operators is illuminating.

 

[vanhees71]

Then the question can not be answered.

 

[JohnH]

I'll admit, it's difficult to follow but I can follow to varying degrees and do appreciate the thorough answer Vanhees gave. I will do some further research on it to get a better understanding if necessary. But I gather the analogy in the video was misguided. But if that is not the explanation, is there a relatively simple explanation for how the wavefunction as a function of x might make this leap from being e^i(kx) to sin(kx). Is there a "naturalistic" interpretation of this? For example, if
ψ = sin kx = [exp(ikx) – exp(-ikx)]
then I could say, ah, okay, so the imaginary parts of exp(ikx) and – exp(-ikx) cancel as in the video and the wavefunction is flattened. But
ψ = sin kx = [exp(ikx) – exp(-ikx)]/2i
and furthermore it seems that thinking of this in terms of momentums canceling is not useful. Do we have any insight into what's "really" going on here? Is it not considered odd that the wavefunction seemingly abandons its e^i(kx) form to take this sin kx form? Are "boundary conditions..." and "solutions to wave equations..." our only insight? Or is it possible to say that this cancelation of momentum in [exp(ikx) – exp(-ikx)]/2i is the reason there is no momentum in the first place and that the factor of 2i is countered by the normalization constant (or some other way of explaining /2i away) such that the naturalistic explanation of the video is valid?

 

[vanhees71]

The simple explanation is just to solve the energy eigenvalue problem as given already in #4. You don't need all this misleading pseudo-didactics of the movie! It's doing more harm than good. Just use sound and solid math and then build physical intuition from the results (e.g., by drawing the wave functions in a plot!).

 

[JohnH]

This answer seems to fit under what is often colloquially referred to as the "shut up and calculate" approach. I honestly didn't realize it was so main stream. And while I hope you understand that I don't find this approach fulfilling, I simultaneously do understand and appreciate how judicious it is.

Thank you for the answers.

 

[PeroK]

This answer seems to fit under what is often colloquially referred to as the "shut up and calculate" approach.

It's not "shut up and calculate", it's "learn to calculate". Any one who understands QM (as far as it can be understood) knows the supporting mathematics and how to calculate using that mathematics. Being able to calculate is an essential element of that understanding. In fact, that is where much of the understanding comes from.

It's only when you can support your ideas with hard calculations that you have anything substantial and unambiguous.

Moreover, learning QM as an academic subject, including the requisite mathematics, is a thousand times harder than watching a YouTube video! Quite literally, it might take several hundred hours to work through a solid introduction to QM from an undergraduate textbook. That's a different level of intellectual commitment from watching a five-minute video.

 

[weirdoguy]

This answer seems to fit under what is often colloquially referred to as the "shut up and calculate" approach.

I would say it's not "shut up and calculate". It's, well, "physics". That's the way physics is. Its only and ultimate language is mathematics. Trying to avoid it will lead you astray and you will waste a lot of time without learning much. Especially when it comes to quantum physics. When we start to learn those topics we have zero intuitions about them. The only way to built that intuitions properly is to learn all the maths behind it.

 

[JohnH]

"Shut up and calculate" is how Feynman described the Copenhagen interpretation. I'm a bit frustrated with the reluctance on this forum to deviate from that interpretation, but I'm honestly grateful for the facts and information shared because I agree that the ability to get the correct predictions is primary. I'm not trying to dispute any of the numerous mistakes I've made on this forum.

 

[PeroK]

"Shut up and calculate" is how Feynman described the Copenhagen interpretation. I'm a bit frustrated with the reluctance on this forum to deviate from that interpretation, but I'm honestly grateful for the facts and information shared because I agree that the ability to get the correct predictions is primary.

This thread is about "wave mechanics", which is not tied to any particular interpretation. In any case, PF is explicitly interpretation-agnostic. There is no standard here to insist on the Copenhagen interpretation - which the reverse, in fact. There's even a QM Interpretations sub-forum.

 

[weirdoguy]

I'm a bit frustrated with the reluctance on this forum to deviate from that interpretation

All interpretations share the same math - the one you were presented here. I guess you are frustrated with people here using math in general, but again - there is no way around that. Math is THE language of physics. You want to learn physics properly - you must learn its language. Every natural-language description of physics is based na math, not the other way around.

 

[vanhees71]

"Shut up and calculate" is how Feynman described the Copenhagen interpretation. I'm a bit frustrated with the reluctance on this forum to deviate from that interpretation, but I'm honestly grateful for the facts and information shared because I agree that the ability to get the correct predictions is primary. I'm not trying to dispute any of the numerous mistakes I've made on this forum.

If you want "interpretation", there's an extra subforum. The problem with interpretation is that often it is discussed by people without a solid fundament in the scientific part of quantum theory. The YouTube video is an example for how quantum theory is mystified rather than presented in a clear way. Quantum theory is based on solid mathematics of the "rigged Hilbert space". This is the mathematical language you need for all of physics, particularly classical field theory and quantum theory. So if you want to do physics you have to learn it anyway. Without a good foundation you cannot understand, let alone discuss, about what quantum theory as a natural science has to say. Great parts of what's called "interpretation" suffers from philosophical debates without firm ground in the scientific facts, and that makes it so difficult to get some really useful results. Feynman's approach in the famous "Feynman Lectures" (vol. 3) is thus the right one to learn about the facts before entering the slippery terrain of philosophical issues with interpretation.

Among the facts concerning the infinite square well is that there is no way do properly define a momentum observable, and that's ignored in this nonsensical YouTube video. Concerning the Hamiltonian there is no problem, because it is a well-defined self-adjoint operator on the Hilbert space, represented by $\text{L}^2([0,L])$ with "rigid boundary conditions" for the wave functions $\psi(0)=\psi(L)$. Before these mathematical facts and their quite obvious physical meaning is not understood, it's of no use (rather of much harm) to indulge into "interpretational debates". One should indeed be aware that the infinite square well is rather mathematical toy model.

 

[PeterDonis]

"Shut up and calculate" is how Feynman described the Copenhagen interpretation. I'm a bit frustrated with the reluctance on this forum to deviate from that interpretation

Discussion of interpretations belongs in the QM interpretations forum, not this one. As far as this forum (as opposed to the QM interpretations forum) is concerned, discussion should make use only of the basic math of QM, its predictions about experiments, and experimental results, without adopting any particular interpretations.

For more information please see the forum guidelines here:

https://www.physicsforums.com/threads/guidelines-for-quantum-physics-forum.978328/