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If we have a potential of the form $V(x,y)=k_{x}x^{2}+k_{y}y^{2}$ then that potential looks like this: A 3d plot of V against x,y As you can see, this potential is clearly round. When dealing with simple harmonic oscillators (like a pendulum free to swing in 2 directions for small swings) then the behaviour we see is clearly... rounded. What I think I mean by this is if $k_x = k_y$ that the SHO in question needs more energy to reach somewhere based only on the euclidean distance from the equilibrium point and if they are different it's only a question of scaling the dimensions first.

Why is it then, that when I render out the 2D QHO what I get is decidedly rectangular.

Rendering of the QHO in motion

This rendering uses hue for argument and brightness for magnitude.

Since it is possible I am using the wrong approach I will say that the eigenstates I am using are $\Psi_{n_x,n_y}(x,y,t)=X(x)\cdot Y(y)\cdot \phi(t)$ where $X$ and $Y$ are only different in their variables and values of k but are of the form $$\psi_{n}(\sqrt{\alpha}x) = \left(\frac{\alpha}{\pi}\right)^{1/4} \cdot \frac{1}{\sqrt{2^{n}n!}}\cdot H_{n}(\sqrt{\alpha}x) e^{-\frac{\alpha x^{2}}{2}}$$ And $\alpha = \sqrt{\frac{mk}{\hbar^{2}}}$ and $\phi(t)=e^{-itE/\hbar}$

From what I've read it seems like this should behave as though it had the potential described but for any values of k that I tried and for any superposition of states I could think of it still ends up looking rectangular.

Is this what I should expect? If so then why is the quantum case so different? Is my code wrong? If so where in my reasoning is the misstep?

Disgusting
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  • If you'd like to find solutions which exhibit elliptical symmetry, might I suggest solving the Schrodinger equation in elliptical coordinates? https://en.wikipedia.org/wiki/Elliptic_coordinate_system – fewfew4 Jul 17 '21 at 00:29
  • @fewfew4 Shouldn't the solution result in the same behaviour regardless of co-ordinates? – Disgusting Jul 17 '21 at 01:51
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    Nope! Remember there are many linearly independent solutions to the Schrodinger equation, and the general solution is a linear combination of these. It is possible to construct a solution which is elliptical using particular linear combinations of the solutions you used, but it's difficult to determine what the precise linear combination is. Instead, start by solving the Schrodinger equation in elliptical coordinates, and the solutions you find will automatically have that symmetry. – fewfew4 Jul 17 '21 at 01:59
  • This rectangular motion also seems quite strange to me. What superposition are you using, or equivalently what is your initial condition? Also, you’re missing an “i” in your time dependence right? – Ian Jul 17 '21 at 04:34
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    You can try starting with a gaussian shaped wavepacket. The center of such a wavepacket follows the motion of a classical trajectory that starts at that center of the initial gaussian wavepacket, when the potential is harmonic. I would recommend also to look at the magnitude squared, as it is easier to see the motion. – Hans Wurst Jul 17 '21 at 08:13
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    First, nondimensionalize dross by absorbing it into your units, so $k_x=k_y=1$, and the profile is a circle, not ellipse. Then, understand how classical motion can be circular with a π/2 phase difference between the x and y components. Study the similarly phased superposition of a simple state before you mindlessly plot things. – Cosmas Zachos Jul 17 '21 at 11:50
  • @Ian I definitely used the "i" in the code. Thanks for pointing out the typo. – Disgusting Jul 17 '21 at 12:20
  • @HansWurst The coefficients are positive reals and based on a gaussian centered at nx=sqrt(kx) and ny=sqrt(ny) but I have tried many combinations. – Disgusting Jul 17 '21 at 12:20
  • @CosmasZachos I've tried setting kx and ky to 1 and it does not result in a circle. When you say pi/2 phase difference between x and y, how would one accomplish that? If you multiply either X or Y by e^-ipi/2 you just multiply the whole thing? I have found sets of eigenstates where I can get it to start on the right or the left but the behaviour looks more like a diamond. – Disgusting Jul 17 '21 at 12:20
  • @fewfew4 I think your comment is closest to an answer. Could you maybe write one so I can accept it and close the question? – Disgusting Jul 17 '21 at 12:20
  • @Disgusting Where do you expect to see a circle ? Do you mean the motion of the wavefunction ? If so you would need an initial wavefunction that starts with some momentum, like a marble thrown into a bowl. If you simply drop the marble and let it roll, you will get only diagonal motion. I think we see that diagonal motion but don't take my word for it, without the actual parameters/code I cant certify it. I also assume that the "rectangular shape" is due to your grid. You are using a finite x,y grid to sample your function, aren't you ? – Hans Wurst Jul 17 '21 at 13:40
  • Look at the (classical!) center motion of a Schroedinger wave packet you've learned about in class. Arrange $\langle \hat x(t)\rangle=\cos t$ and $\langle \hat y(t)\rangle=\sin t$. Observe the wavefunction and probability. Your solution will rotate uniformly with this rotating center. Remember: the solutions need not have the symmetries of the hamiltonian controlling them . This is the very cornerstone of SSB. – Cosmas Zachos Jul 17 '21 at 14:20

2 Answers2

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When looking for solutions to the Schrodinger equation that exhibit a particular symmetry, it's best to use coordinates which mimic this symmetry. You used Cartesian coordinates, and so the basic solutions you find will inevitably have rectangular symmetry.

As an example for how this works for the 2d quantum harmonic oscillator when $k_x=k_y$, look at @CosmasZachos 's answer to 2D isotropic quantum harmonic oscillator: polar coordinates.

fewfew4
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My bet would be that it's because you're plotting magnitude instead of magnitude squared (probability amplitude), combined with your specific initial conditions of the wave function. Square rooting probability amplitude could amount to something like using a different kind of "norm". Although rectangularity more likely points to something like the maximum norm. It is difficult to say if and what you are doing wrong without seeing your code.

As an illustration for what I said about the initial conditions, consider separating variables $x$ and $y$ which gives you a solution that is a product $$\psi(x,y,t)=\psi_1(x,t)\cdot \psi_2(y,t)$$ So if you choose for both factors a moving wave packet solution, forming the product is a little like moving from left to right and back a paper with a vertical bar cut out, and crossing this with a paper with a horizontal bar cut out, and moving it up-down and back. The result would always look "rectangular", I guess. So it is an artifact of choosing your specific solution, which is what fewfew4 has emphasized in his answer, as far as I have understood.

oliver
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