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Pretty much what it says in the title.

For example, in the ‘particle in an infinite square potential well’ problem, we represent the wave function using sinusoids. As these are zero valued at the bounds.

However, could we also choose a different waveform? A square wave, for example. Or any waveform that fulfils that criteria.

Geodatsci
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    This is the subject of the first week or two of an undergraduate quantum mechanics course. Can you tell a little about your math background? Do you know about differential equations and eigenvalue problems? – rob Jun 13 '22 at 14:26
  • I have undergraduate applied maths to a 1st year level and am familiar with differential equations and eigenvalue problems. It has been a while though! I understand that sinusoids have other nice properties, as per the answer from The Photon, however, I was wondering whether they are the only family of functions that fulfil the necessary criteria to be a solution for the problem? – Geodatsci Jun 13 '22 at 14:36
  • @Geodatsci do you consider the complex exponentials to be "other" functions? – The Photon Jun 13 '22 at 14:42
  • sinusoids are basic solutions of wave equations, and quantum mechanics is based on wave equations, Schrodinger, Dirac ... – anna v Jun 13 '22 at 14:51

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Sinuosoids (and the very closely related exponential function) are eigenfunctions of the derivative operator.

That means if you take the derivative of a sinusoid, the output is another sinusoid of the same frequency.

So if you have an equation that equates sums of derivatives of a function to the function, you are likely to be able to solve it with sinusoids or exponentials.

The same thing is not true of functions like square waves, even though you can construct a square wave from an infinite sum of sinusoids.

The Photon
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  • This makes sense! Thank you. – Geodatsci Jun 13 '22 at 14:30
  • Do any other classes of functions fulfil this criteria? – Geodatsci Jun 13 '22 at 14:31
  • @Geodatsci, That might be a question better asked on math SE. I am pretty sure the answer is "no", but they'd be able to explain why. – The Photon Jun 13 '22 at 14:41
  • @Geodatsci: If you consider a differential operator other than just the derivative operator, then you'll get a different class of functions as eigenfunctions. For example, the eigenfunctions of the operator $d^2/dx^2 + x^2$ are all of the form $e^{-x^2/2} H_n(x)$, where $H_n$ is a Hermite polynomial. This makes them important for the quantum harmonic oscillator. – Michael Seifert Jun 13 '22 at 15:13
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As another answer says, you can use the principles of Fourier decomposition to construct a square wave or any other function you like from a superposition of sinusoids —— or, for that matter, from any other "complete basis" of functions.

However, the physics of the Schrödinger equation is that its eigenvalues correspond to a physical observable, the energy:

$$ \hat H \psi_i = E_i \psi_i $$

This means that a system which is observed to have some particular energy eigenvalue $E_i$ can't be described by the kind of superposition $\psi_0 + \psi_1 + \psi_2 + \cdots$ which you need to construct an arbitrary wavefunction.

The sinusoids (or equivalently, the complex exponentials $e^{-ikx}$ and their linear combinations) are solutions to the eigenvalue equation when the Hamiltonian has form $-\partial_x^2$. The relationship between this form of the Schrödinger equation's Hamiltonian and the Hamilton-Jacobi equations of classical mechanics is complicated. Many intro-quantum books are written for students who haven't gotten that deep into classical mechanics, and take the Schrödinger Hamiltonian as an axiom; this oversimplification elides a quarter-century of vigorous research.

rob
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  • I think an equivalent way of phrasing my question would be: Are the sinusoids (and their complex valued exponential equivalents) the only solutions for when the Hamiltonian has the form −∂2?

    ‘Yes’ would answer my question. In the case of ‘no’, I would then ask ‘so why do we use sinusoids?’

     – Geodatsci Jun 13 '22 at 16:50
  • General questions about the uniqueness of solutions to differential equations are probably better suited for [math.se]. As a commenter elsewhere has already pointed out, a Hamiltonian with a nonzero potential will have a different complete set of solutions. – rob Jun 13 '22 at 16:58
  • thanks, between your answer and The Photon’s I am happy with the explanation. – Geodatsci Jun 14 '22 at 10:03
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Sine and cosines and their related exponentials form a nice set of orthogonal functions. In your quantum well case, that fits the boundary conditions nicely.

If you had an optical fiber, or a nanowire quantum well, and you had a cylindrical geometry, when you work through the differential equation and looked for a nice set of orthogonal functions you would find that Bessel functions have a oscillatory and exponential like functions that would fit the boundary conditions.

In both cases you would find that the Eigen value would be related to the frequency(or wavelength)and the geometry. The width the well or the radius of the nano wire.

Same for spheres: spherical harmonics and Legendre functions depending on the partial differential equation you want to solve.

You can also look at Fourier series works . Usually you make them up out of sines and cosines, so a sine or cosine is more fundamental than a square wave. Since they are orthogonal functions you can also normalize them easily. A square wave can be described by the Fourier series. However you can general the Fourier series to use other functions, but sines and cosines are often convenient since people think about waves in terms of sines and cosines.

UVphoton
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