My chemistry textbook introduces the wave function as

$$\psi(x)= A \sin\left(\frac{2\pi x}{\lambda}\right)$$ Therefore, the Schrödinger Equation is:

$$\frac{d^2\psi(x)}{dx^2} = -~\left(\frac{2\pi}{\lambda}\right)^{2}\psi(x)$$

and by multiplying each side by $\frac{-h^2}{8\pi^2m}$ you get:

$$-~\frac{h^2}{8\pi^2m} \frac{d^2\psi(x)}{dx^2} = \frac{p^2}{2m}\psi(x) = (T)(\psi(x))$$ where $T$ is kinetic energy.

Now, I don't fully understand that last bit, and I am willing to see an explanation on it, but my main question is that, when the book transitions from discussing the Schrödinger Equation in the context of a "particle-in-a-box" to the context of a one-electron atom, it seemingly changes the wave function to

$$\psi_{(n,\ell,m)}(r,\theta,\phi) = R_{n,\ell}(r) Y_{\ell,m}(\theta,\phi)$$

where n = principle quantum number, $\ell$ = angular momentum number, m = magnetic quantum number, r = radius (distance between nucleus and electron), R = the "radial part" of the equation, Y = the "angular part" of the equation, and $\theta$ and $\phi$ are some angles that I still don't know how to get.

I guess my question is did the wave function $\psi$ change? Why is suddenly so different and why do r,$\theta$ and $\psi$ suddenly come into the picture? Is it because you are going from the one-dimensional, Cartesian "x" version of the wave function to the three-dimensional, polar coordinate version of it?

  • $\begingroup$ Don't worry too much about the wave function at this point. I think what they did in the third equation is the application of the momentum operator. Operators are a fundamental concept of quantum mechanics. So be sure that you are familiar with them. $\endgroup$ Nov 16, 2016 at 8:17
  • $\begingroup$ The third equation is, as you mention, the wave function of an electron in an atom. $\endgroup$ Nov 16, 2016 at 8:18
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    $\begingroup$ This wave function, is different to the particle in the box wave function since the potentials of the two systems are different. They are two different systems! As such, the variables that need to be considered are different (hence why the form of the wave functions are different) $\endgroup$ Nov 16, 2016 at 8:19
  • $\begingroup$ I think your textbook is trying to prove to you that some waves are eigenfunctions of the momentum operator in the first equations. $\endgroup$ Nov 16, 2016 at 8:20
  • $\begingroup$ Wave function is still just a function. Yes, it is just that we switched to a 3D version, and also to polar coordinates, and then turned on the Coulomb potential, so the function looks pretty different now. Other than that, everything is the same. $\endgroup$ Nov 16, 2016 at 8:43

1 Answer 1


The first function you have there

$\psi(x)=A\sin(\frac{2\pi x}{\lambda})$,

is very similar to the function of a particle in a monodimensional box.

This function is a very helpful example in order to understand how does the quantum mechanics works.

The other function is

$\psi_{(n,m,l)}(r,\theta,\phi)=R_{n,l}(r) Y_{m,l}(\theta,\phi)$.

This function represents how an electron moves in a hydrogen atom and the full expression is quite different to the first function. Just see the representations:

Wavefunction representation for a particle in a box Representations for Y with different values

Ok. The functions do not mean the same concept. So... Why do we use "$\psi$" in both?

In math we use $f(x)$ to speak about any function. In quantum mechanics we use $\psi(x)$ for the same: represent a function easily.

Yeah, but... Why do i have to study the first function if it's just a lie and not the full story?

Remember it's only a very good example. However, the particle in a box function can be used to determine the energy of an electron in a conjugated system as beta-carotene and it's energy for the first excited level. If you subtract the second energy to the first one it gives you the energy for a photon in the visible region: the carrot's orange color.

This means that the particle in a box is a good example if you start to study quantum chemistry and it even works as a good approximation for some systems.

Ok. Too much information but I want to know more: How can I get the second function from the first?

Well, that's... Impossible...

To obtain the functions we use the Hamiltonian operator, which contains the kinetic energy operator and the potential energy operator.

For the particle in a box, we only use the kinetic energy operator because we don't have any potential into the box. This leads to the first function you have.

For the hydrogen atom we must use both operators (all the Hamiltonian) which means to include a function for the potential and we must work in 3D. The potential function can be obtained using the angular momentum. The spherical coordinates makes easier to work in 3D. This leads to radium and angular functions.

Summarizing, we have a different wave function for every different situation. This situation is what determines the function and it's conditions:

  • The particle in a box → Quantization.
  • The free particle → No quantization.
  • The electron in a hydrogen atom → Quantization + quantum numbers + rotation +... → Atomic orbitals.

Ok, seems legit, but it's hard to understand

I recommend you the books "Physical Chemistry" from Atkins, Levine or Engel to understand better the ideas and see the complete development.

In case you hate textbooks you can visit hyperphysics web page for a general looking.

  • $\begingroup$ thank you! so the first wave function (the one for the particle in the box) is basically considering other things than the second wave function (the one for the electron in the hydrogen atom) but because the potential energy is 0 inside the box it doesn't matter but in an atom it does, so you have to include that, and then convert to 3 dimensions and then convert to polar coordinates, making the whole thing look very different, right? $\endgroup$
    – eryk
    Nov 16, 2016 at 21:09

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