I'm taking a class on QM and we're simulating the wave function of an electron in a box at the lowest energy level and I'm supposed to change the simulation to show the wave function for the next energy level.

The problem is that I don't quite understand the relationship between position and energy level. I'm fine with the formula for the energy levels, but I don't see how to relate it to $\psi(x)$, as it doesn't even involve $x$.

And yet, I know there is a relationship -- if I understand correctly, electron orbitals are just PDFs of position obtained from normalizing and squaring the position wave functions of electrons with different quantum numbers, one of which is energy level, and these orbitals are very obviously different, as can be seem from the many diagrams of them. So, clearly energy level does effect the position function, and, for that matter, so do the other quantum numbers.

But what's the specific mathematical relationship, either for a theoretical electron in a box or for an electron in an atom? Like, say I already have an expression for $\psi(x)$ at an arbitrary energy level. How would I modify it to model an electron at a different energy level?

I did see these questions:

Relationship between Quantum Numbers and the Wave-function

Relationship between Energy Level and electron position



but none of them answer the question of what the specific mathematical relationship is in a useful form.

I'm not really sure if this is a physics or chemistry question, but I'm picturing electron orbitals in atoms so I'm leaning towards it being more chemistry.

  • $\begingroup$ Electron in a box has a totally different wavefunction from an electron in an atom because the conditions are totally different. But I'm sure they taught you how to figure out the lowest energy level (or at least prove that it is one and gave you some idea why it works - not how to discover it from scratch). It has to be a sine wave with zeroes at the edges of the box, yes? $\endgroup$ Commented Feb 21, 2023 at 22:32
  • $\begingroup$ @user233751 Isn't the particle in a box a simple model of a one electron atom though? And then there's the Kronig-penney model for electrons in a semiconductor, which IIRC, is basically a sequence of interconnected finite energy wells. I took semiconductor physics last semester and we spent a while with that model and the approach to finding wavefunctions for electrons with this model was very similar to how we did it for the particle in a box problem. $\endgroup$ Commented Feb 21, 2023 at 22:43
  • $\begingroup$ @user253751, right but my question is, if I have the formula for the position wavefunction at, say, n=1, how do I find the formula for the wavefunction at some other value of n? $\endgroup$ Commented Feb 21, 2023 at 22:47
  • $\begingroup$ I don't think there is a way to do that. $\endgroup$ Commented Feb 21, 2023 at 22:47
  • $\begingroup$ Often, the one-electron functions with higher energies have more nodes. Often, the functions are of the same type, just with different parameters. So knowing one solution of the problem might help to find others by analogy. $\endgroup$
    – Karsten
    Commented Feb 23, 2023 at 23:27

3 Answers 3


You need to go back to the very start. Here, you're kind of asking: I have a solution $\psi_0$, how do I get the next solution $\psi_1$? The answer is to look at how $\psi_0$ was obtained, and it turns out that that same process which yielded $\psi_0$ will give you all the $\psi_i$'s along with their associated energies $E_i$.

Physical systems tend to admit a series of wavefunctions*

$$\{\psi_0(x), \psi_1(x), \cdots\},$$

which satisfy the time-independent Schrödinger equation

$$\hat{H}\psi_i(x) = E_i \psi_i(x)$$

for all $i$. The way to obtain the wavefunction is to solve the Schrödinger equation. For a particle in a box, you have that

$$\hat{H} = \frac{p^2}{2m} = -\frac{\hbar^2}{2m}\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\right)$$

(for $0 \leq x \leq L$, $L$ being the length of the box) and so to obtain the wavefunctions you need to solve the differential equation

$$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = E\psi, \label{eq:de}\tag{1}$$

which also respect the boundary conditions $\psi(0) = \psi(L) = 0$. It turns out the functions $\psi$ have a general formula of

$$\psi_n = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),$$

with the energies

$$E_n = \frac{n^2\hbar^2\pi^2}{2mL^2}.$$

So, generally speaking, you don't generate one wavefunction from another one: you obtain all of them at the same time by solving the differential equation (eq. $\ref{eq:de}$).

The same is true of pretty much any other system, including the hydrogen atom (which gives you the orbitals). It's just that the Hamiltonian is different, so you get a different differential equation, and ultimately different solutions.

Now... there are some cases where you can go from one stationary state to another. A notable example is the quantum harmonic oscillator, where you can use 'raising' and 'lowering' operators to go between stationary states; however, this isn't something you can conclude by looking at the final wavefunctions, it's something you have to figure out by studying the form of the Schrödinger equation. So again, you have to go from the start.

TLDR: The "specific mathematical relationship" you're looking for is that they all satisfy the Schrödinger equation.

* To be technically correct, these are stationary states; they are 'special' wavefunctions which do not change over time (in that the expectation values of observable quantities like position and momentum are independent of time). Non-stationary states are perfectly permissible too, and can be constructed as linear combinations of stationary states.


The Schrödinger equation is just a normal type of differential equation. We interpret it as describing how the operator for the sum of kinetic and potential energy (the Hamiltonian) produces the total energy. It has the form $H\psi_i=E_i\psi_i$ with lots of solutions $E_i$. The Hamiltonian has the form $\displaystyle H=-\frac{\hbar}{2m}\frac{d^2}{dx^2}+V(x)$ where $V(x)$ is the potential energy, zero in the case of a particle in a box, but $kx^2$ for a harmonic potential as, for example, in a vibration; $k$ is a constant.

However, our description is not complete, because the 'boundary conditions' have not been specified. To solve this differential equation to find $\psi$, integration is needed and is done using normal mathematical methods. We know that in integrating, limits are always involved in one form or another. These limits are called 'boundary conditions' when solving differential equations and you always have to use them to get to any solution. In the case of a particle in a box, these are that $\psi$ is zero at each side of the box, because here the potential energy is infinite by definition. This condition forces integer solutions onto the problem; the wave functions $\psi$ are sine waves that are always zero at the ends of the box. Using the words 'integer solutions' is just another way of saying that quantum numbers are involved. What boundary conditions are used depends on the particular problem you are trying to solve, and changing the boundary conditions will change the solutions.

Finally, notice that we find all values for $E_i$ and $\psi_i$ at the same time, but how we interpret the wave function is terms of probability of being at a certain position $x$, i.e., $\psi(x)^*\psi(x)$ has nothing to do with the maths, but how we choose to interpret these via quantum theory. To illustrate the ordinariness of the maths involved in the Schrödinger equation, a similar equation and solutions are found when a taught wire is plucked, as in a guitar string.

  • 1
    $\begingroup$ @PeterMortensen The spellchecker marks wavefunction as bad spelling but this is common usage. A matter of taste I guess: reddit.com/r/Physics/comments/3vvjsm/… $\endgroup$
    – Buck Thorn
    Commented Feb 23, 2023 at 9:32
  • 2
    $\begingroup$ We do write wavelength and wavenumber, eigenvalue and eigenfunction, perhaps if wavefunction is commonly used it will appear in dictionaries in the future. $\endgroup$
    – porphyrin
    Commented Feb 23, 2023 at 12:35

A quick intuition to understand how energy affects the position distribution can be obtained by looking at the s orbitals.

For orbitals $1s, 2s, 3s, ..., ns$, the increasing principal quantum number means that you have increased energy. Correspondingly, the average distance between the electron and the nucleus also increases. Thus, the higher energy allows the electron to stray further from the nucleus.

Mathematically, all these orbitals must be solutions to the Schrödinger equation. As pointed out in previous answers here, when you solve these equations you don't obtain one solution, but instead, a class of solutions that, once you impose boundary conditions, will result in wave functions for various different energy levels.


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