# What does the finding the eigenvalue of a wavefunction physically mean?

In https://arxiv.org/abs/physics/0602145 (page 8), there is a passage which says that the discrete nature of the energy levels of an electron in a hydrogen atom comes from the fact that the solutions/eigenvalues of a equation (wavefunction?) are discrete themselves.

Isn't the energy levels' probability dictated by the square of the wavefunction, so these solutions should represent something different. What does finding the solution to this equation physically mean?

To clarify: I was asking for the physical implications of finding the eigenvalue and how it differs from finding the square of the wavefunction.

Please don't presuppose that I have knowledge on QM.

• In general please do not link directly to documents where a link to an abstract page is available. On Physics SE we prefer links to abstract pages. Dec 15, 2022 at 12:07

In textbook quantum mechanics (as is the case in solving the hydrogen atom) we deal with the Schrödinger Equation: $$i\hbar\frac{\partial}{\partial t}\Psi(x,t) = \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t) + V(x,t)\Psi(x,t).$$ The solutions to the Schrödinger Equation (a differential equation) are called wave functions. In general, wave functions represent physical states. Physical states contain within them all knowable information (within the scope of the theory) about the thing being described by that physical state.

In particular, if we let $$V = -\frac{e^2}{4\pi\epsilon_0r}$$, i.e., equal the potential felt by an electron in a hydrogen atom, we obtain the Schrödinger Equation whose solutions (or wave functions) describe the possible physical states that the electron can be in.

It turns out that once we plug $$V$$ into the Schrödinger Equation, we can reduce the Schrödinger Equation into an energy eigenvalue problem (and an angular momentum eigenvalue problem). A more detailed description of this process can probably be found in any intro. quantum textbook.

It is in solving this equation that you find out that the electron in a hydrogen atom cannot take on any old physical state. Instead, the electron can only take on states with particular definite energy values (and superpositions of such states by linearity of the Schrödinger Equation). The fact that the electron cannot be found in a physical state with any energy is perhaps what the paper is describing.

Now, to address some of the other things you've said, it might be helpful to set some things straight.

We already briefly covered what the Schrödinger Equation is and that its solutions correspond to the physically allowed states of something. Let us denote an arbitrary state by $$\Psi$$. These states $$\Psi$$ live in a vector space. You may ask: "A vector space you say?! Is there a basis for this vector space so that we can write $$\Psi$$ as an arbitrary linear combination of said basis vectors?" The answer is absolutely, "Yes". To somewhat see this, we must talk about observables and their corresponding operators.

Every observable quantity (think something you can measure) is associated with an operator in Quantum Mechanics. These operators are usually denoted as operators with a little hat. For example, there is a position operator $$\hat{x}$$, a momentum operator $$\hat{p}$$, and an energy operator $$\hat{H}$$. Amazingly, by Spectral Theorem, these operators are diagonalizable and their eigenvectors form a basis for the space in which $$\Psi$$ lives!

More explicitly, each basis vector associated with, say, the energy operator $$\hat{H}$$ are eigenvectors (also called eigenstates) of the operator $$\hat{H}$$. That is to say these special states $$\Psi_E$$ satisfy the equation $$\hat{H}\Psi_E = E\Psi_E$$ where $$E$$ is a constant known as energy in this particular case. Thus, in the very very special case that our arbitrary vector $$\Psi = \Psi_E$$, it is an energy eigenstate.

It is these energy eigenstates that we actually end up solving for when we solve for part of the solution to the hydrogen atom! And, we find out that instead of a continuous spectrum of energy eigenvalues, we only obtain a discrete spectrum of energy eigenvalues.

You notice that I haven't mentioned anything about probability. The concept of expectation values is separate (at least in this discussion) from the concept of the eigenspectrum of an operator.

I would suggest briefly familiarizing yourself with the first few chapters of an introductory quantum mechanics book, it'll really help your understanding of the solutions to the hydrogen atom if that is your goal!

• Thank you for the answer Silly Goose. Yes, I agree with @garyp. I was looking to have a physical understanding of eigenvalues, as well as how it relates to probability, as they seem related yet not? Dec 15, 2022 at 13:09
• Oh, I am curious: how do you prepare such a hydrogen atom? @garyp Dec 15, 2022 at 17:05
• Oh I see I’ve misunderstood the question then :) @Jonathan Dec 15, 2022 at 17:05
• @Jonathan Are you looking for something like: In textbook quantum, an arbitrary wave function can be written as a linear combination of eigenstates of an observable. Eigenstates of energy, for example, are special states in which upon measurement you are certain to get the eigenvalue as the measured value. We accept that upon measurement, a wave function collapses onto one of these eigenstates with probability equal to $|c_n|^2$ where $c_n$ represents the weight of the nth eigenstate in the linear combination… Dec 15, 2022 at 17:14
• …from here we can observe that if we accept that arbitrary states are linear combinations of eigenstates and the process of collapse occurs, then the eigenvalues represent possible outcomes of measurement. Additionally, the probability of getting a particular possible outcomes is given by the “weight” of that particular outcomes’s eigenstate in the linear combination making up the arbitrary state (wave function). Dec 15, 2022 at 17:16

In https://arxiv.org/ftp/physics/papers/0602/0602145.pdf (page 8), there is a passage which says that the discrete nature of the energy levels of an electron in a hydrogen atom comes from the fact that the solutions/eigenvalues of a equation (wavefunction?) are discrete themselves.

The stationary-state solutions for energies lower than zero (which is the value of the potential at infinity) are discrete. This means that such solutions can be labelled by integers. For example, the hydrogen atom ground state wavefunction is usually labelled with three integers ($$n$$=1, $$\ell$$=0, $$m$$=0).

The corresponding energies can also be labeled with integers, in this case the single integer $$n$$. (Since it turns out that the energies themselves don't depend on $$\ell$$ and $$m$$, only $$n$$).

For example, the ground state energy, in units of electron-volts, is: $$E_{n=1} = \frac{-13.6}{1^2} = -13.6$$

For example, the first excited state energy is: $$E_{n=2} = \frac{-13.6}{2^2} = -3.4$$

The bound states (the states with energy less than zero) are discrete and the energies are given by: $$E_n = \frac{-13.6}{n^2}\;,$$ where $$n$$ is any integer from 1 to infinity.

So, you see that these state and their energy values are discrete. They are labelled by integers.

This is not the whole story. The hydrogen atom also supports a continuum of other states, but often we only are interested in the bound states (the ones I just described).

Isn't the energy levels' probability dictated by the square of the wavefunction, so these solutions should represent something different. What does finding the solution to this equation physically mean?

No, not if the system is in a stationary state. If the system is in a stationary state, $$\phi_{n,\ell,m}(\vec r)$$, then its energy is exactly fixed and will always be measured as $$E_n$$. This is different from something like the position $$\vec r$$, which can be described probabilistically by the square of the wave function.

If the system is in some general bound state $$\psi(\vec r)$$ that is not necessarily a stationary state, then the probability to measure the energy $$E_n$$ is: $$p_n = \sum_{\ell,m}|\int d^3r \phi_{n,\ell,m}^*(\vec r)\psi(\vec r)|^2$$

To clarify: I was asking for the physical implications of finding the eigenvalue and how it differs from finding the square of the wavefunction.

The eigenvalues of Hamiltonian are energies. This is completely different from the square of the wave-function.

If the state is $$\phi_{n,\ell,m}(\vec r)$$, then the probability to find the electron within $$d^3r$$ of $$\vec r$$ is: $$P(\vec r)d^3r = |\phi_{n,\ell,m}(\vec r)|^2 d^3r$$ But, in contrast, the probability that the energy is $$E$$ is: $$\delta_{E,E_n}\;,$$ i.e., zero unless $$E=E_n$$, since the state is known to have energy $$E_n$$. This is basically what it means to "be in" a stationary state (stationary states have fixed energy).

I suspect that the eigenvalue gives possible solutions and the square of the wavefunction give the pdf of how often that solution would come up, but need confirmation, and a deeper understanding of both.

No, that doesn't really make sense... unless this is meant in the same sense as the equation for $$p_n$$ written above.

I think it's easier to comprehend this by starting with concrete physics on a human scale. A guitar string is a fine example of an object that is well modeled by a wave equation. The eigenvalues of this equation are the frequencies of the fundamental and overtones of the note that the string is tuned to (possibly with some scaling).

It's the same concept for the hydrogen atom, although the details are changed. We usually formulate Schrödinger's equation in terms of energy, but that's just a scaled frequency by the Planck relation. The confinement of the standing waves is understood to be due to a potential rather than a fret/bridge combination. But, basically, the eigenvalues capture the "tuning" of the electron wave.

Now, you can most certainly take this idea to the far reaches of abstraction and use it to model physics that has little resemblance to the tuning of a guitar. But, as usual in physics, the idea has its origins in ordinary, concrete phenomena.