The Energy Eigenvalue of a Wavefunction

I've been reading an introduction to quantum mechanics online, and while constructing the Schrodinger equation for a free particle, the equation $i\hbar \frac{d \Psi}{dt}=\hbar\omega\Psi$ is obtained.

It is then explained that if $i\hbar \frac{d}{dt}$ is considered as an operator, then the equation means "that when it operates on $\Psi$, the result we get back is the energy eigenvalue of the wave."

It's clear to me that $\Psi$ is an eigenfunction of $i\hbar\frac{d}{dt}$, with the eigenvalue being $\hbar\omega$, but I don't understand why they say that $\hbar\omega\Psi$ is the "energy eigenvalue of the wave."

As far as my understanding goes, $\hbar\omega$ gives the energy of a photon with angular frequency $\omega$, and so it's unclear to me why the energy of a photon multiplied by $\Psi$ would be of particular significance.

Furthermore, I don't think I understand what is meant by the statement "the energy eigenvalue of the wave".

If anyone could help explain these concepts (preferably without presupposing too much knowledge of QM) that would be great.

$E=\hbar \omega$ is the energy of any particle, not just a photon.

The terminology eigenvalue comes from linear algebra. Given a matrix $M$, an eigenvector $v$ of $M$ with eigenvalue $\lambda$ is a solution to the equation

$$Mv = \lambda v$$

In quantum mechanics the wave function is to be thought of as a kind of vector. Observables are represented by (hermitian) operators (which are morally the same as (hermitian) matrices), and the eigenvalues of those operators are the possible values the observable can take.

So setting $M=i\hbar \frac{\partial}{\partial t}$, $v=\Psi$, and $\lambda=\hbar \omega$, you see the equation you wrote is just an eigenvalue equation for the "energy operator" $i\hbar \frac{\partial}{\partial t}$. The eigenvalue is $\hbar \omega$. It's called an energy eigenvalue because the observable is the energy.

Your physical system will be described by a wave function (I prefer to call it a "state") $\Psi$. If $\Psi$ satisfies that eigenvalue equation, then you will measure the system to have the energy $E$ with 100% probability. In general $\Psi$ will not be an eigenvector of the energy operator. However it can be written as a sum of energy eigenvectors due to the magic properties of hermitian operators (note added--I mean the special theorem, https://en.m.wikipedia.org/wiki/Spectral_theorem). In that case if you measure the energy you will measure it to be one of the eigenvalues of the eigenvectors making up $\Psi$, with probability given by the square of the amplitude of that eigenvector.

If the notes you are following did not explain this clearly before diving into talking about energy eigenvalues, I would strongly recommend looking for alternative references to read. There are a million references that explain quantum mechanics at all levels and in many different ways. This point you are asking about is the central concept of all of quantum mechanics, so it's definitely worth reading many sources about this to find the one that has the explanation you find the clearest.

• Great, thanks. I guess most of my confusion came from not realizing that $\hbar\omega$ was the energy of any particle (instead of just photons), but now that that's cleared up this makes sense. Aug 2 '13 at 12:10
• which particular magic property are you referring to here?
– inya
Jan 10 '17 at 20:57
• The spectral theorem. I added a link to Wikipedia. Jan 10 '17 at 21:40

the result we get back is the energy eigenvalue of the wave.

It's not correct to think of "the" energy eigenvalue of a wavefunction.

Rather, a wavefunction is either an energy eigenfunction or it isn't.

If it is an energy eigenfunction, then there is an energy eigenvalue else there isn't.

In other words, your first equation holds only if the $\Psi$ is an energy eigenfunction.

The general wavefunction is not an energy eigenfunction but, it can always be decomposed into a weighted sum of energy eigenfunctions.

• Okay. So what is the significance of the energy term in this equation? That is, even if the eigenvalue does represent energy, it's just the energy of a photon right? So what is the importance of noting that when $i\hbar\frac{d}{dt}$ operates on $\Psi$ the eigenvalue of the eigenevalue equation is the energy of a photon? Aug 2 '13 at 2:28
• @nsanger The point is that the Schrodinger equation $i\hbar\mathrm{d} \psi/\mathrm{d}t = \hat{H} \psi$ is completely general, even when $\psi$ is not an eigenstate and even when you are describing something more complicated than a single photon. Really, fundamentally, this is the definition of energy. Energy is the operator that generates time evolution. They are just giving you a plausibility argument with an example where you already know the answer. The eigenstates are important because their time evolution is particularly simple and you can use them to build more general states. Aug 2 '13 at 4:33