# Eigenvalues of an operator correspond to energy states in quantum mechanics, why?

When finding the discrete energy states of a operator I have been taught to use the time-independent Schrodinger equation which restates the definition of eigenvalues and eigenvectors. What I don’t understand is why the eigenvalues are the energy states, is there firstly a mathematical reason and secondly a physical reason?

Does this arise from Hamiltonian or Lagrangian mechanics which I am not familiar with?

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Sorry I mean eigenvalues of the operator not wave funciton – joshlk Apr 6 '11 at 17:00
Keep in mind that we are only dealing with Hermetian operators, because their eigenvalues are real, and hence correspond to positive definite probabilities. – Matt Calhoun Apr 8 '11 at 16:44

As has been remarked by others and explained clearly, and mathematically, the eigenvalues are important because a) they allow you to solve the time-dependent equation, i.e., solve for the evolution of the system and b) a state which belongs to the eigenvalue $E$, i.e., as we say, a state which is an eigenstate with eigenvalue $E$, has an expectation value of the energy operator which is easy to see has to be $E$ itself. But those explanations are advanced and rely on the maths. And they do not explain why $E$ should be considered 'an energy level'. At some risk, I will try to answer your question more physically.

What is the physical reason why the energy states of a system, e.g., an atom, are the eigenvalues of the operator $H$ that appears in the time-independent Schroedinger equation? Well, first, note that it's absolutely the same $H$ that appears in the time-dependent Schrodinger equation, $$H\cdot \psi = -i{\partial \psi \over \partial t}$$ which controls the rate of change of $\psi$.

The answer doesn't come from the classical Hamiltonian or Lagrangian mechanics, but from the then-new quantum properties of Nature. A non-classical feature of QM is that some states are stationary, which means they do not change in time. E.g., the electron in a Bohr orbit is actually not moving, not orbiting at all, and this solves the classical paradoxes about the atom (why the rotating charge doesn't radiate its energy away and fall into the centre).

The first key point is that an eigenstate is a stationary state: what is the explanation for this? well, Schroedinger's time dependent equation clearly says that, up to a constant of proportionality, the time-rate of change of any state $\psi$ is found by applying the operator $H$ (the Hamiltonian: we do not yet know it is also the energy operator) to it: the new vector or function $H\cdot\psi$ is the change in $\psi$ per unit time. Obviously if this is zero, $\psi$ does not change (this was the only classical possibility). But also if $H\cdot\psi$ is even a non-zero multiple of $\psi$, call it $E\psi$, then $\psi$ plus this rate of change is still a multiple of $\psi$, so as time goes on, $\psi$ changes in a trivial fashion: just to another multiple of itself. In QM, a multiple of the wave function represents the same quantum state, so we see the quantum state does not change.

Now the next key point is that a state with a definite energy value must be stationary. Why? In QM, it is not automatic that a system has a definite value of a physical quantity, but if it does, that means its measurement always leads to the same answer, so there is no uncertainty. So if there is no uncertainty in the energy, by Heisenberg's uncertainty principle there must be infinite uncertainty in something else, whatever is 'conjugate' to energy. And that is time. You cannot tell the time using this system, which implies it is not changing. So it is stationary. (remember, we are not assuming that $H$ is also the energy operator and we are not assuming the formula for expectations).

Thus being an eigenstate of $H$ implies $\psi$ is stationary. And having a definite energy value implies it is stationary. Being physicists, we now conclude that being an eigenstate implies it has a definite energy value, which answers your question, and these are the 'energy levels' of a system such as an atom: a system, even an atom, might not possess a definite energy, but if it doesn't, it won't be stationary, and being microscopic, the time-scale in which it will evolve will be so rapid we are unlikely to be able to observe its energy, or even care (since it won't be relevant to molecules or chemistry). So, 'most' atoms for which we can actually measure their energy must be stationary: this is 'why' the definite values of energy which a stationary state can possess are called the 'energy levels' of the system, and historically were discovered first, before Schroedingers equation. From a human perspective, most atoms that we care about spend most of their time that matters to us in an approximately stationary state.

In case you are wondering why time is the conjugate to energy, whereas Heisenberg's original analysis of his uncertainty principle showed that position was conjugate to momentum, we rely on relativity: time is just another coordinate of space-time, and so is analogous to position. And in relativistic mechanics, momentum in a spatial direction is analogous to energy (or mass, same thing). In the standard relativistic equation $$p^2-m^2=E^2,$$ we see that momentum ($p$) and mass $m$ are symmetric (except for the negative sign) with each other. So since momentum is conjugate to position, $m$ or energy must be conjugate to time. For this reason, Bohr was able to extend Heisenberg's analysis, of the uncertainty relations between measurements of position and measurements of momentum, to show the same relations between energy and time.

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The physics of this is the DeBroglie relation for particles, which relates the energy to the frequency of some wave. The energy of a photon is the frequency of the emitted electromagnetic wave.

When a quantum mechanical atom is weakly interacting with the photon field, and goes from a state with frequency f to a state with frequency f', it can only emit photons with frequency f-f'. The reason is that the transition process is only resonant with waves of frequency equal to the beat frequency \Delta f= f-f'. The atomic relative phases during the transition process recurs with time $1\over \Delta f$, and for any outgoing wave whose frequency does not match this, the process will be cancelling at long times, and no wave will be emitted.

This means that atomic transitions from f to f' are accompanied by a loss of energy of $h\Delta f$, so that one must identify the frequency with the energy in general quantum systems. The Schrodinger waves of definite frequency are the solutions of the time independent problem, since when

$$i{d\over dt} \psi = H \psi$$

and $H\psi = E\psi$, that is, if $\psi$ is an eigenvector of H, then $\psi(t) = e^{-iEt} \psi(0)$, so the time dependence of the wave has a definite frequency. I am giving a physical argument here, because the notion that energy is frequency is engrained into the foundation of quantum mechanics, and it is hard to argue that it is true using a formalism built upon this as a foundation.

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The basic experimental fact the inventors of QM had to deal with was the uncertainty principle. The mathematics behind this principle has two major parts, one involving linear algebra and another involving Fourier analysis.

In other words, the operator algebra of QM is necessary in order to have a theory which obeys the uncertainty principle, and if you want to know why this is true, you have to study the mathematics.

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I think you should specify you answer. – Self-Made Man Nov 14 '13 at 14:29

You seem to be confusing two things, namely the eigenstates of an operator and Schrödinger's equation. A priori, these two have nothing to do with each other.

In Quantum Mechanics, measurable quantities are represented by (hermitian) operators on a Hilbert space. For instance there is an operator $P$ corresponding to the momentum. In general, when measuring the momentum of a state $|\psi \rangle$, the result will not be deterministic. However, the average over several measurements will be equal to the expection value

$$\langle \psi | P | \psi \rangle$$

However, when $|\psi\rangle$ is an eigenvector of the operator, $P|\psi\rangle = \lambda |\psi\rangle$, then the measurement will always be the same value $\lambda$.

In particular, there is an operator corresponding to the total energy, the Hamiltionian $H$. The form of this operator can be obtained from classical physics if you replace momentum and location by their corresponding operators. For instance, the Hamiltonian of an electron in an electric potential $V$ is

$$H = \frac1{2m} P^2 + eV(X) .$$

Thus, the expectation value for the energy of a state $|\psi\rangle$ is $\langle \psi|H|\psi\rangle$.

Now, the Hamiltionian is a very interesting operator because it features prominently in the equation of motion, the Schrödinger equation.

$$i\hbar \partial_t |\psi(t)\rangle = H |\psi(t)\rangle .$$

What does this have to do with the eigenvalues of the Hamiltionian? A priori nothing, but the point is that knowing the eigenvectors and -values of $H$ allow you to solve this equation. Namely, if you have an eigenvator $|\psi_n\rangle$, then you have

$$i\hbar \partial_t |\psi_n(t)\rangle = H |\psi_n(t)\rangle = E_n|\psi(t)\rangle$$

which can be solved to

$$|\psi_n(t)\rangle = e^{-\frac{i}{\hbar} E_nt} |\psi(0)\rangle$$

To summarize, the eigenvalues of an operator tells you something about what happens when you perform measurements, but in addition, the eigenvalues of the energy operator help you solve the equations of motion.

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Both eigenvalues and eigenstates belong to some operator. In your case, this is the Hamiltonian operator $\hat H$. It's fundamental because of many reasons. First is that it is indeed an operator that represents energy in a sense that possible energy levels are encoded in its spectrum (i.e. a set of eigenvalues). The second important reason is that it is the operator that can be found in Schrodinger equation $i \hbar \partial_t \left | \psi(t) \right > = \hat H \left | \psi(t) \right >$. This equation can then be solved by writing $\left | \psi(t) \right >$ as superposition of eigenstates of $\hat H$: $\left | \psi(t) \right > = \sum_n c_n(t) \left | \psi_n \right >$. If we can find these states, we are done as $c_n(t) = \exp({-iE_n t \over \hbar}) c_n(t=0)$ solves the equation (and it also shows the importance of these eigenstates because they are preserved by time evolution).

So this means the problem of time evolution in quantum mechanics can be reduced to the problem of finding the eigenvalues and eigenstates of $\hat H$, the equation for that being $\hat H \left | \psi_n \right> = E_n \left| \psi_n \right>$.

Note: the above assumes that $\hat H$ is time-independent. If it's not (as is the case in basically all practical applications, using perturbation theory) then we use different techniques, e.g. of path integration, or various scattering formulas.

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The reason why it is the eigenvalues of the Hamiltonian and not some other operator that will give you the energy states is that in classical Mechanics, the Hamiltonian function is just the energy of your system, expressed as a function of position $x$ and momentum $p$. As a simple example, the Hamiltonian for a harmonic oscillator is $$H(x,p) = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2$$ Note that this really is just the sum of kinetic and potential energy, so we could write $$H(x,p) = E$$.

To get to quantum mechanics, one now performs what is called canonical quantization. There is no mathematically rigorous reason why this will give you a correct quantum mechanics. Since quantum isn't classical, we cannot really expect to find a seamless and watertight derivation of the former from the latter. To my knowledge, this approach has, however, always given correct results.

So, in canonical quantization, what one does is to replace the variables of the Hamiltonian, i.e., $x$ and $p$, with their operator versions, $\hat x$ and $\hat p$. Now we cannot simply write $H(x,p) = E$ anymore, since the energy is a scalar, but the Hamiltonian $H$ is now an operator. Operators are functions that take a wavefunction and modify it in some way and give you a new wavefunction. Now, another postulate of quantum mechanics is that you get the expectation value of an operator $\hat A$ in a given state $\Psi$ by calculation the integral $$\int dx \Psi^*(x) \hat A \Psi(x)$$ Hence, we get the expectation value of the energy by calculating $$\int dx \Psi^*(x) H \Psi(x)$$ Obviously, if $H\Psi(x) = E\Psi(x)$, then the expectation value yields $E$, and it is not hard to show that for such an eigenstate, the variance of $E$ will be $0$, i.e. every measurement of the energy in state $\Psi$ will yield the same value $E$.

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A transformation from one set to another can be regarded as a matrix if we define a particular basis. Likewise, an operator can be thought of as a matrix. What is the matrix equation that relates eigenvalues and eigenvectors? You are solving an eigenvalue problem when you are solving the time independent Schrodinger equation.

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