# Quantum Bound States: the Hydrogen Atom

In quantum mechanics, bound states can be put together by choosing a potential energy and solving the a wave equation. The first non toy example that students learn is the hydrogen atom (or any other single electron atom) studied using Schroedginer’s wave equation. Our interest in this blog is more general bound state problems, but there is a lot we can learn by examining the hydrogen problem and its solutions.

The primary motivation for studying the hydrogen atom was to find an explanation for the light that was given off or absorbed when the atom switched from one energy state to another. For this reason, it was natural to look for the bound states as classified by their energies. That is, we will be looking for wave functions that correspond to sharp energies. If we wanted a more general solution, we can always combine different energy solutions by linear superposition.

The equation we wish to solve is $H\psi_n(x,y,z,t) = E_n\psi_n(x,y,z,t)$ where “H” is the quantum operator for energy (which is a sum of an operator for kinetic energy and one for potential energy), $E_n$ is the energy, $\psi_n$ is the wave function, and n =1, 2, 3, … is an index that distinguishes different energy solutions. It turns out the energy will be proportional to $n^{-2}$. Our solutions do not depend on time so we will leave off the t. And instead of writing (x,y,z), we will just write (x), or even leave it off completely like this: $H\psi = E\psi$.

The method one uses for solving Schroedinger’s wave equation is to write it in spherical coordinates $(r,\theta,\phi)$, and then separate the equation into three equations using separation of variables. In this method of solving differential equations, we guess that the solution can be written as a product of three contributions which depend on $r, \theta$, and $\phi$, but not on combinations of these. This is a very restrictive assumption. It works handily because the energy will depend only on r.

Energy is generally not enough to specify a solution to the equation. The separation of variables tells us this. To completely choose a solution, we can look for other operators that commute with H and each other. In quantum mechanics, when we find as many such operators as we can, they are called a “complete set of commuting operators.” The usual choice to complete H is the squared total angular momentum $L^2$, and the angular momentum in the z-direction $L_z$. These three operators, $H, L^2, L_z$, commute and completely specify the solutions to the wave equation.

The eigenvalues of $L^2$ are numbered with l = 0, 1, 2, …, n, and are proportional to l(l+1)\$. And the eigenvalues of $L_z$ are numbered with m = -l, -l+1, …, l-1, l, and are proportional to m. So a completely specified solution to the equation is given by $\psi_{nlm}$ and leaving off the unimportant proportionality constants (don’t do this at home), we have: Since these operators are Hermitian or self adjoint, we can take the complex conjugate of the above equations to get: Orthogonality

Given any operator Q and any wave function $\psi$, the expectation value of the operator is given by the usual quantum mechanics formula: In the above, I’ve written “ $\pm \infty$” as the limits of integration. By that I mean all of space. All of our integrals are going to be over all space. In the rest of this post I’ll leave off the limits but all the integrals in this post are definite integrals over all space.

Since the wave function $\psi_{nlm}$ is an eigenstate of the three operators $E, L^2, L_z$, if we replace Q with any of these three operators, the expectation value will be, respectively, -1/n^2, l(l+1), or m. Furthermore, if we consider two different $\psi_{nlm}$, then they will differ in at least one of these three expectation values.

The expectation value of the constant 1 is = 1, so we have: Since the squared magnitude of the wave function gives the probability density for the electron, the above equation also corresponds to computing the probability of the electron is anywhere.

These equations for expectation values have all involved two copies of the wave function. As functions of the wave function, they are quadratic. We can consider what happens to the expectation computation when we use two arbitrary wave functions in the two places where it is used. This makes the expectation function into a bilinear function of the two wave functions.

If the two arbitrary wave functions happen to be the same one, we’ve got the situation where we’re computing the expectation value of the operator 1. This gives 1. If the two wave functions are different (and are from our complete set of eigenstates), then they have to have different eigenvalues for at least one of the three operators. The three operators all define unique eigenvalues, that is two different quantum numbers imply two different eigenvalues. Without loss of generality, let’s suppose that they have different energy eigenvalues. This gives two different results for this mixed expectation value equation. We have two different ways of making the calculation: The last two lines are obtained just by moving the parentheses around so they are the same. But if $E_n \neq E_n'$ they would be different, unless the two states are orthogonal.

We’ve just shown that the bound states are orthonormal. Furthermore, we didn’t use any particular details about the hydrogen atom. The result is perfectly general. A complete set of bound states are orthogonal to each other. Note that this does not say that any two different arbitrary bound states must be orthogonal. They are orthogonal here because we’ve taken them from the same complete basis set. We could take two bound states, from two different basis sets, (for instance they might have the same energy but have angular momentum in slightly different directions), and they would not have to give zero. But so long as we choose eigenstates created by the same complete basis set, that is, with a complete set of commuting operators, they will be orthonormal.

Density Matrix Formalism

The orthogonality calculation was done by switching from the quadratic version of the expectation value to a bilinear version. We took the two places where one inserts the wave function and replaced them with two different wave functions. But those two wave functions shared the same position variable. To get a density matrix form for the wave function, we sort of do the reverse. Instead of using two different wave functions, we will use the same wave function, but we will give it two different position variables. And we also apply the complex conjugate to the right hand wave function instead of the left: The reason for switching the complex conjugate around is to allow us to do density matrix expectation value calculations without having to commute the wave functions. While these are just complex functions, when we add spin to the mix they will become 2×1 spinors and will not commute. But for now, we leave them in this order so that it makes the calculations sweet. The above density matrix form is not general. It is the form for a pure density matrix, that is, one that can be created from a state vector (wave function) which is what we just did.

In the density matrix formalism, the assumption is made that the density matrices are the fundamental quantum object. To obtain a wave function from a density matrix, first find a pair of positions (u,u’) where $\rho(u,u') \neq 0$. Then set $\psi(x) = \rho(x,u')$, and normalize if desired.

The orthogonality relation of the wave functions is a little different in the pure density matrices: In both cases the computation gives zero for different states. But while the computation gives 1 for the wave functions, in density matrix form it gives back the density matrix.

The above suggests that we define a multiplication on density matrices. Given two density matrices, define the product as the above. This means that we won’t bother keeping track of the independent variables x and x’. Instead, we know where they go, and we could put any arbitrary variables in there. So in this notation, the orthogonality relation becomes: The notation works because there really isn’t any other reasonable way to interpret the product of two density matrices. And the notation is compatible with the notation for matrix multiplication so it will be quite natural when we move to the matrix version of density matrices.

We can also insert an operator between two copies of a density matrix: For a pure density matrix as defined above, the above is just the expectation value of Q multiplied by the density matrix. The multiplication by the density matrix is reminiscent of the orthogonality relation which gave back the density matrix and this occurs for the same reason. So it is natural to write it in multiplication form as: With wave functions we wrote a formula for the expectation value but in writing the simplest way of doing the same thing with density matrices we end up with an answer given by the product of the expectation value with the density matrix itself. Of course this is no big deal. We could pick any nonzero place on the density matrix and use that value to factor out the density matrix from the expectation value. What is more commonly done, however, is to define a trace “tr” function linear on the density matrices so that $tr(\rho) = 1$. For matrices, this is just the trace function. For the density matrices, the rule would be to integrate over the coordinates: With this form, we can get rid of one of the density matrices (by relying on the orthonormality of the pure density matrices) and write: The above form also works for mixed density matrices so it is the form you will see used almost universally.

Bound States as Matrices

The solutions to the hydrogen bound state problem could be characterized by the requirements that their density matrices are unchanged when they are squared: $\;\rho\;\rho = \rho$ (which mathematicians call “idempotency” ), and their traces are 1: $\;tr(\rho) = 1$ (which mathematicians say make them “primitive” idempotents). Solutions with different quantum numbers multiplied to zero (mathematicians would call such a set “annihilating primitive idempotents” ).

We can apply these same requirements on the matrices and see what we come up with. Let’s start easy with the 2×2 matrices. The idempotency equation gives: The above gives four quadratic equations in a, b, c, and d; the equations for a and d are on the diagonal, the equations for b and c are off diagonal. The trace = 1 requirement means that a+d=1. That solves the two off diagonal equations leaving only the two for a and d. These can be solved in terms of the product bc by the quadratic equation to give the general solution as: The signs on the b and c are chosen so that the two choices of signs, + and -, give two solutions that annihilate each other as the reader can quickly verify. Note that the two solutions also add to unity.

The above matrices may not seem familiar, but if we require that b and c be Hermitian conjugates $b = c^*$, and small enough that $4bc = 4|b|^2 \leq 1$, we find that the above matrices are the density matrices for pure spin-1/2 states. In particular, putting b=c=0 gives the density matrices (projection operators) for spin in the z-direction. Putting b=c=1/2 gives spin in the x-direction while b=i/2, and c=-i/2 gives spin in the y-direction.

So, is this evidence that spin is the result of a bound state? Well, complete sets of operators always do this sort of thing. Instead, the process of moving from a general wave function to a matrix that reduces the number of degrees of freedom certainly reminds one of the way that spin-1/2 was discovered as a generalization of angular momentum. It would not be unexpected to find that bound states, especially deeply bound states where the particles have no positional freedom, should have a small matrix representation.

In fact, Koide’s mass formula for the charged leptons was extended to the neutrino case using primitive idempotent 3-matrices. If one were to interpret this as evidence for the leptons being (single particle) bound states, we would be looking at a particle that was restricted to being found in one of three possible positions (or orientations, which is where the Pauli MUB comes about).

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### 8 responses to “Quantum Bound States: the Hydrogen Atom”

1. carlbrannen

I guess I should add that part of the reason for writing this up is that I’m extending my book on density matrix formalism to handling the case of wave functions. So I’m going to be putting little chunks of writing up.

2. strangerep

Bound states of the Hydrogen atom are
more elegantly handled using representation
theory of SO(4,2). See, e.g., Wybourne’s
book.

3. carlbrannen

Yes, that is much more elegant, at least in some ways, but I’m writing this for an introductory book on quantum mechanics and want to make it accessible at the junior level.

4. Everything is energy

please look at http://www.everythingisenergy.info as this is the key law

5. Vladimir Kalitvianski

Even if the atom is in a pure state, the electron and the nucleus are always in mixed states, see “On Probing Small Distances …” at http://groups.google.com/group/qed-reformulation.

6. Carl Brannen

Vladimir, I think I agree completely with your ideas for eliminating the point particles and this does imply that the electron and nucleus must be in mixed states. And of course this is completely compatible with the concept that the bound object can be in a pure state so I see our work as being very compatible. When I get around to publishing this I plan on including citations to your work as it so vividly illustrates an alternative way of looking at the same thing.

7. Vladimir Kalitvianski

Thanks, Carl, but getting a citation is not my purpose. It is not my invention that a couple of interacting particles (r1,r2) can be describes with two decoupled equations for quasi-particles (R,r) in pure states, and at the same time each particle depends (being in interaction) on the other one so the wave function expressed via personal particle variables r1 and r2 is not factorized. Thus each particle is in a mixed sate.