Schrödinger's equation relation to quantum numbers

Question

I tried searching for an answer everywhere but the equation is either solved from the physical point of view or they just cite it in chemistry textbooks without even showing how it is made to talk about orbitals.

What I was curious about, is how, specifically, quantum numbers describing an electron/ orbitals are related to the Schrödinger's equation.

Moreover I saw that there are different types of it, time dependent/time independent, which of these are used for atoms?

Ps : I don't have any deep understanding about differential equations.

Answer 1

When solving a differential equation, such as the Schrödinger equation (or the Helmholtz equation), one would find that the equation has many distinct solutions. These solutions are distinguished by some index or quantity. In the context of quantum mechanics these indices or quantities are interpreted as quantum numbers.

One can consider the Helmholtz equation in free space (to avoid the complicated math associated with solving the Schrödinger equation for the atom). In Cartesian coordinates the solutions would be plane waves. These plane waves are distinguished by their propagation vectors. In the quantum context these propagation vectors can be interpreted as the ``quantum numbers'' of the plane waves. In cylindrical coordinates, on the other hand, the solutions are Bessel modes. They are distinguished (among other things) by integers representing the orders of the Bessel functions. These can also be interpreted as quantum numbers.

In an analogous way, one would have different solutions (sometimes called modes) for the Schrödinger equation applied to the atom. The quantities that distinguish the different solutions represent the quantum numbers for those solutions.

The choice of whether one uses the time-dependent or time-independent equation is determined by the specific problem under investigation. In the case of an atom that remains in a steady state, which is independent of time, one would use the time-independent Schrödinger equation.

Answer 2

The quantum numbers come from symmetries. This is very fundamental, and deeply mathematical so it will be difficult to describe without math, but I will try.

Quantum numbers are labels related to symmetries and conservation laws, so first one has to understand symmetry and conservation laws.

Let's start with trivial example: parity. Suppose the hamiltonian of the system describes a quantum well with potential $V(x)=kx^2$. The potential function is even. Time evolution will be according to TD-Scrödinger Eq. as $\partial_t \psi=-iH\psi$. Because H is even, if $\psi$ is even then $even*even=even$ and $\psi$ will always remain even. If $\psi$ is odd, $even*odd=odd$ and $\psi$ will always remain odd. Alas, we found that because of symmetry in Hamiltonian $V(x)=-V(x)$ we find that there is a conservation law (parity of the function is conserved during the evolution). And finally, we conclude that parity is a good quantum number. It makes sense to talk about even or odd functions. In practice, that quantum number is usually labeled as g and u.

In principle, for every symmetry, there will be a quantum number. If the symmetry is finite (say like in previous example) the quantum number has finite set of values. If the symmetry is continuous, the quantum number will have infinite number of values (either countable or uncountable). Let's take a few examples.

Atom

Atom has spherical symmetry. The total energy (Hamiltonian) of atom is invariant to arbitrary rotations. This creates the two quantum numbers l and m. Why 2 and what they mean is out of scope of the answer, but l=0, m=0 is easy to understand. It means fully spherical symmetric solution. It has the same symmetry as the Hamiltonian. But like in parity example for odd functions, also here, the full symmetry of the Hamilonian comserved also other l and m values.

One can probably already guess the conserved quanity because of rotation symmetry: angular momentum. Thus, l and m are quantum numbers describing angular momentum and they are conserved during time evolution of the system.

Let's dig deeper: The Hamiltonian of atom is also invariant to arbitrary time translations $H(t)=H(t+T)$. (In classical mechanics one says that Hamiltonian is the infinitesimal generator for time translation. In Schrödinger equation this is similar, Hamiltonian describes how things move forward in time) The time translation symmetry is trivially found, because Hamiltonian doesn't depend on time. This creates the principal quantum number n. Thus, the time translation symmetry of Hamiltonian predicted quantized stationary states with discrete quantum numbers. If hamiltonian would depend on time (say in electric field of a laser), stationary states wouln't exists. And what quantity does time translation invariant Hamiltonian conserve: of course energy! That is why one diagonalizes Hamilonian so often. To obtain a set of states which are stationary in time!

Free particle

Hamiltonian is both space and time translation invariant. This implies conservation of momentum and energy. This will yield quantum numbers k and $\omega$. Due to dispersion relation, relationship between k and $\omega$ is fixed, so momentum $\hbar k$ fully already describes the system.

Periodic system

In lattices, say Silicon bulk, there is lattice translation symmetry of the Hamiltonian. The quantum number coming out from this symmetry is the k-vector. It implies conservation of lattice momentum. That is why people are as interested in band structures as people are of energy levels of states in atoms. Both give the energy of states as a function of relevant quantum numbers.

Answer 3

Moreover I saw that there are different types of it, time dependent/time independent, which of these are used for atoms?

I am assuming you know what the potential $V(x)$ in the S. Equation is, its the pull of the nucleus on the electrons. Or, depending on how far you have covered, it could be even easier, a 1 dimensional square well.

If the potential is only dependent on $x $ , then use the time independent S. Equation, if the potential varies with time, as well as $x $, $V (x,t) $ then use time dependent S. Equation.

Go to this site Hyperphysics and read down the page, it's straightforward enough and it covers most of what you are studying .

What I was curious about, is how, specifically, quantum numbers describing an electron/ orbitals are related to the Schrödinger's equation.

When you solve the s.e. you should get a solution that the energy (n, the first quantum mumber ) is in, if you know the differential equation for the movement of a spring, you are halfway there. If you recognize $\sqrt {k/m }$, that is a big step towards the energy.

Another link to a YouTube video, and look around YouTube for Dr. Physics, I think he does this kinda stuff Quantum Numbers

Best of luck with it.

Answer 4

I know you said you don't have a deep knowledge of differential equations, but if you at least understand what they are, you can have a much better appreciation for what's going on here.

Essentially, the schrodinger equation in spherical polar coordinates for a single particle in couloumb potential (which is what you have in the niave picture of an atom) is the same as the Spherical Bessel Equation. The space of solutions of that equation is interpretable as the allowed wavefunctions of the system. So the quantum numbers n,l,m, are simply a way to index the solutions of that equation.

Just like the differential equation

$$ \frac{dx}{dy} = 0 $$

has the set of solutions

$$ x(y) = c $$

has a space of solutions that can be indexed by a line, namely the set of all c. The spherical Bessel equation has a space of solutions that can be indexed by those "quantum numbers."

Answer 5

I want to give a counterpoint to all those answers focussing on the Schrödinger equation being a "differential equation". The appearence of "quantum numbers" labelling states is fundamentally a phenomenon of linear algebra, not of the theory of differential equations as such, and happens even for finite-dimensional quantum systems where no description through differential equations is chosen.

The significance of quantum numbers can only be understood once you stop viewing the Schrödinger equation as the only object of interest in quantum mechanics, and start realizing that all quantum observables, not only the Hamiltonian appearing in the Schrödinger equation, are of relevance.

One of the goals when examining a quantum system is to "map" the space of states. The most straightforward attempt to do so is to solve the time-independent Schrödinger equation. Now the significant thing about this Schrödinger equation is not that it is a differential equation (that's just a consequence of how we chose to represent our operators), but that it is an eigenvalue equation for the Hamiltonian operator.

By the spectral theorem, the eigenvectors of a self-adjoint operator form a basis of the space of states (modulo some issues with unbounded operators). Finding an explicit basis is a very good way to "map" a vector space. But sometimes, these eigenvectors are degenerate, i.e. there's more than one eigenvector to a given eigenvalue, and also, the values may be "inconvenient" to label the stats with. Labelling all the eigenstates of the Hamiltonian with their energy works, but it's a bit messy, and may label some different states with the same energy.

So to arrive at the traditional quantum numbers of the hydrogen (and other) atoms, we choose a simpler complete set of commuting observables:

• The Hamiltonian without the angular momentum contribution. We label the lowest eigenvalue as the "first" ($n=1$) eigenvalue, and then simply count up. This is degenerate since there are states of different angular momentum with the same $n$.

• The orbital angular momentum operator, counting $\ell$ as multiples of $\hbar$. We start at $\ell = 0$, and it turns out that states with a fixed $n$ can only exist up to $\ell < n$, and that $\ell$ goes in integer steps. This is still degenerate because states can have the same total angular momentum but it can be oriented differently.

• So we pick a random spatial axis (usually then call the $z$-axis) and look at the operator $L_z$ measuring the component of angular momentum along that direction. For a state with total momentum $\ell$, the eigenvalues $m_z$ go in integer steps from $-\ell$ to $\ell$.

• Finally, the fundamentally quantum property of spin (see the first question ever asked on this site) means that there are two distinct states with the same $n,\ell,m_z$ labeled by $s_z= -1/2,1/2$.

Answer 6

In order to extract the eigenvalues, and resultingly the quantum numbers, from the Hamiltonian, one must apply boundary conditions to the system. This is extremely important, for that application is what distinguishes a physically possible system from all the other mathematical possibilities.

First, the functions which solve the SWE in a physically possible way must be square-integrable and finite. This limitation drastically reduces the set of possible solutions, and is the main reason the primary (energy) quantum number appears. The continuity, finiteness, and differentiability of the angular solutions is another important boundary condition, and that sets up the integer nature of the orbital ($m$ and $\ell$), and consequently, energy ($n$) quantum numbers.

Without the boundary conditions there are no quantized eigenstates.