Quantum Physics Articles

Article / Updated 09-22-2022

Each quantum state of the hydrogen atom is specified with three quantum numbers: n (the principal quantum number), l (the angular momentum quantum number of the electron), and m (the z component of the electron’s angular momentum, How many of these states have the same energy? In other words, what’s the energy degeneracy of the hydrogen atom in terms of the quantum numbers n, l, and m? Well, the actual energy is just dependent on n, as you see in the following equation: That means the E is independent of l and m. So how many states, |n, l, m>, have the same energy for a particular value of n? Well, for a particular value of n, l can range from zero to n – 1. And each l can have different values of m, so the total degeneracy is The degeneracy in m is the number of states with different values of m that have the same value of l. For any particular value of l, you can have m values of –l, –l + 1, ..., 0, ..., l – 1, l. And that’s (2l + 1) possible m states for a particular value of l. So you can plug in (2l + 1) for the degeneracy in m: And this series works out to be just n2. So the degeneracy of the energy levels of the hydrogen atom is n2. For example, the ground state, n = 1, has degeneracy = n2 = 1 (which makes sense because l, and therefore m, can only equal zero for this state). For n = 2, you have a degeneracy of 4: Cool.

Cheat Sheet / Updated 02-14-2022

In dabbling in quantum physics, you come across spin operators, commutation relationships, and many formulae and principles. You also learn about various effects named for people, such as the Hamiltonian, the Heisenberg Uncertainty Principle, the Schrödinger Equation, and the Compton Effect. This Cheat Sheet provides a quick reference to some of the main equations used in quantum physics.

Article / Updated 03-26-2016

The Schrödinger equation is one of the most basic formulas of quantum physics. With the Schrödinger equation, you can solve for the wave functions of particles, and that allows you to say everything you can about the particle — where it is, what its momentum is, and so on. In the following version of the Schrödinger equation, the first term represents the kinetic energy and the second term represents the potential energy: where

Article / Updated 03-26-2016

Don’t think quantum physics is devoid of anything but dry science. The fact is that it’s full of relationships, they’re just commutation relationships — which are pretty dry science after all. In any case, among the angular momentum operators Lx, Ly, and Lz, are these commutation relations: All the orbital angular momentum operators, such as Lx, Ly, and Lz, have analogous spin operators: Sx, Sy, and Sz. And the commutation relations work the same way for spin:

Article / Updated 03-26-2016

At some point, your quantum physics instructor may ask you to find the eigenfunctions of Lz in spherical coordinates. In spherical coordinates, the Lz operator looks like this: which is the following: And because this equation can be written in this version: Cancelling out terms from the two sides of this equation gives you this differential equation: This looks easy to solve, and the solution is just where C is a constant of integration. You can determine C by insisting that be normalized — that is, that the following hold true: (Remember that the asterisk symbol [*] means the complex conjugate. A complex conjugate flips the sign connecting the real and imaginary parts of a complex number.) So this gives you You are now able to determine the form of which equals

Article / Updated 03-26-2016

In quantum physics, you can find the eigenvalues of the raising and lowering angular momentum operators, which raise and lower a state’s z component of angular momentum. Start by taking a look at L+, and plan to solve for c: L+| l, m > = c | l, m + 1 > So L+ | l, m > gives you a new state, and multiplying that new state by its transpose should give you c2: To see this equation, note that On the other hand, also note that so you have What do you do about L+ L–? Well, you assume that the following is true: So your equation becomes the following: Great! That means that c is equal to So what is Applying the L2 and Lz operators gives you this value for c: And that’s the eigenvalue of L+, which means you have this relation: Similarly, you can show that L– gives you the following:

Article / Updated 03-26-2016

Because spin is a type of built-in angular momentum, spin operators have a lot in common with orbital angular momentum operators. As your quantum physics instructor will tell you, there are analogous spin operators, S2 and Sz, to orbital angular momentum operators L2 and Lz. However, these operators are just operators; they don’t have a differential form like the orbital angular momentum operators do. In fact, all the orbital angular momentum operators, such as Lx, Ly, and Lz, have analogs here: Sx, Sy, and Sz. The commutation relations among Lx, Ly, and Lz are the following: And they work the same way for spin: The L2 operator gives you the following result when you apply it to an orbital angular momentum eigenstate: And just as you’d expect, the S2 operator works in an analogous fashion: The Lz operator gives you this result when you apply it to an orbital angular momentum eigenstate: And by analogy, the Sz operator works this way: What about the raising and lowering operators, L+ and L–? Are there analogs for spin? In angular momentum terms, L+ and L– work like this: There are spin raising and lowering operators as well, S+ and S–, and they work like this:

Article / Updated 03-26-2016

When you have the eigenvalues of angular momentum states in quantum mechanics, you can solve the Hamiltonian and get the allowed energy levels of an object with angular momentum. The eigenvalues of the angular momentum are the possible values the angular momentum can take. Here’s how to derive eigenstate equations with Note that L2 – Lz2 = Lx2 + Ly2, which is a positive number, so That means that And substituting in and using the fact that the eigenstates are normalized, gives you this: So there’s a maximum possible value of which you can call You can be clever now, because there has to be a state such that you can’t raise any more. Thus, if you apply the raising operator, you get zero: Applying the lowering operator to this also gives you zero: And because that means the following is true: Putting in gives you this: At this point, it’s usual to rename You can say even more. In addition to a there must also be a such that when you apply the lowering operator, L–, you get zero, because you can’t go any lower than And you can apply L+ on this as well: From you know that which gives you the following: And comparing this equation to gives you Note that because you reach by n successive applications of you get the following: Coupling these two equations gives you Therefore, can be either an integer or half an integer (depending on whether n is even or odd). Because and n is a positive number, you can find that So now you have it: The eigenstates are | l, m >. The quantum number of the total angular momentum is l. The quantum number of the angular momentum along the z axis is m. For each l, there are 2l + 1 values of m. For example, if l = 2, then m can equal –2, –1, 0, 1, or 2. You can see a representative L and Lz in the figure. L and Lz. L is the total angular momentum and Lz is the projection of that total angular momentum on the z axis.

Article / Updated 03-26-2016

In quantum physics, when you look at the spin eigenstates and operators for particles of spin 1/2 in terms of matrices, there are only two possible states, spin up and spin down. The eigenvalues of the S2 operator are and the eigenvalues of the Sz operator are You can represent these two equations graphically as shown in the following figure, where the two spin states have different projections along the z axis. Spin magnitude and z projection. In the case of spin 1/2 matrices, you first represent the eigenstate like this: And the eigenstate looks like this: Now what about spin operators like S2? The S2 operator looks like this in matrix terms: And this works out to be the following: Similarly, you can represent the Sz operator this way: This works out to Using the matrix version of Sz, for example, you can find the z component of the spin of, say, the eigenstate Finding the z component looks like this: Putting this in matrix terms gives you this matrix product: Here’s what you get by performing the matrix multiplication: And putting this back into ket notation, you get the following: How about the raising and lowering operators S+ and S–? The S+ operator looks like this: And the lowering operator looks like this: Here it is in matrix terms: Performing the multiplication gives you this: Or in ket form, it’s Cool.

Article / Updated 03-26-2016

In quantum physics, you can break the three-dimensional Schrödinger equation into three one-dimensional Schrödinger equations to make it easier to solve 3D problems. In one dimension, the time-dependent Schrödinger equation (which lets you find a wave function) looks like this: And you can generalize that into three dimensions like this: Using the Laplacian operator, you can recast this into a more compact form. Here’s what the Laplacian looks like: And here’s the 3D Schrödinger equation using the Laplacian: To solve this equation, when the potential doesn’t vary with time, break out the time-dependent part of the wave function: Here, is the solution of the time-independent Schrödinger equation, and E is the energy: So far, so good. But now you’ve run into a wall — the expression is in general very hard to deal with, so the current equation is in general very hard to solve. So what should you do? Well, you can focus on the case in which the equation is separable — that is, where you can separate out the x, y, and z dependence and find the solution in each dimension separately. In other words, in separable cases, the potential, V(x, y, z), is actually the sum of the x, y, and z potentials: V(x, y, z) = Vx(x) + Vy(y) + Vz(z) Now you can break the Hamiltonian in into three Hamilitonians, Hx, Hy, and Hz: where When you divide up the Hamiltonian as in you can also divide up the wave function that solves that equation. In particular, you can break the wave function into three parts, one for x, y, and z: Where X(x), Y(y), and Z(z) are functions of the coordinates x, y, and z and are not to be confused with the position operators. This separation of the wave function into three parts is going to make life considerably easier, because now you can break the Hamiltonian up into three separate operators added together: E = Ex + Ey + Ez So you now have three independent Schrödinger equations for the three dimensions: This system of independent differential equations looks a lot easier to solve than In essence, you’ve broken the three-dimensional Schrödinger equation into three one-dimensional Schrödinger equations. That makes solving 3D problems tractable.