Quantum Physics Articles

Cheat Sheet / Updated 05-10-2024

This Cheat Sheet is intended to supplement Quantum Physics For Dummies, 3rd edition, by Andrew Zimmerman Jones. It begins by reviewing some useful operators used in quantum mechanics calculations. Then it covers a useful method for solving the Schrödinger equation for the quantum wave function, and then how you can use that wave function to calculate probabilities in quantum physics. Finally, it lays out some key moments in one of the most important discoveries of quantum physics: the understanding of wave-particle duality.

Article / Updated 10-06-2022

At some point, your quantum physics instructor may want you to find the x, y, and z equations for three-dimensional free particle problems. Take a look at the x equation for the free particle, You can write its general solution as where Ay and Az are constants. Because you get this for where A= Ax Ay Az. The part in the parentheses in the exponent is the dot product of the vectors That is, if the vector a = (ax, ay, az) in terms of components and the vector b = (bx, by, bz), then the dot product of a and b is So here’s how you can rewrite the

Article / Updated 10-06-2022

At some point, your quantum physics instructor may want you to add time dependence and get a physical equation for a three-dimensional free particle problem. You can add time dependence to the solution for if you remember that, for a free particle, That equation gives you this form for Because the equation turns into In fact, now that the right side of the equation is in terms of the radius vector r, you can make the left side match: That’s the solution to the Schrödinger equation, but it’s unphysical. Why? Trying to normalize this equation in three dimensions, for example, gives you the following, where A is a constant: (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. The limits on the integral just mean to integrate over all of space, like this: Thus, the integral diverges and you can’t normalize as written here. So what do you do here to get a physical particle? The key to solving this problem is realizing that if you have a number of solutions to the Schrödinger equation, then any linear combination of those solutions is also a solution. In other words, you add various wave functions together so that you get a wave packet, which is a collection of wave functions of the form such that The wave functions interfere constructively at one location. They interfere destructively (go to zero) at all other locations. Look at the time-independent version: However, for a free particle, the energy states are not separated into distinct bands; the possible energies are continuous, so people write this summation as an integral: So what is It’s the three-dimensional analog of That is, it’s the amplitude of each component wave function. You can find from the Fourier transform of like this: In practice, you choose yourself. Look at an example, using the following form for which is for a Gaussian wave packet (Note: The exponential part is what makes this a Gaussian wave form): where a and A are constants. You can begin by normalizing to determine what A is. Here’s how that works: Okay. Performing the integral gives you which means that the wave function is You can evaluate this equation to give you the following, which is what the time-independent wave function for a Gaussian wave packet looks like in 3D:

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.

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.