5.2: The Ket Vector

To represent the state of a quantum particle, or a quantum system, we introduce the “ket vector”

\(\ |\psi\rangle\)

This is just an abstract mathematical notation, a compact way of saying “the state of this particle”. The name “ket” is the latter half of the word “braket”, a misspelling of bracket. We will later learn about “bra” vectors, which are written as \(\ \langle\psi|\). However, don’t worry about those for now; let’s focus on Schrödinger's kets.

Why the \(\ \psi\) inside the ket vector? It’s traditional to use the Greek letter “psi” for the state of a system. However, you could put anything you want inside the vertical bar and right angle bracket. It’s similar to the convention of using \(\ x\) as the variable for the horizontal axis of a graph in algebra or geometry; people do it a lot, but you can use any letter you want. Sometimes, we will use other Greek letters. Sometimes, we will use something that gives useful information about just what this state is. However, even if we just use something that doesn’t tell you anything, like \(\ \psi\), remember that this ket vector is just a way of representing the state so that we can talk about it, so that we can get a handle on it, and so that we can perform mathematical operations with it.

Why do we call it a vector? This is potentially a source of confusion. This is not a vector in there-dimensional space, the way angular momentum, velocity, momentum, or displacement are. In fact, it’s a vector in an abstract mathematical space called a “Hilbert space”. However, for now, don’t worry about that. We will see later the ways in which the state vector behave sort of like three-dimensional vectors like velocity. For now, take it as an idiom that when we talk about the “state vector”, we’re just talking about a mathematical representation of the state of a quantum particle or a quantum system.

Consider, for example, the sequence of Stern-Gerlach machines shown below:

Consider an electron going into the first SGz machine. If its angular momentum can be oriented in any direction, we would say that the electron is “unpolarized”. We don’t know the state of the electron, so we will just pick a name for it, and call it \(\ |\psi\rangle\).

Now consider an electron coming out of the first SGz machine. If it is measured to have a \(\ z\) spin of \(\ +\hbar / 2\), then it will come out of the + output. At that point, we know the \(\ z\) angular momentum of the electron, so let’s wisely choose a representation for the state that will make it easy for us to remember: \(\ |+z\rangle\). Note that the “\(\ +z\)” inside the notation doesn’t mean anything about adding any variable named \(\ z\), nor have we defined a variable \(\ z\). It’s just a name. I could just as easily have named the state vector \(\ \text { |Fred}\rangle\). That would have allowed us to carry it around in equations, talk about it, and perform mathematical operations with it. However, for us humans reading the equations, it’s convenient of the name is something that reminds us what we know about the state. So, we’ll choose \(\ |+z\rangle\) as our name so that we remember that, aha, this is an electron whose \(\ z\) spin is known to be positive.

Similarly, an electron coming out of the − output from the SGz machine will be in the state \(\ |-z\rangle\). Again, there’s no subtraction, or any multiplying by negative one going on here. It’s just a name, the same way \(\ x\) is just a name for a (possibly unknown) variable in algebra.

Now move on to the second SG machine, the SGx machine. An electron going into this machine is in state \(\ |+z\rangle\); we know that, because all of these electrons are coming out of the + output of an SGz machine. However, as this electron goes through the SGx machine, its state changes. If it comes out the + output of the SGx machine, it’s in a state we shall choose to call \(\ |+x\rangle\). If it comes out of the − output of the SGx machine, it’s in a state we shall choose to call \(\ |-x\rangle\).

Remember what happened in the previous chapter when we then took an electron in state \(\ |+x\rangle\) — that is, an electron whose \(\ x\) spin was measured to be positive— and put it back into a second SGz machine. It had a 50% chance of being measured with positive \(\ z\) spin, and a 50% chance of being measured with a negative \(\ z\) spin. In other words, the electron is no longer in state \(\ |+z\rangle\), nor is it in state \(\ |-z\rangle\); the state \(\ |+x\rangle\) is different from both of those \(\ z\) states.

It turns out, however, that you can describe the state \(\ |+x\rangle\) in terms of the states \(\ |+z\rangle\) and \(\ |-z\rangle\). Remember the algebraic equation \(\ 2 x+y=b\), which allowed us to figure out that \(\ x=(b-y) / 2\). The variables \(\ b\) and \(\ y\) are abstract representations of numbers, and \(\ x\) is an abstract representation of another number. The equation \(\ x=(b-y) / 2\) tells us that the thing that \(\ x\) represents is not independent from \(\ b\) and \(\ y\); were \(\ b\) or \(\ y\) to change, \(\ x\) would have to change along with it. It also tells us how to figure out \(\ x\) in terms of \(\ b\) and \(\ y\). If we have rules for doing things to \(\ b\) and \(\ y\), we can then apply those rules to the right side of the equation to figure out how \(\ x\) changes.

Bearing that in mind, it turns out that you can represent the \(\ x\) spin states of an electron in terms of the \(\ z\) spin states as follows:

\(\ \begin{aligned}
|+x\rangle &=\frac{1}{\sqrt{2}}|+z\rangle+\frac{1}{\sqrt{2}}|-z\rangle \\
|-x\rangle &=\frac{1}{\sqrt{2}}|+z\rangle-\frac{1}{\sqrt{2}}|-z\rangle
\end{aligned}\)

In a future chapter, we will see why this seemingly odd combination, with all of its square roots of two, makes sense. We will also see why being able to represent the \(\ x\) states in terms of the \(\ z\) states is useful. For now, however, recognize this as the first rule about performing mathematical operations on these state vectors: you can multiply a state vector by a number. Don’t worry about how you would actually calculate something from that. In algebra, you can write down \(\ x / 2\), it has meaning even if you don’t know \(\ x\) and can’t calculate a number for \(\ x / 2\). Or, if you’re doing algebra with 3-vectors in space, you could write down \(\ \vec{v}_{1}=\vec{v}_{2} / 2\). Even if you don’t have numbers for all three components of \(\ \vec{v}_{2}\), and thus can’t calculate numbers for all three components of

\(\ \vec{v}_{1}\), this equation is still meaningful. Similarly, it’s a valid mathematical operation to multiply a ket vector by a number. For now, we’ll just leave it written out as that number multiplied by the ket vector, and bear in mind that when you multiply a number (a complex number— it doesn’t have to be real!) by a ket vector, you get another ket vector as a result.

You can also add two ket vectors together, and the result of that addition is yet a third ket vector. Here, we see that particular combinations of constants times \(\ |+z\rangle\) and \(\ |-z\rangle\) turn out to be equal to \(\ |+x\rangle\). This is a mathematical expression, within the theory of quantum mechanics, that represents a truth about reality. It’s very similar to the Pythagorean Theorem, \(\ a^{2}+b^{2}=c^{2}\), which expresses a truth about triangles (if \(\ c\) is a representation of the length of the hypotenuse of the triangle, and \(\ a\) and \(\ b\) are representations of the lengths of the legs of the triangle).

Let’s go through the next step of our sequence of devices above. After the SGx machine, the two beams (one in state \(\ |+x\rangle\) and one in state \(\ |-x\rangle\)) are recombined together into a single beam. Let us pretend that we don’t know what the state of the electron coming out of the recombining apparatus is. Just as in algebra, when we have a quantity we don’t know, we’ll give this state a name; let’s call it \(\ \left|\psi_{\mathrm{RC}}\right\rangle\), with “RC” standing for “recombined”. If we knew the all of the rules for figuring out how quantum states evolve as they pass through these SG machines, we could figure out what this state is by performing calculations on the previous states, based on where the beams have been. However, we don’t yet know these rules. Instead, what we do is perform one more thought experiment: we send this electron, in state \(\ \left|\psi_{\mathrm{RC}}\right\rangle\), through an SGz machine. We discover that 100% of the time, the electron comes out of the + output of the SGz machine. From this experiment, we’ve figured out that

\(\ \left|\psi_{\mathrm{RC}}\right\rangle=|+z\rangle\)

In upcoming chapters, we’ll learn how to figure out theoretically that this is the state of that electron.