4 
Entanglement

In this chapter we study the mathematics of entanglement. To do this, we need to introduce one more idea from linear algebra: the tensor product. We start by looking at two systems with no interaction between them. Since there is no interaction, we could study each system by itself, without any reference to the other system, but we will show how we can combine the two systems using tensor products. Then we introduce the tensor product of two vector spaces and show that most of the vectors in this product represent what are referred to as entangled states.

Throughout this chapter there will be two qubits. Alice has one, and Bob has another. We will begin our study by examining a case where there is no interaction between Alice’s and Bob’s systems. This analysis initially might seem to make something that is very simple look somewhat complicated, but once we have described everything in terms of tensor products it becomes fairly straightforward to extend the underlying ideas to the general entangled case.

The approach we take, however, is not the approach we have taken so far. Instead of presenting physical experiments and then deriving a mathematical model, we proceed in the other direction. We will extend our model in the simplest way possible and then see what the model predicts should be found when we perform experiments. We find that the model predicts the experiments accurately, but the conclusions are quite surprising.

Alice and Bob’s Qubits Are Not Entangled

We suppose that Alice is measuring using the orthonormal basis and Bob is measuring with orthonormal basis . A typical qubit for Alice is , and for Bob is . We can combine these two state vectors using a new type of product that we call a tensor product, giving us a new vector denoted by .

Now . How do we multiply these two terms using this new product? Well, we do it the most natural way possible. We expand it in the usual way we multiply out algebraic expressions of the form . We write

If you are familiar with the FOIL method, you should recognize that this is exactly what we have done. To make the terminology even simpler we will use juxtaposition of two kets to mean the tensor product, so will be denoted as .

Though this is just the standard way of multiplying out two expressions, there is one thing that we have to be very careful about: The first ket in the tensor product belongs to Alice, and the second ket belongs to Bob. For example, means that belongs to Alice and that belongs to Bob. The product means that belongs to Alice and that belongs to Bob. So, in general, will not equal . The technical term for this is to say the tensor product is not commutative.

Alice is measuring with her orthonormal basis and Bob is measuring with his orthonormal basis . We are describing both Alice’s and Bob’s qubits using tensor notation. This description involves the four tensor products that come from the basis vectors: , , , and . These four products form an orthonormal basis for the tensor product of Alice’s and Bob’s systems: Each of these products is a unit vector, and they are orthogonal to one another.

At this stage, though we have introduced new notation, we have not introduced anything new in terms of concepts. It is just the information that we already knew, but in a different package. For example, the number is a probability amplitude. Its square gives the probability that when both Alice and Bob measure their qubits Alice’s qubit jumps to , that is, she reads 0, and Bob’s qubit jumps to , that is, he reads 0. But we already knew that the probability that Alice’s qubit would jump to is and the probability that Bob’s would jump to is . So we really knew that the probability that both would occur is , which is, of course, the same as . In a similar way, the numbers , , and give the probabilities that Alice and Bob read 01, 10, and 11, respectively. (Remember that Alice’s bit is always listed before Bob’s.)

Next we will replace these probability amplitudes using just one symbol instead of two. We will let , , , and , so . We know that , because they are probability amplitudes. We also know that ru = st, because both ru and st equal Now we come to the new idea. We are going to describe the states of Alice’s and Bob’s qubits by tensors of the form . Again we stipulate that , so that we can treat r, s, t, and u as probability amplitudes. But we no longer insist that ru = st. We allow any values of r, s, t, and u just as long as the sum of their squares is 1.

Given a tensor of the form with there are two cases. The first case is when ru = st. In this case we say that Alice’s and Bob’s qubits are not entangled. The second case is when . In this case we say that Alice’s and Bob’s qubits are entangled. This rule is easy to remember if the terms are written out with the subscripts in the order we have presented them: 00, 01, 10, 11. In this order, ru are the outer terms and st are the inner, so the qubits are not entangled if the product of the outer terms is equal to the product of the inner ones, and they are entangled if the products are not equal.

We will look at examples that illustrate both of these cases.

Unentangled Qubits Calculation

Suppose we are told that Alice and Bob’s qubits are given by

.

We quickly calculate the products of the outer and inner probability amplitudes. Both products equal , so the qubits are unentangled.

The probability amplitudes tell us what happens when Alice and Bob both make measurements. They will get 00 with probability 1/8, 01 with probability 3/8, 10 with probability 1/8 and 11 with probability 3/8.

What is slightly trickier is to see what happens if just one of them makes a measurement. We start by assuming that Alice is going to make a measurement, but Bob is not. To do this we begin by pulling out common factors from Alice’s perspective. We rewrite the tensor product as

.

Next, we want the expressions in the parentheses to be unit vectors, so we divide by their lengths inside the parentheses and multiply by their lengths outside, which gives us

.

We can then pull out the common factor in the parentheses. (But remember that it’s Bob’s, so we must keep it on the right.)

.

Written in this form, it becomes explicit that the states are not entangled. We have a tensor product of a qubit that belongs to Alice with a qubit that belongs to Bob.

From this we can deduce that if Alice measures first she will obtain 0 and 1 with equal probability. This measurement has no effect on the state of Bob’s qubit. It was and remains

.

We can also read from the factored expression that if Bob measures first he will get 0 with probability 1/4 and 1 with probability 3/4. Again, it is clear that Bob’s measurement has no effect on Alice’s qubit.

When the qubits are unentangled, a measurement of one of the qubits has absolutely no effect on the other qubit. The situation is completely different with entangled qubits. If qubits are entangled, the measurement of one will have an effect on the other one. We will illustrate this with an example.

Entangled Qubits Calculation

Suppose we are told that Alice’s and Bob’s qubits are given by

.

We quickly calculate the products of the outer and inner probability. The product of the outer terms is 0. Since the product of the inner terms is not equal to 0, the two qubits are entangled.

Usually both Alice and Bob will make measurements. As in the previous example, we can use the probability amplitudes to tell us what happens when Alice and Bob both measure their qubits. They will get 00 with probability 1/4, 01 with probability 1/4, 10 with probability 1/2, and 11 with probability 0. Notice that there is nothing strange going on. This is exactly the same calculation as in the unentangled case.

We will now see what happens if just one of them makes a measurement. We start by assuming that Alice is going to make a measurement, but Bob is not. To do this we begin by pulling out common factors from Alice’s perspective. We rewrite the tensor product as

.

As before, we want the expressions in the parentheses to be unit vectors, so we divide by their lengths inside the parentheses and multiply by their lengths outside, which gives us

.

In the previous example, the terms in the parentheses were the same, and we could pull this common term out as a common factor. But in this case the terms in the parentheses are different. This is what it means to be entangled.

The probability amplitudes in front of Alice’s kets tell us that when she measures she will get 0 and 1 with equal probability. But when Alice gets 0, her qubit jumps to . The combined system jumps to the unentangled state , and Bob’s qubit is no longer entangled with Alice’s. It is . When Alice gets 1, again Bob’s qubit is no longer entangled with Alice’s. It becomes .

The result of Alice’s measurement affects Bob’s qubit. If she gets 0, Bob’s qubit becomes . If she gets 1, his qubit becomes . This does seem strange. Alice and Bob could be far apart. As soon as she makes a measurement Bob’s qubit becomes unentangled, but exactly what it is depends on Alice’s outcome.

For completeness, we will see what happens when Bob measures first.

We start with the initial tensor product.

Rewriting from Bob’s perspective gives

.

As always, we want the expressions in the parentheses to be unit vectors, so we divide by their lengths inside the parentheses and multiply by their lengths outside, which gives us

.

When Bob measures his qubit he gets 0 with probability 3/4 and 1 with probability 1/4. When Bob gets 0, Alice’s qubit jumps to state . When he gets 1, her qubit becomes .

When the first person measures her or his qubit, the second person’s qubit immediately jumps to one of two states. These states depend on the result of the first person’s measurement. This is quite unlike our everyday experience. Later we will see clever ways of exploiting entangled qubits, but first we consider superluminal communication.

Superluminal Communication

Superluminal communication is communication faster than the speed of light. Two apparently contradictory inferences seem to be able to be deduced concerning this. The first is that Einstein’s special theory of relativity tells us that as you travel faster, approaching the speed of light, time slows down. If you could travel at the speed of light, time stops. And if you travel faster than the speed of light, time would go backward. The theory also tells us that as we approach the speed of light our mass increases without bound, which means that we can never reach that speed. Also, it seems unlikely that we could go back in time. If we could, then we run into all the science fiction scenarios in which we can prevent some history-changing event from happening. Time travel seems to lead to contradictions. It’s not just physical travel that seems to be ruled out, but also communication. If we could send messages back in time, we could still change the course of history—we could still design scenarios in which the communication causes some dramatic change in the present—for example, prevents us being born. So, one of our immediate thoughts is that superluminal communication should not be possible.

On the other hand, suppose that Alice and Bob are on opposite sides of the universe and have a number of entangled qubits. These are electrons whose spin states are entangled. Alice has one of each entangled pair of electrons in her possession, and Bob has the other one. (Though we are talking about entangled electrons, we should be clear that the actual electrons are totally separate. It’s their spin states that are entangled.)

When Alice makes a measurement on one of her electrons, the spin state of the corresponding electron in Bob’s possession instantaneously jumps into one of two distinct states. Instantaneous is clearly faster than the speed of light! Can’t entanglement be used for instantaneous communication?

Let’s suppose that each pair of the entangled electrons is in the entangled spin state that we have just studied:

.

Suppose that Alice measures the spins of her electrons before Bob measures the spin of their partners. We have seen that she gets a random string of 0s and 1s, with each symbol occurring with equal probability.

Suppose instead that Bob measures his spins before Alice. Then Alice measures the spins. What does she now get? When Alice makes her measurement, they will both have made measurements, so we can use the probability amplitudes of the initial expression. We know that they will obtain 00 and 01 with probability 1/4, 10 with probability 1/2 and 11 with probability 0. Consequently, Alice will get 0 with probability of , and 1 with probability . So, Alice gets a random string of 0s and 1s, with each symbol occurring with equal probability. But this is exactly the same situation as when she measured first. So Alice cannot tell from her measurements whether they were made before or after Bob’s. All entangled states behave this way. If there is no way of Alice and Bob being able to tell from their measurements who went first, there certainly can be no way of sending any information from one to the other.

We have shown that Alice and Bob cannot send information when their qubits have a particular entangled state, but the argument generalizes to any entangled state. No matter what states Alice’s and Bob’s qubits have, it is impossible for them to send information by solely measuring their qubits.

Now that we have seen that superluminal communication is not possible, we turn to the more prosaic task of writing tensor products using standard bases. But afterward we will return to our exploration of entangled qubits using the quantum clock example from the previous chapters.

The Standard Basis for Tensor Products

The standard basis for is . When both Alice and Bob use the standard basis, tensor products have the form:

Therefore, the standard ordered basis for is

.

Since it has four vectors in the basis, it is a four-dimensional space. The standard four-dimensional space is with ordered basis is

.

We identify the basis vectors in with those in making sure to respect the ordering.

      

The easiest way to remember this is by the following construction.

.

Notice also that the subscripts follow the standard binary ordering: 00, 01, 10, 11.

How Do You Entangle Qubits?

This book is about the mathematics that underlies quantum computing. It is not about how to physically create a quantum computer. We are not going to spend much time on the details of physical experiments, but the question of how physicists create entangled particles is such an important one that we will briefly address it. We can represent entangled qubits by either entangled photons or electrons. Though we often say the particles are entangled, what we really mean is that the vector describing their states, a tensor in , is entangled. The actual particles are separate and, as we have just noted, can be very far apart. That said, the question remains: How do you go about creating a pair of particles whose state vector is entangled? First, we look at how physical experiments create entangled particles. Then we look at how quantum gates create entangled qubits.

The most commonly used method at this time involves photons. The process is called spontaneous parametric down-conversion. A laser beam sends photons through a special crystal. Most of the photons just pass through, but some photons split into two. Energy and momentum must be conserved—the total energy and momentum of the two resulting photons must equal the energy and momentum of the initial photon. The conservation laws guarantee that the state describing the polarization of the two photons is entangled.

In the universe, electrons are often entangled. At the start of the book we described Stern and Gerlach’s experiment on silver atoms. Recall that the electron spins in the inner orbits canceled, leaving the lone electron in the outer orbit to give its spin to the atom. The innermost orbit has two electrons. These are entangled so that their spins cancel. We can think of the state vector describing the spin of these electrons as

.

Entangled electrons also occur in superconductors, and these electrons have been used in experiments. However, often we want to have entangled particles that are far apart—as we will see later when we talk about the Bell test.

The main problem with using entangled electrons that are near one another and then separating them is that they have a tendency to interact with the environment. It is difficult to separate them without this happening. On the other hand, entangled photons are much easier to separate, though more difficult to measure. It is possible, however, to get the best of both worlds. This has been done by an international team based at the Delft University of Technology in what they describe as a loophole free Bell test. They used two diamonds separated by 1.3 kilometers. Each diamond had slight imperfections—nitrogen atoms altered the carbon atom lattice structure in places. Electrons become trapped in the defects. A laser excited an electron in each of the diamonds in such a way that both electrons emitted photons. The emitted photons were entangled with the spins of the electrons that they were emitted from. The photons then traveled toward one another through a fiber optic cable and met in a beam splitter—a standard piece of equipment that is usually used to split a beam of photons in two, but here it is used to entangle the two photons. The photons were then measured. The result was that the two electrons were now entangled with one another.* (We will explain why the team was doing this experiment in the next chapter.)

In quantum computing, we will usually input unentangled qubits and entangle them using the CNOT gate. Later we will explain exactly what gates are, but the actual computations involve just matrix multiplication. We briefly look at this.

Using the CNOT Gate to Entangle Qubits

We leave the actual definition of what a quantum gate is until later, but we’ll just make the comment now that they correspond to orthonormal bases or, equivalently, to orthogonal matrices.

The standard ordered basis for four-dimensional space is is

.

The CNOT gate comes from interchanging the order of the last two elements. This results in the matrix for the CNOT gate.

This gate acts on pairs of qubits. To use the matrix, everything must be written using four-dimensional vectors. We look at an example.

We start by taking the unentangled tensor product

.

When we send qubits through the gate, they are changed. The resulting qubits are obtained by multiplying by the matrix.

This last vector corresponds to a pair of entangled qubits—the product of the inner amplitudes is zero, which is not equal to the product of the outer amplitudes. This can be rewritten as

.

We will often use entangled qubits in this state. It has the very nice property that if Alice and Bob measure in the standard basis, they will both get , corresponding to 0, or they will both get corresponding to 1. The two cases are equally probable.**

We examine this further with a quantum clock analogy.

Entangled Quantum Clocks

Recall the quantum clock metaphor. We can ask only about whether the hand is pointing in a certain direction, and the clock will answer either that it is or that it is pointing in the opposite direction.

We let the vector correspond to pointing to twelve, and to pointing to six. Consider a pair of clocks in the entangled state In fact, consider one hundred pairs of clocks, each pair of which is in this state. Suppose that you have one hundred of these clocks, and I have the hundred partners. We are both going to ask the same question repeatedly: Is the hand pointing toward twelve?

In the first scenario, we don’t contact one another. We just go through the clocks one at a time and ask the question. Each time the clock will answer either yes or no. We will write 1 if it is yes, and 0 if it is no. After we have finished asking questions, we have a string of 0s and 1s. I analyze my string and you analyze yours. Both strings are a random sequence of 0s and 1s. Both digits occur about the same number of times. We now contact one another and compare strings. Both your string and my string are identical. In all one hundred places the strings agree.

In the second scenario, we again each have one hundred clocks. This time we make an agreement that you will measure first. You will ask your question on the hour, and I will ask mine half an hour later. During these half-hours between our questions you will call me and tell me what my clock’s answer will be. At the end of the experiment we both have a string of 0s and 1s. Both strings agree in every place. Every time you called me and told me what the result of my measurement was going to be you were exactly right. Can we conclude that your measurements were affecting mine?

Well, suppose that I now tell you that I was cheating. I didn’t follow the rules. In fact, I was asking the questions of the clock half an hour before you asked yours. I knew your answer before you did. Your calls were just confirming what I knew.

There is no way from the data that you can tell whether or not I was following the rules or I was cheating. There is no way you can tell whether I am asking my questions before or after you asked yours.

There is no causation here, just correlation. As we saw earlier, we cannot use these entangled clocks to send messages between us. But the process is still mysterious. Albert Einstein described entanglement as implying spooky action at a distance. Nowadays many people would say that there is no action, just correlation. Of course, we can quibble about the definition of “action,” but even if we agree that there is no action, there seems to be something spooky going on.

Suppose that you and I have a pair of the entangled quantum clocks, and we are talking on the phone to one another. Neither of us has asked our clock a question, so they are still entangled. In this state, if you were to ask your clock the question, you would have an equal chance of getting an answer that the hand was pointing to twelve or six. But as soon as I ask my clock a question, you no longer have an equal chance of getting one of the two answers. You will get exactly the same answer as mine.

This correlation would not be spooky if when our clocks were entangled it was decided, but unknown to us, whether both our hands were pointing at either twelve or six. We had to wait until one of us asked the question, and as soon as one of us knows the answer so does the other.

But this is not what our model describes. Our model says that the decision on which direction our hands are pointing is not made beforehand. It’s made only when the first of us asks our question. This is what makes it spooky.

In the next chapter we will look at this in detail. We will look at a model that incorporates correlation in an intuitive and nonspooky way. Unfortunately, it is wrong. John Stewart Bell came up with an ingenious test that shows that the simple explanation is not correct and that the mysterious spookiness has to remain.

* There is a short video on this at https://www.youtube.com/watch?v=AE8MaQJkRcg/.

** In the next chapter, we will see that Alice and Bob don’t need to stick to the standard basis. If they both use the same orthonormal basis, no matter which one, they will still get exactly the same results.