3 
Spin and Qubits

The first chapter described measurements involving the spin of an electron. We saw that if you measure the spin in the vertical direction, you don’t obtain a continuum of values, but just two of them: Either the electron has its north pole vertically upward, or it is vertically downward. If we measure the spin first in the vertical direction and then once more in the same direction, we obtain exactly the same result for both measurements. If the first measurement shows that the electron has its north pole upward, then so will the second measurement. We also saw that if we measure first in the vertical direction and then in the horizontal direction, the electrons will have spin N and S in direction 90° each with probability of one half. It doesn’t matter what the first measurement is; the second measurement will give a random choice of either N or S. The second chapter introduced the mathematics of linear algebra. The goal of this chapter is to combine these previous two chapters, giving a mathematical model that describes the measurement of spin. We will then show how this relates to qubits. But before we start this description, we introduce the mathematics of probability.

Probability

Imagine that we have a coin and we repeatedly toss it, counting both the number of tosses and the number of times it comes up heads. If the coin is fair—equally likely to land heads up as tails up—the ratio of the number of heads to the number of tosses, after tossing it a large number of times, will be close to one half. We say that the probability of the outcome “heads” is 0.5.

In general, we perform an experiment—often we will call it making a measurement—that has a finite number of possible outcomes. We will denote these by . The underlying assumption is that the result of the experiment, or measurement, will be one and only one of these n outcomes. Associated with outcome is a probability . Probabilities must be numbers between 0 and 1 that sum to 1. In the case of tossing a coin, the two outcomes are getting a head and getting a tail. If the coin is fair, the probability of each event is 1/2.

We return to the experiments involving the spin of a particle from the first chapter using a slightly more formal notation to describe them. Suppose that we are going to measure the spin in direction 0°. There are two possible outcomes that we will denote as N and S. Both of these outcomes will have an associated probability. We will denote by the probability of obtaining N, and the probability of obtaining S. If we already know that our electron has spin N in direction 0°, then we know that when we measure again in this direction we will get the same result, so, in this case, and . On the other hand, if we know our electron has spin N in direction 90° and we now measure in direction 0°, then we are equally likely to obtain N and S as the outcome, so, in this case .

Mathematics of Quantum Spin

We will now present the mathematical model that describes quantum spin. It uses both probabilities and vectors.

The basic model is given by a vector space. When we make a measurement there will be a number of possible outcomes. The number of outcomes determines the dimension of this underlying vector space. For spin, there are just two possible outcomes from any measurement, so the underlying vector space is two-dimensional. We will take the space to be —this is the standard two-dimensional plane with which we are all familiar. This is fine for our purposes because we are only rotating our measuring apparatus in the plane. If we also wanted to consider all possible three-dimensional rotations of the apparatus, the underlying space would still be two-dimensional—two is still the number of possible outcomes for each measurement—but instead of using vectors with real coefficients, we would have to use vectors that involve complex numbers. The underlying vector space would then be the two-dimensional complex space denoted . For the reasons listed in the previous chapter, is fine for our needs.

We will not consider all of the vectors in , just the unit vectors. For kets, this means we are restricting to kets of the form , where

Choosing a direction to measure spin corresponds to choosing an ordered, orthonormal basis . The two vectors in the basis correspond to the two possible outcomes for the measurements. We will always associate N with the first basis vector and S with the second. Before we measure the spin, the particle will be in a spin state given by a linear combination of and , that is, it has the form . We will sometimes refer to this as a state vector and sometimes just call it a state. After we measure, its state vector will jump to either or . This is one of the major ideas in quantum mechanics: Measurement causes the state vector to change. The new state is one of the basis vectors associated with the measurement. The probability of getting a particular basis vector is given by the initial state. The probability of its being is ; the probability of is . The numbers and are called the probability amplitudes. It’s important to remember that the probability amplitudes are not probabilities. They can be positive or negative. It’s the squares of these numbers that are probabilities. To make everything more concrete, we will return to the experiments where we measured in spin the vertical and horizontal directions.

As we mentioned in the previous chapter, the ordered orthonormal basis corresponding to measuring spin in the vertical direction is given by , where and The first vector listed in the basis corresponds to the electron having spin N in direction 0° and the second vector to S in direction 0°.

The spin in the horizontal direction is given by the ordered orthonormal basis , where and . The first vector listed in the basis corresponds to the electron having spin N in direction 90° and the second vector to S in direction 90°.

We first measure spin in the vertical direction. Initially, we might not know the spin state of the incoming electron, but it must be a unit vector and so can be written as , where . We now perform the measurement. Either the electron is diverted upward in which case the state jumps to or it is diverted downward in which case its state jumps to . The probability of it being diverted upward is and the probability of it being diverted downward is

We now repeat exactly the same experiment, measuring the spin once more in the vertical direction. Suppose that the electron was deflected upward by the first set of magnets. We know it is in spin state . When we measure again, the state jumps to with probability , or to with probability . This just means that it remains in state , and so is deflected upward once more.

Similarly, if the electron was deflected downward it will be in state . No matter how many times we measure it in the vertical direction it will remain in this state, telling us that however many times we repeat the experiment the electron will keep being deflected downward. As we noted in the first chapter, if we repeat exactly the same experiment, we get exactly the same outcome.

Instead of repeatedly measuring spin in the vertical direction, we will first measure spin in the vertical direction and then measure spin in the horizontal direction. Suppose that we have just performed the first measurement. We have measured spin in the vertical direction, and let us suppose that the electron has spin N in direction 0°. Its state vector is now . Since we are next going to measure spin in the horizontal direction, we have to write this vector in terms of the orthonormal basis that corresponds to this direction, which means we must find the values of and that solve . We know how to do this: It’s the second tool in the toolbox listed at the end of the last chapter.

First construct the matrix A by stacking the kets that form the orthonormal basis side by side.

Then calculate to get the probability amplitudes with respect to the new basis.

This tells us that .

When we measure in the horizontal direction the state will jump to with probability , or it will jump to with probability . This tells us that the probability that the electron has spin Nin the 90° direction is equal to the probability that it has spin S in the 90°direction; both probabilities are exactly one-half.

Notice that we didn’t really need to calculate the matrix A to do this calculation. The matrix that we need to use is . We can calculate this by stacking the bras that correspond to the orthonormal basis on top of one another. We must, of course, keep things in the same order. The left to right ordering of kets corresponds to the top to bottom ordering of bras, so the first element of the basis is the topmost bra.

In the first chapter we measured the spin three times. The first and third measurements were in the vertical direction, the second was in the horizontal direction. We will describe the mathematics that corresponds to the third measurement. After the second measurement, the state vector of our electron will have one of two values. It will be either or . We are now going to measure the spin in the vertical direction, so we need to express these as linear combinations of the vertical orthonormal basis. This gives and . In either case, when we measure spin in the vertical direction the state vector will jump to either or to , each occurring with probability one-half.

Equivalent State Vectors

Suppose that we are given a number of electrons and are told that their spins are given by either or by . Can we distinguish between the two cases? Is there any measurement that we can perform that would tell them apart? The answer is that there is not.

To see this, let’s suppose that we choose a direction in which to measure spin. This is equivalent to choosing an ordered, orthonormal basis. We will denote this basis by .

Suppose our electron has state . We have to find the values of a and b that solve the equation . When we perform the measurement, the probability of the spin being N is , and the probability of the spin being S is .

Suppose our electron has state . For exactly the same values of a and b, we have . When we perform the measurement the probability of the spin being N is and the probability of the spin being S is .

We get exactly the same probabilities for both cases, so there is no measurement that can distinguish electrons with state vectors of form from those of .

Similarly, given electrons with state there is no way to distinguish them from electrons with state . Since these states are indistinguishable, they are considered equivalent. Saying that an electron has spin given by means exactly the same as saying that it has spin given by .

To help illustrate this point further, consider these four kets:

By the preceding remarks, we know that and are equivalent, and that and are equivalent. So, these four kets describe at most two distinguishable states. But what about and ? Do these describe the same state, or are they distinguishable?

We do have to be a little careful. If we choose to measure the spin in the vertical direction, these two kets are not distinguishable. In both cases, we get or each occurring with probability of a half. But we know that and . Consequently, if we measure in the 90° direction, we will obtain S for the first ket and N for the second. This choice of basis does distinguish them, and so they are not equivalent.

One thing that is probably not clear at the moment is how the basis associated with a direction of measurement is chosen. We have seen that the basis associated with measuring in the vertical (0°) direction is and with the horizontal (90°) direction is .

But where did these bases come from? Later, when we come to Bell’s theorem, we will need the bases associated with 120° and 240°. What are these? We answer these questions in the next section.

The Basis Associated with a Given Spin Direction

We begin with our measurement apparatus. We take the vertical direction as the starting point and start rotating in the clockwise direction. As we have already noted, when it has been rotated through 90°, we are measuring in the horizontal direction. By the time it’s rotated through 180°, we are measuring the vertical direction once more. An electron that has spin N in direction 0° will have spin S in direction 180°, and an electron that has spin S in direction 0° will have spin N in direction 180°. Clearly, saying a magnet has its north pole in one direction conveys exactly the same information as saying the magnet has its south pole in the opposite direction, and consequently we need only to rotate our apparatus through an angle between 0° and 180° to cover all possible directions.

We will now consider bases. We take the standard basis as our starting point. This can be pictured as two vectors in the plane, as shown in figure 3.1.

Figure 3.1 The standard basis.

Now we rotate these vectors. The general picture, with rotation of α° is depicted in figure 3.2. The vector rotates to , and rotates to .

Figure 3.2 The standard basis rotated by α°.

Rotating through α° changes our initial ordered, orthonormal basis from to .

If the basis is rotated through 90° it becomes , which simplifies to As we previously noted, is equivalent to , so rotating through 90° brings us back to a basis equivalent to the original one, except that the order of the basis elements has been interchanged (i.e., N and S have been interchanged).

We will let θ denote the angle we are rotating our measurement apparatus and α the angle we rotate our basis vectors. We have seen that we get a complete set of directions as θ goes from 0° to 180°, and that we get a complete set of rotated bases as α goes from 0° to 90°. Once we reach θ = 180° or equivalently α = 90°, N and S measured in direction 0° are interchanged.

We make the natural definition that θ = 2α. Consequently, the basis associated with rotating our apparatus by θ is . Figure 3.3 illustrates this.

Figure 3.3 Rotating measuring apparatus by θ°.

Rotating the Apparatus through 60°

As an example to illustrate our formula, we look at what happens when we rotate our measuring apparatus by 60°. Suppose that we first measured our electron to have spin N in direction 0°. We will measure it again using the apparatus turned through 60°. What is the probability that it gives a result of N?

In this case the associated basis to the rotated apparatus is which simplifies to .

Since the electron initially was measured to have spin N in direction 0°, its state vector after the initial measurement was . We must now express this as a linear combination of the new basis vectors. To get the coordinates relative to the new basis we can multiply the state vector on the left by the matrix consisting of the bras of the basis. This gives:

,

telling us that

.

So the probability of getting N when we measure in the 60° direction is

The Mathematical Model for Photon Polarization

In most of the book we will restrict our attention to measuring spin of electrons, but in the first chapter we said that we could rewrite everything in terms of the polarization of photons. In the next few sections we will explain the analogy between electron spin and photon polarization and give the mathematical model of polarization.

We start by associating the angle of 0° with a polarized filter in the vertical direction, that is, a filter that lets through photons that are polarized vertically, which means that horizontally polarized photons are absorbed by the filter. As with the spin of electrons, we associate the standard basis to the angle of 0°. The vector corresponds to a vertically polarized photon and the vector to a horizontally polarized one.

We will rotate the filter through an angle β°. It now lets through photons that are polarized in direction β° and blocks photons that are polarized perpendicularly to β°.

The mathematical model follows that for the spin of electrons. For each direction, there is an ordered orthonormal basis associated with making a polarization measurement in this direction. The ket corresponds to a photon that is polarized in the given direction—that is, that passes through the filter. The ket corresponds to a photon that is polarized orthogonally to the given direction—that is absorbed by the filter.

A photon has a polarization state given by a ket, . This can be written as a linear combination of the vectors in the basis:.

When the polarization is measured in the direction given by the ordered basis, the result will be that the photon is polarized in the given direction with probability and polarized perpendicularly with probability ; that is, the probability the photon passes through the filter is , and the probability it is absorbed is .

If the result of the measurement is that the photon is polarized in the given direction—it passes through the filter—then the state of the photon becomes .

The Basis Associated with a Given Polarization Direction

Recall that if we start with our standard basis and rotate these vectors though an angle α, we obtain the new orthonormal basis Also recall that rotating through an angle of 90° brings us back to the same basis as the original, except that the order of the basis elements has been interchanged.

Now consider rotating a polarized filter through an angle β. When β is 0°, we are measuring in the vertical and horizontal direction. The vertically polarized photons pass through the filter, and the horizontally polarized photons are absorbed. Once β reaches 90° we will be measuring photons in the vertical and horizontal direction, but now the vertically polarized photons are absorbed and the horizontally polarized ones pass through. In this case, β = 90° corresponds to α = 90°, and, in general, we can take α = β.

In conclusion, the ordered orthonormal basis associated with rotating a polarized filter through an angle β is .

The Polarized Filters Experiments

Using our model, we describe the experiments that we looked at in the first chapter.

In the first experiment we have two polarized squares. One measures polarization in direction 0° and the other in direction 90°. No light is let through the region of overlap, as depicted in figure 3.4.

Figure 3.4 Two polarized squares.

The basis associated with 0° is the standard orthonormal basis. The basis associated with 90° is the same, except the order of the elements has been changed. A photon that passes through the first filter has had a measurement made—it is vertically polarized—and so is now in state . We now measure it with the second filter. This lets through photons with state vector and absorbs photons with state vector . Consequently, any photon that passes through the first filter is absorbed by the second.

In the three-filter experiment we have the two filters arranged as above. We take the third sheet and rotate it through 45°, and slide this sheet between the other two. Some light comes through the region of overlap of all three squares. This is depicted in figure 3.5.

Figure 3.5 Three polarized squares.

The ordered bases for the three filters are , and . A photon that passes through all three filters will have had three measurements made. Photons that pass through the first filter will be in state .

The second measurement corresponds to passing through the filter rotated by 45°. We need to rewrite the state of the photon using the appropriate basis.

The probability of a photon passing through the second filter once it has gone through the first is . Consequently, half the photons that pass through the first filter will pass through the second filter. Those that do will now be in state .

The third filter corresponds to making a measurement using the third basis. We must rewrite the state of our photon using this basis.

The third filter lets through photons corresponding to state . The probability of this is . Consequently, half the photons that pass through the first two filters will pass through the third filter.

We have shown how the mathematical model relates the spin of an electron to the polarization of a photon. This model is also exactly what we need to describe qubits.

Qubits

A classical bit is either 0 or 1. It can be represented by anything that has two mutually exclusive states. The standard example is a switch that can be in either the on or off position. In classical computer science the measurement of bits does not enter the picture. A bit is a bit. It is either 0 or it is 1, and that is all there is to it. But for qubits the situation is more complicated, and measurement is a crucial part of the mathematical description.

We define a qubit to be any unit ket in Usually, given a qubit, we will want to measure it. If we are going to measure it, we also need to include a direction of measurement. This is done by introducing an ordered orthonormal basis . The qubit can be written as a linear combination—often called a linear superposition—of the basis vectors. In general, it will have the form . After we measure, its state will jump to either or . The probability of its being is ; the probability of is . This is exactly the same model we have been using, but now we connect the classical bits 0 and 1 to the basis vectors. We will associate the basis vector with the bit 0 and the basis vector with the bit 1. So when we measure the qubit we will obtain 0 with probability and 1 with probability .

Since a qubit can be any unit ket and there are infinitely many unit kets, there are infinitely many possible values for a qubit. This is quite unlike classical computation, where we just have two bits. It is important, however, to notice that to get information out of a qubit we have to measure it. When we measure it we will get either 0 or 1, so the result is a classical bit.

We will give some illustrative examples using Alice, Bob, and Eve.

Alice, Bob, and Eve

Alice, Bob, and Eve are three characters that often appear in cryptography. Alice wants to send a confidential message to Bob. Unfortunately, Eve wants to eavesdrop with evil intent. How should Alice encrypt her messages so that Bob can read them but Eve cannot? This is the central question of cryptography. We will look at it later. But for the moment we will just concentrate on Alice sending Bob a stream of qubits.

Alice measures qubits using her orthonormal basis, which we will denote as . Bob measures the qubits that Alice sends to him using his orthonormal basis .

Suppose that Alice wants to send 0. She can use her measuring apparatus to sort qubits into either state or . Since she wants to send 0, she sends a qubit in state . Bob is measuring with respect to his ordered basis. To calculate what happens we must write as a linear combination of Bob’s basis vectors. It will have the form . When Bob measures the qubit one of two things happens: Either it jumps to state with probability and he writes down 0, or it jumps to state with probability and he writes down 1.

You might be wondering why Bob and Alice don’t choose to use the same basis. If they did this, Bob would receive 0 with certainty whenever Alice sent 0 and receive 1 with certainty whenever Alice sent 1. This is true, but remember Eve. If she also chooses the same basis, then she too will receive exactly the same message as Bob. We will see later that there are good reasons for why Alice and Bob might choose different bases to thwart Eve.

For an example, Alice and Bob might to choose to measure their qubits using either the basis or the basis . The calculations are exactly as before, where we were considering spin in the vertical and horizontal directions. The only change is that we replace N with 0 and S with 1. Only if Alice and Bob choose to use the same basis will Bob end up with exactly the bit that Alice wanted to send. If they choose to use different bases, then half of the time Bob gets the correct bit, but half of the time he gets the wrong one. This might not seem very useful, but we will see at the end of this chapter that Alice and Bob can use these two bases to secure their communications.

A couple of chapters from now, Alice and Bob will each choose one of three bases at random. These correspond to measuring the spin of an electron in the directions of 0°, 120°, or 240°. We will need to analyze all the possibilities, but now, to give a concrete example, we will have Alice measure in the 240° direction and Bob in the 120° direction.

We know the orthonormal basis in direction θ is . Consequently, Alice’s basis is and Bob’s is . Since multiplying a ket by –1 gives an equivalent ket, we can simplify Alice’s basis to . (Notice that this is the basis for direction 60° that we looked at earlier, with the order of the basis vectors switched. There is nothing surprising about this. In fact, that’s exactly what we expect. Measuring N in direction 240° is exactly the same as measuring S in direction 60°.)

If Alice wants to send 0, she sends the qubit . To calculate what Bob measures we need to write this as a linear superposition of his basis vectors. We can get the probability amplitudes by forming the matrix consisting of the bras of his basis vectors and then multiplying the qubit by this matrix.

.

This tells us that

.

This means that when Bob measures the qubit, he gets 0 with probability 1/4 and 1 with probability 3/4. Similarly, it can be checked that if Alice sends 1, Bob will get 1 with probability 1/4 and 0 with probability 3/4.

It can also be checked, and it is an excellent exercise, that if Alice and Bob choose from the three bases, where the third is the standard basis, and end up with different bases, then Bob always gets the correct bit with probability 1/4.

Probability Amplitudes and Interference

If you drop a stone into a pond, waves propagate outward from where the stone hits the water. If you drop two stones, the waves propagating form one stone can interfere with the waves coming from the other one. If the waves are in phase—the peaks or the troughs coincide—then you get constructive interference: The amplitude of the resulting wave increases. If the waves are out of phase—the peak of one meets the trough of the other—then you get destructive interference: The amplitude of the resulting wave decreases.

A qubit has the form , where and are the probability amplitudes. The square of these numbers gives the probabilities that the qubit jumps to the corresponding basis vector. Probabilities are not allowed to be negative, but probability amplitudes can be. This fact allows both constructive and destructive interference to take place.

As an example, consider the qubits that we are denoting by and If we measure either of them in the standard basis, they will jump to either or . Each has a probability of 1/2 of occurring. If we are translating this back to bits, we will get either a 0 or a 1 with equal probability. We now take a superposition of the original two qubits, . If we were to measure in the horizontal direction, we would get either or with equal probability. But if we measure in the vertical direction we get 0 with certainty, because

.

The terms in and that give 0 have interfered constructively, and the terms that give 1 have interfered destructively.

This will be important when it comes to quantum algorithms. We want to choose linear combinations carefully so that terms that we are not interested in cancel, but terms that we are interested in are amplified.

There are a very limited number of things that we can do with one qubit, but one thing we can do is to enable Alice and Bob to communicate securely.

Alice, Bob, Eve, and the BB84 Protocol

We often want to send secure messages. All Internet commerce depends on it. The standard way that messages are encrypted and decrypted uses two steps. The first step is when first contact is made. The two parties agree on a key—a long string of binary digits. Once they both have the same key, they then use it to both encode and decode messages from one another. The security of the method comes from the key. It is impossible to decode the messages between the two parties without knowing the key.

Alice and Bob want to communicate securely. Eve wants to eavesdrop. Alice and Bob want to agree on a key, but they need to be sure that Eve does not know it.

The BB84 protocol derives its name from its inventors, Charles Bennett and Gilles Brassard, and the year that it was invented, 1984. It uses two sets of ordered, orthonormal bases: the standard one, , that we used for measuring spin in the vertical direction, and so is denoted by V, and that we used for measuring spin in the horizontal direction, and so is denoted by H. In both cases, the classical bit 0 will correspond to the first vector in the ordered basis and 1 to the second.

Alice chooses the key that she wants to send to Bob. This is a string of classical bits. For each bit, Alice chooses one of the two bases V and H at random and with equal probability. She then sends Bob the qubit consisting of the appropriate basis vector. For example, if she wants to send 0 and chooses V, she will send , if she chooses H, she will send . She follows the same process for each bit, keeping a record of which basis she has used for each bit. If the string is 4n binary digits long, she will end up with a string of length 4n consisting of Vs and Hs. (The reason we are using 4n and not n will become clear in a moment, but n should be a fairly large number.)

Bob also chooses between the two bases at random and with equal probability. He then measures the qubit in his chosen basis. Bob does this for each bit, and he keeps a record of which basis he has used. At the end of the transmission he also ends up with two strings of length 4n, one consisting of 0s and 1s from his measurements, the other consisting of Vs and Hs corresponding to the bases he chose.

Alice and Bob are choosing the basis for each bit at random. Half the time they end up using the same basis, while half the time they use different bases. If they both choose the same basis, then Bob will obtain the bit that Alice is sending with certainty. If they choose different bases, then half the time Bob gets the right bit, but half the time it is the wrong bit—no information is transmitted when they choose different bases.

Alice and Bob now compare their strings of Vs and Hs over an unencrypted line. They keep the bits corresponding to the times when they both used the same basis and erase the bits that correspond to times that they used different bases. If Eve is not intercepting the message, they both end up with the same string of binary digits that has length about 2n. They now must check to see if Eve was listening in.

If Eve intercepts the qubit on the way from Alice to Bob, she would really like to clone it, sending one copy on to Bob and measuring the other qubit. Unfortunately for Eve, this is impossible. To obtain any information, she has to measure the qubit that Alice has sent, and this could change the qubit—it will end up as one of the basis vectors in the basis with which she chooses to measure. The best she can do is to choose one of the two bases at random, measure the qubit, and then send the qubit on to Bob. Let’s see what happens.

Alice and Bob are interested only in the measurements where they chose the same basis. We will restrict our attention to these times. When Alice and Bob agree on the basis, half the time Eve will also agree, and half the time she will choose the other basis. If all three agree on the basis, then they will all get the same bit as the measurement. If Eve chooses the wrong basis, then she will send a qubit that is in a superposition of Bob’s basis states. When Bob measures this qubit he will get 0 and 1 with equal probability; he will get the right bit one half of the time.

We now return to Alice and Bob and their strings of bits of length, at the moment, of 2n. They know that if Eve is not intercepting qubits, these strings will be identical. But they know that if Eve is intercepting qubits, she is going to choose the wrong basis half the time, and in these cases Bob will end up with the wrong bit half of the time. So, if Eve is intercepting qubits, a quarter of Bob’s bits will disagree with Alice’s. They now compare half of the 2n bits over an unencrypted line. If they agree on all of them, they know Eve is not listening in and can use the other n bits as the key. If they disagree on a quarter of the bits, they know that Eve is intercepting their qubits. They know that they need to find another way to secure their communication.

This is a nice example of sending one qubit at a time. There are, however, very few things that we can do with qubits that don’t interact with one another. In the next chapter we look at what happens when we have two or more qubits. In particular, we look at yet another phenomenon that is not part of our classical worldview but that plays an essential part of the quantum world: entanglement.