2003 - Quantum Computation Lecture 3

Duration: 00:43:09

Transcript

David Deutsch

00:01:05 - 00:03:22

Today I’m going to talk about the quantum theory of measurement. Measurement is an important type of quantum computation. It’s also, of course, what links quantum theory with experiment, but for historical reasons, unfortunately, the quantum theory of measurement also plays a prominent role in what we could call quantum mechanics folklore: the informal misunderstandings of quantum physics that range from careless remarks in textbooks through bad philosophy and all the way down to complete nonsense. You may have heard that the measurement process in quantum theory is deeply mysterious and spooky, that it has a special role for the conscious observer. You may have heard that when you make a measurement in one place, there’s an instantaneous effect on quantum systems in a different place. You may have heard that measurements are inherently irreversible processes, even though the laws of motion in fundamental physics are all reversible. As you’ll see in this lecture and later lectures, none of that is true. In fact, quantum measurement theory is pretty well understood and has been a powerful tool for understanding quantum physics. Nowadays, it’s part of the quantum theory of computation. The most elementary but important thing to understand about measurements is that they are physical processes that we understand using the same theory and laws of physics as we do for all other physical processes. A measurement is a process in which one physical system, a measuring instrument, interacts with another physical system that’s being measured. Some observable of the measuring instrument is affected by some observable of the system being measured and ends up ideally having the same value.

00:03:25 - 00:06:28

Let’s confine our attention for the moment to the useful idealization of a perfect measurement. We can define a perfect measurement process by the effect it has when the observable being measured is sharp, has a single value. First, the outcome of a perfect measurement of a sharp observable is the value of that observable, and second, at the end of the measurement, the observable being measured is still sharp with the same value it had before. So a perfect measurement records sharp values accurately and leaves them unchanged. The simplest possible measurement is the measurement of a Boolean observable where the outcome is stored in the second Boolean observable. Let’s think about that first in the case of a classical computation. Let’s make it a reversible classical computation partly to substantiate what I said just now about measurement not being inherently irreversible, but mainly because in fact classical reversible computations are a special case of quantum computation. Imagine that we have a single bit that we’re going to measure, and the bit has an unknown value that’s either plus one or minus one. Call that unknown value a, and we have a second bit somewhere in an apparatus in which we want to record a copy of the value of the first bit. The second bit is called the target bit. So the target bit has to end up with the value a, and the bit being measured has to keep the value a. Considered as a computation, perfect measurement is a process of faithfully copying information, so that where there was one copy of the sharp value of the observable, there are now two, the second one being in the measuring apparatus. As far as the definition of a perfect measurement is concerned, we’re not interested in what the initial value of the target bit was so long as the final value is a, but in a reversible computation, different inputs must always produce different outputs. So it follows that if the computation only involves those two bits, it can only be a perfect measurement for one initial value of the target bit.

00:06:28 - 00:09:01

Let’s say for the value $+1$ . So it’s not a perfect measurement when the target bit starts at minus one. What does happen if it’s minus one? It depends on the law of motion. One interesting law of motion is that of a very useful operation called controlled not. The ordinary not operation is the single bit or single qubit operation that figured in the interference experiment. I discussed last time. It flips plus one to minus one and vice versa. Control not means flip the value of the target bit only if the first bit is minus one. The first bit is now called the control bit if the control bit is plus one the target bit remains unaffected. So we can summarize this controlled not operation like this: if the input values of the control and target bits are a and B, then the outputs are a and AB. A computer like this that performs a simple computation on a fixed number of bits and completes it in a fixed time is called a computational gate also known as a logic gate, and in this case, it’s a reversible logic gate because it’s performing a reversible computation: the controlled not operation. Let me just summarize some of the ways in which we can think about what the controlled not gate does. First of all, if B equals plus one, then considering it as a physical process, it’s a perfect measurement: the first bit starts with an unknown value a, and the second bit ends up holding the measured value of the first bit which should also be a. Second, considered as a computation, if B equals plus one then the process copies information, the information a, which starts out in only one of the bits ends up in both of them.

00:09:04 - 00:12:56

Third, for an arbitrary B, well, that’s the controlled not operation. The target bit is flipped if and only if the control bit is minus one. A fourth interpretation: this gate is the reversible version of the classical exclusive or gate. If we consider plus one as standing for false and minus one for true, then a times B is the same thing as the exclusive or of A and B, and the target bit ends up as the exclusive or of the two inputs. And as I said like all classical reversible gates, the controlled not gate has a quantum implementation that works on two qubits instead of two bits. That’s the quantum controlled not gate. In some implementations there’s a physical object you can identify as the gate as with the mirror in the interference experiment last time which was a not gate and the beam splitter which was a fractional power of not but speaking more precisely the computational gate is not the object that makes the process happen, but the dynamical process itself that’s undergone by the qubit or qubits that physically realize the process and that actually satisfies a rule that defines each output in terms of the inputs and nothing else. Now to describe the quantum physical system consisting of two qubits passing through a quantum controlled not gate, I’m going to make the Z observable of the control qubit control the Z observable of the target qubit. To describe that quantum system, I have to tell you as always the static constitution, the dynamics, and on each particular occasion when the experiment is done the state which summarizes how the qubits were prepared as the input. I’ll start as usual with the static constitution. For this I have to tell you the algebra of all the observables at any one time, which will then be the same as the algebra at any other time because the algebra summarizes the time invariant features of a quantum system. So here we have a two qubit quantum system. We already know what the algebra of the observables of either one of the two qubits is because that algebra is itself an invariant. It’s the same for all qubits and regardless of what interactions the qubit happens to be undergoing so as in the one qubit interference experiment, the first qubit will have observables x y and z which have the same algebra as the Pauli matrices, together with the unit observable and linear combinations of those with constant coefficients. Let me call these observables $X_1$ , $Y_1$ , and $Z_1$ now, where the suffixes indicate that they are observables of qubit number one the control qubit of the quantum controlled not gate. We don’t have to put a suffix on the unit observable because, as you will remember, the unit observable can be measured without even referring to the system.

00:12:56 - 00:23:29

The second qubit the target qubit will have different observables $X_2$ , $Y_2$ , $Z_2$ and so on but they’ll also have the Pauli algebra. Now I have to specify all the additional algebraic relations that may hold involving observables from both qubits. And here’s a universal rule that defines the algebra of any composite quantum system given the algebras of its constituent systems. It just says that the observables of different quantum systems commute with each other. That is to say if a1 is an observable of one system and b2 is an observable of another, then $A_1B_2=B_2A_1$ . I told you last time that two by two matrix representations of a qubit algebra wouldn’t always be enough. Here’s why: this set of algebraic relations can’t be faithfully represented by two by two matrices. This set by itself can, so can this set. But because this commutation relation for the combined system has to be represented as well, qubits one and two can’t use the same set of matrices. Take the $Z_2$ for instance. It has to commute with every observable of qubit one but the only two by two matrices that commute with every two by two matrix are multiples of the unit matrix, and $Z_2$ can’t be represented by a multiple of the unit matrix because it’s a Boolean observable. It has to have two distinct eigenvalues and the multiples of the unit matrix have only one eigenvalue. So there is no two by two matrix representation of this algebra as a whole. The simplest representation turns out to be a four by four representation, and the way it is constructed applies to any two quantum systems not just a pair of qubits that you want to consider as a single system. It uses the tensor product of matrices. The tensor product of two matrices of dimensions M and N is an M N dimensional matrix consisting of all possible products of pairs of elements one from the first matrix and one from the second like this: A B C D with E F G H is the four by four matrix A E A F and so on. That symbol denotes the tensor product. So the tensor product of a pair of two by two matrices is a matrix of all 16 possible products consisting of an element of the first matrix multiplied by an element of the second. You can verify in the worked examples that the tensor product of a pair of Hermitian matrices is a Hermitian matrix. So the tensor product of two observables one from each system is an observable of the combined system. What observable is it? Well, if you have two systems and a is a matrix representing an observable of system one by itself and B is a matrix representing an observable of system two by itself, then the tensor product $A\otimes B$ is an observable that you measure by measuring a on the first system and B on the second and then multiplying the results together. You can verify that the tensor product operation is associative. But it’s not commutative. In other words $A\otimes B$ is not the same as $B\otimes A$ . You can also verify that the tensor product has the following property with regard to ordinary matrix multiplication: $(A\otimes B)(C\otimes D)=AC\otimes BD$ . Now it follows from all that that, using the tensor product, we can make a four by four representation of the algebra of observables of qubit one as follows. Just multiply each matrix of any two by two representation by the two by two unit matrix on the right. These are now four by four matrices, and because of this property, it’s easy to show that they do indeed represent the standard algebra for one qubit. For qubit two we can do a similar thing this time multiplying by the unit matrix on the left. They too form a four by four representation of the standard algebra for a single qubit. The object of the exercise: any observable of the first qubit commutes with any observable of the second. For instance X1 Y2 at time zero is $\sigma_X\otimes I$ times $I\otimes\sigma_Y$ , which equals $\sigma_X\otimes\sigma_Y$ and taking it the other way around $Y_2X_1=(I\otimes\sigma_Y)(\sigma_X\otimes I)$ , which also equals $\sigma_X\otimes\sigma_Y$ by the way. Like in the single qubit two by two case, we need never work with the actual components of these matrices. Every four by four Hermitian matrix can be expressed as a linear combination of tensor products of the form $\sigma\otimes\sigma$ , $\sigma\otimes I$ , $I\otimes\sigma$ , and $I\otimes I$ which is the four by four unit matrix. So again, we can just express all four by four matrices and so all observables in terms of Pauli matrices and do all matrix arithmetic in terms of the algebra of Pauli matrices. So I’ve told you the static constitution of a pair of qubits, which would be the same for any two qubit gate. Now the dynamics of the controlled not gate in particular. As usual we’ll imagine that it performs its operation in one unit of time one computational step. For the moment, we’re considering the gate as an elementary quantum computation, so we’re not interested in its internal workings. We’re only interested in how the output depends on the input. So for present purposes the dynamics of the gate just means how the observables at time one after the gate has acted depend on the observables at time zero. Well, the dynamics of a controlled not gate are defined by a set of equations. Most of whose details are not relevant for present purposes, but I’ll put them on screen anyway, just so that I can point out the properties of the gate being a controlled not gate and highlight the parts that are relevant. These six equations tell us how six representative observables change: three of them from one qubit and three from the other. We don’t have to list the unit observable explicitly because it never changes. Using these equations, we can find out how any other observable behaves as well by expressing it at time zero in terms of these representative observables and then using the fact that algebraic relations between observables at a given time don’t change with time. These equations are actually redundant. We could make do with just four of them because for instance the static constitution for any qubit already defines its observable $Z$ at any given time in terms of its observables X and Y at that time. Now take a look at the equations for the time evolution of just the $Z$ observables. You can see that $Z_1$ and $Z_2$ at time one depend only on $Z_1$ and $Z_2$ at time zero. The X’s and Y’s don’t affect the evolution of the $Z$ observables. So the $Z$ observables almost form a physical system in their own right a little subsystem of the two qubit system that evolves independently of the rest. It doesn’t quite count as a separate physical system because you can never change $Z$ without changing X or Y. Nevertheless in a controlled not computation, the $Z$ observables are autonomous.

00:23:29 - 00:26:52

They’re evolving independently of all other observables. I’ve drawn your attention to that because it will come up again in later lectures, but for the moment its only significance is that it makes it easy for us to check that this gate really does perform a perfect measurement. Remember a perfect measurement interaction is defined by what it does when the observable being measured is sharp. So now it’s time for me to specify a convenient state for our two qubits in which the $Z$ observables of both qubits are sharp. There are only four possible cases of this, so we may as well deal with all four of them wholesale. Let the observable $Z_1$ be sharp with the value a and let $Z_2$ be sharp with the value B. A and B are each plus or minus 1. So to specify the state or actually four possible states, we have to specify the expectation values of all the observables. And I’ve just said that the expectation value of $Z_1(0)=a$ and the expectation value of $Z_2(0)=b$ . As I mentioned in the last lecture and you will have proved in the worked examples $Z$ being maximally sharp implies that X and Y are both minimally sharp. In other words, that their expectation values are both 0. This in turn tells us how to find expectation values of Pauli matrices in this state. In the representation we’ve chosen for the observables, we have expectation value of $\sigma_Z\otimes I$ equals a and expectation value of $I\otimes\sigma_Z$ equals B. By linearity this tells us all the expectation values of matrices of the form something cross 1 and 1 cross something. You’ll see in the worked examples that these relations also tell us that in this state the expectation values of any tensor product of the form $A\otimes B$ where a and B are two by two matrices is the product of the expectation values. With these rules, we can find expectation values of general matrices and hence of general observables in this state. So what do we predict for the output of the controlled not gate where the inputs were sharp? Well, the control observable $Z_1$ has the same expectation value at time 1 as it did at time 0, namely a, and since a is plus or minus 1, that means that $Z_1$ is still sharp with the value a, which is correct. $Z_2$ is sharp as well, but its value has changed.

00:26:52 - 00:29:09

It’s now $aB$ , just like the target bit of a classical control not gate. Okay, the next obvious thing to work out is what happens if we perform a perfect measurement of an observable that’s not sharp. Well, you can prove in the worked examples that if $Z_1$ isn’t sharp at time 0, it’ll be just as unsharp at time 1, and also $Z_2$ will have become just as unsharp as well. So the measurement interaction propagates unsharpness from one system to the other. But there’s more. Consider the observable $Z_1Z_2$ . That is an observable because $Z_1$ and $Z_2$ commute at any one time So $Z_1Z_2$ is Hermitian, and from what I’ve said its operational meaning is that it’s the observable for what you’d get if you measured $Z_1$ and $Z_2$ and multiplied the outcomes together In other words. It’s a Boolean observable whose eigenvalue minus 1 means that the outcomes of measuring $Z_1$ and $Z_2$ would be different, and plus 1 means that those outcomes would be the same. Now here’s a remarkable thing Starting in a state where $Z_1$ is not sharp at time 0 and hence both $Z_1$ and $Z_2$ are unsharp at time 1, calculate the expectation value of $Z_1Z_2$ . You’ll find it’s plus 1, which means that the observable $Z_1Z_2$ is sharp In other words $Z_1$ and $Z_2$ are equal at the end of the measurement, sharply equal even though neither of them has a sharp value? How can two things be perfectly equal without either of them being sharp?

00:29:09 - 00:33:17

Well like this, of course Finally Let’s look at the case where this measurement of an unsharp observable Takes place in the middle of an attempted interference experiment Let’s work out what happens when we combine The analysis of the previous lecture of the interference experiment with this time’s analysis of the dynamics of measurement Let’s take a qubit based on the photon’s direction of motion and Pass it through a beam splitter to make $Z_1(1)$ unsharp So from the equations of motion $Z_1$ of 1 takes this form In the 2 by 2 representation of last time But now we’re going to measure the direction of motion So we need a second qubit and therefore a 4 by 4 representation like this The second qubit will be a subsystem of some instrument that detects the direction of motion in principle That could be another subatomic particle But equally it could be a pair of photon detectors like the ones we actually used in the interference experiment The essence of any such instrument is that somewhere in there is an observable That’s going to be the target of a controlled not or perfect measurement operation a boolean observable Measuring the Z observable as we defined it for the photon’s direction of motion So We can just analyze those two qubits and forget all the other degrees of freedom just as we did before The details again are in the worked examples. We find that the expectation value of $Z_1$ The photon direction of motion at time 2 just after the measurement is 0 just as it was in the original experiment Because in half the universes the photon travels on one path and in the other half it travels on the other and So the expectation value of $Z_2$ of 2 is also 0 Just as we would expect from something that has measured the correct value of $Z_1$ in each universe then we let the photon go on and do exactly what it did before namely bounce off the mirrors and Strike the second beam splitter Which previously had the effect of making the direction of motion sharp again But look not in this experiment The expectation value of $Z_1$ is still 0 at the end of the experiment The fact of having made a measurement Even a perfect measurement of the value of $Z_1$ at an intermediate stage of the experiment has spoiled the interference phenomenon This is why we don’t see very clear examples of quantum interference in everyday life It’s because undergoing an interaction in which even one qubit from the outside is affected by a physical system is enough to suppress interference and It doesn’t have to be a measurement interaction.

00:33:17 - 00:35:08

You’ll see that almost any interaction affecting an outside qubit will do Specifically what prevents interference is when something carries off information about the system and The reason why that makes a difference is that any process that transfers information out of a system always changes the system itself if The process is a perfect measurement. It doesn’t change the observable being measured The observable being measured say Z but it changes other observables like X or Y that are inextricably linked with it by the uncertainty principle and by the dynamics of quantum physics and that can make the system subsequently behave differently a Process in which information is carried off in this way is known as a decoherence process in practical implementations Decoherence is the great enemy of quantum computation and I’ll be saying more about that So far When I’ve described the dynamics of quantum systems to you They’ve all been quantum gates I’ve always just told you the laws of motion of the gate by fiat Just a set of equations for how the representative Observables of the qubits are changed by passage through the gate in the next lecture I’ll show you a general framework for quantum dynamics and I’ll tell you in principle Which sets of such equations describe processes that can occur in nature and which can’t?

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