Quantum circuits

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Introduction

This lesson introduces the quantum circuit model of computation, which is a standard description of quantum computations that we'll use throughout this series.

We'll also introduce a few important mathematical concepts, including inner products between vectors, the notions of orthogonality and orthonormality, and projections and projective measurements, which generalize standard basis measurements. Through these concepts, we'll derive fundamental limitations on quantum information, including the no-cloning theorem and the impossibility to perfectly discriminate non-orthogonal quantum states.

Circuits

In computer science, circuits are models of computation in which information is carried by wires through a network of gates, which represent operations that transform the information carried by the wires. Quantum circuits are just one example of a model of computation based on this more general concept.

Although the word "circuit" often refers to a circular path, circular paths aren't actually allowed in the most common circuit models of computation. That is to say, we usually study acyclic circuits when we're thinking about circuits as computational models. Quantum circuits follow this pattern; although we could run a quantum circuit as many times as we like, a quantum circuit itself represents a finite sequence of operations that cannot contain feedback loops.

Boolean circuits

Here is an example of a (classical) Boolean circuit, where the wires carry binary values and the gates represent Boolean logic operations:

The flow of information along the wires goes from left to right: the wires on the left-hand side of the figure labeled $\mathsf{X}$ and $\mathsf{Y}$ are input bits, which can each be set to whatever binary value we choose, and the wire on the right-hand side is the output. The intermediate wires take whatever values are determined by the gates, which are evaluated from left to right.

The gates are AND gates (labeled $\wedge$), OR gates (labeled $\vee$), and NOT gates (labeled $\neg$). The functions computed by these gates will likely be familiar to many readers, but here they are represented by tables of values:

The two small circles on the wires just to the right of the names $\mathsf{X}$ and $\mathsf{Y}$ represent fanout operations, which simply create a copy of whatever value is carried on the wire on which they appear, so that this value can be input into multiple gates. Fanout operations are not always considered to be gates in the classical setting — sometimes they are treated as if they are "free" in some sense — but when we discuss how ordinary Boolean circuits can be converted into equivalent quantum circuits, we must classify fanout operations explicitly as gates and account for them correctly.

Here is the same circuit illustrated in a style more common in electrical engineering, which uses conventional symbols for the AND, OR, and NOT gates:

We won't use this style or these particular gate symbols further, but we do use different symbols to represent gates in quantum circuits, which we'll explain as we encounter them.

The particular circuit in this example computes the exclusive-OR (or XOR for short), which is denoted by the symbol $\oplus$:

In the next diagram we consider just one choice for the inputs: $\mathsf{X}=0$ and $\mathsf{Y}=1.$ Each wire is labeled by value it carries so you can follow the operations. The output value is $1$ in this case, which is the correct value for the XOR: $0 \oplus 1 = 1.$

You can check the other three possible input settings in a similar way.

Other types of circuits

As suggested above, the notion of a circuit in computer science is very general. For example, circuits whose wires carry values other than $0$ and $1$ are sometimes studied, as are gates representing different choices of operations.

In arithmetic circuits, for instance, the wires may carry integer values and the gates may represent arithmetic operations, such as addition and multiplication. The following figure depicts an arithmetic circuit that takes two variable input values ($x$ and $y$) as well as a third input set to the value $1.$ The values carried by the wires, as functions of the values $x$ and $y,$ are shown in the figure.

We can also consider circuits that incorporate randomness, such as ones where gates represent probabilistic operations.

Quantum circuits

In the quantum circuit model, wires represent qubits and gates represent operations acting on these qubits. We'll focus for now on operations we've encountered so far, namely unitary operations and standard basis measurements. As we learn about other sorts of quantum operations and measurements, we'll enhance our model accordingly.

Here's a simple example of a quantum circuit:

In this circuit, we have a single qubit named $\mathsf{X},$ which is represented by the horizontal line, and a sequence of gates representing unitary operations on this qubit. Just like in the examples above, the flow of information goes from left to right — so the first operation performed is a Hadamard, the second is an $S$ operation, the third is another Hadamard, and the final operation is a $T$ operation. Applying the entire circuit therefore applies the composition of these operations, $THSH,$ to the qubit $\mathsf{X}.$

Sometimes we wish to explicitly indicate the input or output states to a circuit. For example, if we apply the operation $THSH$ to the state $\vert 0\rangle,$ we obtain the state $\frac{1+i}{2}\vert 0\rangle + \frac{1}{\sqrt{2}} \vert 1 \rangle.$ We may indicate this as follows:

Quantum circuits often have all qubits initialized to $\vert 0\rangle,$ as we have in this case, but there are also cases where we wish to set the input qubits to different states.

Now let's see how we can specify this circuit in Qiskit, beginning with the imports needed for the current section.

To begin, we can create the circuit as follows, by sequentially adding gates from left to right.

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Output:

The default names for qubits in Qiskit are $\mathsf{q_0},$ $\mathsf{q_1},$ $\mathsf{q_2},$ etc., and when there is just a single qubit like in our example, the default name is $\mathsf{q}$ rather than $\mathsf{q_0}.$ If we wish to choose our own name we can do this using the QuantumRegister class like this:

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Output:

Here is another example of a quantum circuit, this time with two qubits:

As always, the gate labeled $H$ refers to a Hadamard operation, while the second gate is a two-qubit gate: it's the controlled-NOT operation, where the solid circle represents the control qubit and the circle resembling the symbol $\oplus$ denotes the target qubit.

Before examining this circuit in greater detail and explaining what it does, it is imperative that we clarify how qubits are ordered in quantum circuits, which connects with the convention that Qiskit uses for naming and ordering systems mentioned briefly in the previous lesson.

Qiskit's qubit ordering convention for circuits

In Qiskit, the topmost qubit in a circuit diagram has index $0$ and corresponds to the rightmost position in a tuple of qubits (or in a string, Cartesian product, or tensor product corresponding to this tuple). The second-from-top qubit has index $1,$ and corresponds to the position second-from-right in a tuple, and so on, down to the bottommost qubit, which has the highest index, and corresponds to the leftmost position in a tuple. In particular, Qiskit's default names for the qubits in an $n$-qubit circuit are represented by the $n$-tuple $(\mathsf{q_{n-1}},\ldots,\mathsf{q_{0}}),$ with $\mathsf{q_{0}}$ being the qubit on the top and $\mathsf{q_{n-1}}$ on the bottom in quantum circuit diagrams.

Further information on this ordering convention can be found on the Bit-ordering in Qiskit documentation page.

Although we will often deviate from the specific default names $\mathsf{q_{n-1}},\ldots,\mathsf{q_{0}}$ used for qubits by Qiskit, we will follow this ordering convention throughout this course for interpreting circuit diagrams.

Thus, our interpretation of the circuit above is that it describes an operation on a pair of qubits $(\mathsf{X},\mathsf{Y})$ — and if the input to the circuit is a quantum state $\vert \psi\rangle \vert \phi\rangle,$ then this means that the lower qubit $\mathsf{X}$ starts in the state $\vert \psi\rangle$ and the upper qubit $\mathsf{Y}$ starts in the state $\vert \phi\rangle.$ To understand what the circuit does, we can go from left to right through its operations.

1. The first operation is a Hadamard operation on $\mathsf{Y}$:

   

   When applying a gate to a single qubit like this, nothing happens to the other qubits — and nothing happening is equivalent to the identity operation being performed. In our circuit there is just one other qubit, $\mathsf{X},$ so the dotted rectangle in the figure above represents this operation:

   $$\mathbb{I}\otimes H = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0\\[2mm] \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0\\[2mm] 0 & 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\[2mm] 0 & 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{pmatrix}.$$

   Note that the identity matrix is on the left of the tensor product and $H$ is on the right, which is consistent with Qiskit's ordering convention.

2. The second operation is the controlled-NOT operation, where $\mathsf{Y}$ is the control and $\mathsf{X}$ is the target:

   

   The controlled-NOT gate's action on standard basis states is as follows:

   

   Given that we order the qubits as $(\mathsf{X}, \mathsf{Y}),$ the matrix representation of the controlled-NOT gate is this:

   $$\begin{pmatrix} 1 & 0 & 0 & 0\\[2mm] 0 & 0 & 0 & 1\\[2mm] 0 & 0 & 1 & 0\\[2mm] 0 & 1 & 0 & 0 \end{pmatrix}.$$

The unitary operation of the entire circuit, which we'll call $U,$ is the composition of the operations:

In particular, recalling our notation for the Bell states,

we get that

So, this circuit gives us a way to create the state $\vert\phi^+\rangle$ if we run it on two qubits initialized to $\vert 00\rangle.$ More generally, it gives us a way to convert the standard basis to the Bell basis. (The $-1$ phase factor on the last state, $-\vert \psi^-\rangle,$ could be eliminated if we wanted, but this would require a change to the circuit. For instance, we could add a controlled-$Z$ gate at the beginning or a swap gate at the end to eliminate the minus sign without affecting the circuit's action on the other three standard basis states.)

In general, quantum circuits can contain any number of qubit wires. We may also include classical bit wires, which are indicated by double lines like in this example:

In this circuit we have a Hadamard gate and a controlled-NOT gate on two qubits $\mathsf{X}$ and $\mathsf{Y},$ just like in the previous example. We also have two classical bits, $\mathsf{A}$ and $\mathsf{B},$ as well as two measurement gates. The measurement gates represent standard basis measurements: the qubits are changed into their post-measurement states, while the measurement outcomes are overwritten onto the classical bits to which the arrows point.

Here is an implementation of this circuit using Qiskit:

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Output:

The circuit can be simulated using the Sampler primitive.

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Output:

Sometimes it is convenient to depict a measurement as a gate that takes a qubit as input and outputs a classical bit (as opposed to outputting the qubit in its post-measurement state and writing the result to a separate classical bit). This means the measured qubit has been discarded and can safely be ignored thereafter.

For example, the following circuit diagram represents the same process as the one in the previous diagram, but where we ignore $\mathsf{X}$ and $\mathsf{Y}$ after measuring them:

As the series continues, we will see many more examples of quantum circuits, which will usually be much more complicated than the simple examples above. Here are some symbols for common gates:

• Single-qubit gates are generally shown as squares with a letter indicating which operation it is, like this:

  

  Not gates (also known as $X$ gates) are also sometimes denoted by a circle around a plus sign:

  

• Swap gates are denoted as follows:

  

  Controlled-gates, meaning gates that describe controlled-unitary operations, are denoted by a filled-in circle (indicating the control) connected by a vertical line to whatever operation is being controlled. For instance, controlled-NOT gates, controlled-controlled-NOT (or Toffoli) gates, and controlled-swap (Fredkin) gates are denoted like this:

  

• Arbitrary unitary operations on multiple qubits may be viewed as gates. They are depicted by rectangles labeled by the name of the unitary operation. For instance, here is a depiction of an (unspecified) unitary operation $U$ as a gate, along with a controlled version of this gate:

  

Inner products, orthonormality, and projections

To better prepare ourselves to explore the capabilities and limitations of quantum circuits, we now introduce some additional mathematical concepts — namely the inner product between vectors (and its connection to the Euclidean norm), the notions of orthogonality and orthonormality for sets of vectors, and projection matrices, which will allow us to introduce a handy generalization of standard basis measurements.

Inner products

Recall from Lesson 1 that when we use the Dirac notation to refer to an arbitrary column vector as a ket, such as

the corresponding bra vector is the conjugate transpose of this vector:

Alternatively, if we have some classical state set $\Sigma$ in mind, and we express a column vector as a ket, such as

then the corresponding row (or bra) vector is the conjugate transpose

We also observed that the product of a bra vector and a ket vector, viewed as matrices either having a single row or a single column, results in a scalar. Specifically, if we have two (column) vectors

so that the row vector $\langle \psi \vert$ is as in equation $(1),$ then

Alternatively, if we have two column vectors that we have written as

so that $\langle \psi \vert$ is the row vector $(2),$ we find that

where the last equality follows from the observation that $\langle a \vert a \rangle = 1$ and $\langle a \vert b \rangle = 0$ for classical states $a$ and $b$ satisfying $a\neq b.$

The value $\langle \psi \vert \phi \rangle$ is called the inner product between the vectors $\vert \psi\rangle$ and $\vert \phi \rangle.$ Inner products are critically important in quantum information and computation. We would not get far in understanding quantum information at a mathematical level without this fundamental notion.

Let us now collect together some basic facts about inner products of vectors.

1. Relationship to the Euclidean norm. The inner product of any vector

   $$\vert \psi \rangle = \sum_{a\in\Sigma} \alpha_a \vert a \rangle$$

   with itself is

   $$\langle \psi \vert \psi \rangle = \sum_{a\in\Sigma} \overline{\alpha_a} \alpha_a = \sum_{a\in\Sigma} \vert\alpha_a\vert^2 = \bigl\| \vert \psi \rangle \bigr\|^2.$$

   Thus, the Euclidean norm of a vector may alternatively be expressed as

   $$\bigl\| \vert \psi \rangle \bigr\| = \sqrt{ \langle \psi \vert \psi \rangle }.$$

   Notice that the Euclidean norm of a vector must always be a nonnegative real number. Moreover, the only way the Euclidean norm of a vector can be equal to zero is if every one of the entries is equal to zero, which is to say that the vector is the zero vector.

   We can summarize these observations like this: for every vector $\vert \psi \rangle$ we have

   $$\langle \psi \vert \psi \rangle \geq 0,$$

   with $\langle \psi \vert \psi \rangle = 0$ if and only if $\vert \psi \rangle = 0.$ This property of the inner product is sometimes referred to as positive definiteness.

2. Conjugate symmetry. For any two vectors

   $$\vert \psi \rangle = \sum_{a\in\Sigma} \alpha_a \vert a \rangle \quad\text{and}\quad \vert \phi \rangle = \sum_{b\in\Sigma} \beta_b \vert b \rangle,$$

   we have

   $$\langle \psi \vert \phi \rangle = \sum_{a\in\Sigma} \overline{\alpha_a} \beta_a \quad\text{and}\quad \langle \phi \vert \psi \rangle = \sum_{a\in\Sigma} \overline{\beta_a} \alpha_a,$$

   and therefore

   $$\overline{\langle \psi \vert \phi \rangle} = \langle \phi \vert \psi \rangle.$$

3. Linearity in the second argument (and conjugate linearity in the first). Let us suppose that $\vert \psi \rangle,$ $\vert \phi_1 \rangle,$ and $\vert \phi_2 \rangle$ are vectors and $\alpha_1$ and $\alpha_2$ are complex numbers. If we define a new vector

   $$\vert \phi\rangle = \alpha_1 \vert \phi_1\rangle + \alpha_2 \vert \phi_2\rangle,$$

   then

   $$\langle \psi \vert \phi \rangle = \langle \psi \vert \bigl( \alpha_1\vert \phi_1 \rangle + \alpha_2\vert \phi_2 \rangle\bigr) = \alpha_1 \langle \psi \vert \phi_1 \rangle + \alpha_2 \langle \psi \vert \phi_2 \rangle.$$

   That is to say, the inner product is linear in the second argument. This can be verified either through the formulas above or simply by noting that matrix multiplication is linear in each argument (and specifically in the second argument).

   Combining this fact with conjugate symmetry reveals that the inner product is conjugate linear in the first argument. That is, if $\vert \psi_1 \rangle,$ $\vert \psi_2 \rangle,$ and $\vert \phi \rangle$ are vectors and $\alpha_1$ and $\alpha_2$ are complex numbers, and we define

   $$\vert \psi \rangle = \alpha_1 \vert \psi_1\rangle + \alpha_2 \vert \psi_2 \rangle,$$

   then

   $$\langle \psi \vert \phi \rangle = \bigl( \overline{\alpha_1} \langle \psi_1 \vert + \overline{\alpha_2} \langle \psi_2 \vert \bigr) \vert\phi\rangle = \overline{\alpha_1} \langle \psi_1 \vert \phi \rangle + \overline{\alpha_2} \langle \psi_2 \vert \phi \rangle.$$

4. The Cauchy–Schwarz inequality. For every choice of vectors $\vert \phi \rangle$ and $\vert \psi \rangle$ having the same number of entries, we have

   $$\bigl\vert \langle \psi \vert \phi \rangle\bigr| \leq \bigl\| \vert\psi \rangle \bigr\| \bigl\| \vert \phi \rangle \bigr\|.$$

   This is an incredibly handy inequality that gets used quite extensively in quantum information (and in many other fields of study).

Orthogonal and orthonormal sets

Two vectors $\vert \phi \rangle$ and $\vert \psi \rangle$ are said to be orthogonal if their inner product is zero:

Geometrically, we can think about orthogonal vectors as vectors at right angles to each other.

A set of vectors $\{ \vert \psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ is called an orthogonal set if every vector in the set is orthogonal to every other vector in the set. That is, this set is orthogonal if

for all choices of $j,k\in\{1,\ldots,m\}$ for which $j\neq k.$

A set of vectors $\{ \vert \psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ is called an orthonormal set if it is an orthogonal set and, in addition, every vector in the set is a unit vector. Alternatively, this set is an orthonormal set if we have

for all choices of $j,k\in\{1,\ldots,m\}.$

Finally, a set $\{ \vert \psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ is an orthonormal basis if, in addition to being an orthonormal set, it forms a basis. This is equivalent to $\{ \vert \psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ being an orthonormal set and $m$ being equal to the dimension of the space from which $\vert \psi_1\rangle,\ldots,\vert\psi_m\rangle$ are drawn.

For example, for any classical state set $\Sigma,$ the set of all standard basis vectors

is an orthonormal basis. The set $\{\vert+\rangle,\vert-\rangle\}$ is an orthonormal basis for the $2$-dimensional space corresponding to a single qubit, and the Bell basis $\{\vert\phi^+\rangle, \vert\phi^-\rangle, \vert\psi^+\rangle, \vert\psi^-\rangle\}$ is an orthonormal basis for the $4$-dimensional space corresponding to two qubits.

Extending orthonormal sets to orthonormal bases

Suppose that $\vert\psi_1\rangle,\ldots,\vert\psi_m\rangle$ are vectors that live in an $n$-dimensional space, and assume moreover that $\{\vert\psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ is an orthonormal set. Orthonormal sets are always linearly independent sets, so these vectors necessarily span a subspace of dimension $m.$ From this we immediately conclude that $m\leq n$ because the dimension of the subspace spanned by these vectors cannot be larger than the dimension of the entire space from which they're drawn.

If it is the case that $m<n,$ then it is always possible to choose an additional $n-m$ vectors $\vert \psi_{m+1}\rangle,\ldots,\vert\psi_n\rangle$ so that $\{\vert\psi_1\rangle,\ldots,\vert\psi_n\rangle\}$ forms an orthonormal basis. A procedure known as the Gram–Schmidt orthogonalization process can be used to construct these vectors.

Orthonormal sets and unitary matrices

Orthonormal sets of vectors are closely connected with unitary matrices. One way to express this connection is to say that the following three statements are logically equivalent (meaning that they are all true or all false) for any choice of a square matrix $U$:

1. The matrix $U$ is unitary (i.e., $U^{\dagger} U = \mathbb{I} = U U^{\dagger}$).
2. The rows of $U$ form an orthonormal set.
3. The columns of $U$ form an orthonormal set.

This equivalence is actually pretty straightforward when we think about how matrix multiplication and the conjugate transpose work. Suppose, for instance, that we have a $3\times 3$ matrix like this:

The conjugate transpose of $U$ looks like this:

Multiplying the two matrices, with the conjugate transpose on the left-hand side, gives us this matrix:

If we form three vectors from the columns of $U,$

then we can alternatively express the product above as follows:

Referring to the equation (3), we now see that the condition that this matrix is equal to the identity matrix is equivalent to the orthonormality of the set $\{\vert\psi_1\rangle,\vert\psi_2\rangle,\vert\psi_3\rangle\}.$

This argument generalizes to unitary matrices of any size. The fact that the rows of a matrix form an orthonormal basis if and only if the matrix is unitary then follows from the fact that a matrix is unitary if and only if its transpose is unitary.

Given the equivalence described above, together with the fact that every orthonormal set can be extended to form an orthonormal basis, we conclude the following useful fact: Given any orthonormal set of vectors $\{\vert\psi_1\rangle,\ldots,\vert\psi_m\rangle\}$ drawn from an $n$-dimensional space, there exists a unitary matrix $U$ whose first $m$ columns are the vectors $\vert\psi_1\rangle,\ldots,\vert\psi_m\rangle.$ Pictorially, we can always find a unitary matrix having this form:

Here, the last $n-m$ columns are filled in with any choice of vectors $\vert\psi_{m+1}\rangle,\ldots,\vert\psi_n\rangle$ that make $\{\vert\psi_1\rangle,\ldots,\vert\psi_n\rangle\}$ an orthonormal basis.

Projections and projective measurements

Projection matrices

A square matrix $\Pi$ is called a projection if it satisfies two properties:

1. $\Pi = \Pi^{\dagger}.$
2. $\Pi^2 = \Pi.$

Matrices that satisfy the first condition — that they are equal to their own conjugate transpose — are called Hermitian matrices, and matrices that satisfy the second condition — that squaring them leaves them unchanged — are called idempotent matrices.

As a word of caution, the word projection is sometimes used to refer to any matrix that satisfies just the second condition but not necessarily the first, and when this is done the term orthogonal projection is typically used to refer to matrices satisfying both properties. In this series, however, we will use the terms projection and projection matrix to mean matrices satisfying both conditions.

An example of a projection is the matrix

for any unit vector $\vert \psi\rangle.$ We can see that this matrix is Hermitian as follows:

Here, to obtain the second equality, we have used the formula

which is always true (for any two matrices $A$ and $B$ for which the product $AB$ makes sense).

To see that the matrix $\Pi$ in $(4)$ is idempotent, we can use the assumption that $\vert\psi\rangle$ is a unit vector, so that it satisfies $\langle \psi \vert \psi\rangle = 1.$ Thus, we have

More generally, if $\{\vert \psi_1\rangle,\ldots,\vert \psi_m\rangle\}$ is any orthonormal set of vectors, then the matrix

is a projection. Specifically, we have

and

where the orthonormality of $\{\vert \psi_1\rangle,\ldots,\vert \psi_m\rangle\}$ is used just for the second-to-last equality.

In fact, this exhausts all of the possibilities: every projection $\Pi$ can be written in the form $(5)$ for some choice of an orthonormal set $\{\vert \psi_1\rangle,\ldots,\vert \psi_m\rangle\}.$ (The zero matrix $\Pi=0,$ which is a projection, is a special case: to fit it into the general form (5) we have to allow the possibility that the sum is empty, resulting in the zero matrix.)

Projective measurements

As has already been mentioned, the notion of a measurement of a quantum system is more general than just standard basis measurements. Projective measurements are measurements that are described by a collection of projections whose sum is equal to the identity matrix. In symbols, a collection $\{\Pi_0,\ldots,\Pi_{m-1}\}$ of projection matrices describes a projective measurement if

When such a measurement is performed on a system $\mathsf{X}$ while it is in some state $\vert\psi\rangle,$ two things happen:

1. For each $k\in\{0,\ldots,m-1\},$ the outcome of the measurement is $k$ with probability equal to

   $$\operatorname{Pr}\bigl(\text{outcome is $k$}\bigr) = \bigl\| \Pi_k \vert \psi \rangle \bigr\|^2.$$

2. For whichever outcome $k$ the measurement produces, the state of $\mathsf{X}$ becomes

   $$\frac{\Pi_k \vert\psi\rangle}{\bigl\|\Pi_k \vert\psi\rangle\bigr\|}.$$

We can also choose outcomes other than $\{0,\ldots,m-1\}$ for projective measurements if we wish. More generally, for any finite and nonempty set $\Sigma,$ if we have a collection of projection matrices

that satisfies the condition

then this collection describes a projective measurement whose possible outcomes coincide with the set $\Sigma,$ where the rules are the same as before:

1. For each $a\in\Sigma,$ the outcome of the measurement is $a$ with probability equal to

   $$\operatorname{Pr}\bigl(\text{outcome is $a$}\bigr) = \bigl\| \Pi_a \vert \psi \rangle \bigr\|^2.$$

2. For whichever outcome $a$ the measurement produces, the state of $\mathsf{X}$ becomes

   $$\frac{\Pi_a \vert\psi\rangle}{\bigl\|\Pi_a \vert\psi\rangle\bigr\|}.$$

For example, standard basis measurements are equivalent to projective measurements, where $\Sigma$ is the set of classical states of whatever system $\mathsf{X}$ we're talking about and our set of projection matrices is $\{\vert a\rangle\langle a\vert:a\in\Sigma\}.$

Another example of a projective measurement, this time on two qubits $(\mathsf{X},\mathsf{Y}),$ is given by the set $\{\Pi_0,\Pi_1\},$ where

If we have multiple systems that are jointly in some quantum state and a projective measurement is performed on just one of the systems, the action is similar to what we had for standard basis measurements — and in fact we can now describe this action in much simpler terms than we could before. To be precise, let us suppose that we have two systems $(\mathsf{X},\mathsf{Y})$ in a quantum state $\vert\psi\rangle,$ and a projective measurement described by a collection $\{\Pi_a : a\in\Sigma\}$ is performed on the system $\mathsf{X},$ while nothing is done to $\mathsf{Y}.$ Doing this is then equivalent to performing the projective measurement described by the collection

on the joint system $(\mathsf{X},\mathsf{Y}).$ Each measurement outcome $a$ results with probability

and conditioned on the result $a$ appearing, the state of the joint system $(\mathsf{X},\mathsf{Y})$ becomes

Implementing projective measurements using standard basis measurements

Arbitrary projective measurements can be implemented using unitary operations, standard basis measurements, and an extra workspace system, as will now be explained.

Let us suppose that $\mathsf{X}$ is a system and $\{\Pi_0,\ldots,\Pi_{m-1}\}$ is a projective measurement on $\mathsf{X}.$ We can easily generalize this discussion to projective measurements having different sets of outcomes, but in the interest of convenience and simplicity we will assume the set of possible outcomes for our measurement is $\{0,\ldots,m-1\}.$ Let us note explicitly that $m$ is not necessarily equal to the number of classical states of $\mathsf{X}$ — we'll let $n$ be the number of classical states of $\mathsf{X},$ which means that each matrix $\Pi_k$ is an $n\times n$ projection matrix. Because we assume that $\{\Pi_0\ldots,\Pi_{m-1}\}$ represents a projective measurement, it is necessarily the case that

Our goal is to perform a process that has the same effect as performing this projective measurement on $\mathsf{X},$ but to do this using only unitary operations and standard basis measurements.

We will make use of an extra workspace system $\mathsf{Y}$ to do this, and specifically we take the classical state set of $\mathsf{Y}$ to be $\{0,\ldots,m-1\},$ which is the same as the set of outcomes of the projective measurement. The idea is that we will perform a standard basis measurement on $\mathsf{Y},$ and interpret the outcome of this measurement as being equivalent to the outcome of the projective measurement on $\mathsf{X}.$ We will need to assume that $\mathsf{Y}$ is initialized to some fixed state, which we'll choose to be $\vert 0\rangle.$ (Any other choice of fixed quantum state vector could be made to work, but choosing $\vert 0\rangle$ makes the explanation to follow much simpler.)

Of course, in order for a standard basis measurement of $\mathsf{Y}$ to tell us anything about $\mathsf{X},$ we will need to allow $\mathsf{X}$ and $\mathsf{Y}$ to interact somehow before measuring $\mathsf{Y},$ by performing a unitary operation on the system $(\mathsf{Y},\mathsf{X}).$ First consider this matrix:

Expressed explicitly as a block matrix, this matrix looks like this:

(Each $0$ in this matrix represents an $n\times n$ matrix filled entirely with zeros.)

Now, $M$ is certainly not a unitary matrix (unless $m=1,$ in which case $\Pi_0 = \mathbb{I},$ giving $M = \mathbb{I}$ in this trivial case) because unitary matrices cannot have any columns (or rows) that are entirely $0;$ unitary matrices have columns that form orthonormal bases, and the all-zero vector is not a unit vector. However, it is the case that the first $n$ columns of $M$ are orthonormal, and we get this from the assumption that $\{\Pi_0,\ldots,\Pi_{m-1}\}$ is a measurement. To verify this claim, notice that for each $j\in\{0,\ldots,n-1\},$ column number $j$ of $M$ is this vector:

Taking the inner product of column $i$ with column $j$ (still assuming we're talking about the first $n$ columns, so $i,j\in\{0,\ldots,n-1\}$) gives

which is what we needed to show.

Thus, because the first $n$ columns of the matrix $M$ are orthonormal, we can replace all of the remaining zero entries by some different choice of complex number entries so that the entire matrix is unitary:

(If we are given the matrices $\Pi_0,\ldots,\Pi_{m-1},$ we can compute suitable matrices to fill in for the blocks marked $\fbox{?}$ in the equation — using the Gram–Schmidt process — but it will not matter specifically what these matrices are for the sake of this discussion.)

Finally we can describe the measurement process: we first perform $U$ on the joint system $(\mathsf{Y},\mathsf{X})$ and then measure $\mathsf{Y}$ with respect to a standard basis measurement. For an arbitrary state $\vert \phi \rangle$ of $\mathsf{X},$ we obtain the state

where the first equality follows from the fact that $U$ and $M$ agree on their first $n$ columns. When we perform a projective measurement on $\mathsf{Y},$ we obtain each outcome $k$ with probability

in which case the state of $(\mathsf{Y},\mathsf{X})$ becomes

Thus, $\mathsf{Y}$ stores a copy of the measurement outcome and $\mathsf{X}$ changes precisely as it would had the projective measurement described by $\{\Pi_0,\ldots,\Pi_{m-1}\}$ been performed directly on $\mathsf{X}.$

Limitations on quantum information

Despite sharing a common underlying mathematical structure, quantum and classical information have some key differences. As we continue on in this series, we will see many examples of tasks that quantum information allows, but classical information does not.

Before doing this, however, we should note some important limitations on quantum information. Understanding things quantum information can't do helps us identify the things it can do.

Irrelevance of global phases

The first limitation we'll cover — which is really more of a slight degeneracy in the way that quantum states are represented by quantum state vectors, as opposed to an actual limitation — concerns the notion of a global phase.

What we mean by a global phase is this. Suppose that $\vert \psi \rangle$ and $\vert \phi \rangle$ are unit vectors representing quantum states of some system, and assume moreover that there exists a complex number $\alpha$ on the unit circle (meaning that $\vert \alpha \vert = 1,$ or alternatively $\alpha = e^{i\theta}$ for some real number $\theta$) such that

The vectors $\vert \psi \rangle$ and $\vert \phi \rangle$ are then said to differ by a global phase. We also sometimes refer to $\alpha$ as a global phase, although this is context-dependent: any number on the unit circle can be thought of as a global phase when multiplied to a unit vector.

Now consider what happens when a system is in one of two quantum states that differ by a global phase, $\vert\psi\rangle$ and $\vert\phi\rangle,$ and the system undergoes a standard basis measurement. In the first case, in which the system is in the state $\vert\psi\rangle,$ the probability of measuring any classical state $a$ is

In the second case, in which the system is in the state $\vert\phi\rangle,$ the probability of measuring any classical state $a$ is

because $\vert\alpha\vert = 1.$ That is, the probability of an outcome appearing is the same for both states.

Let's consider what happens when we apply an arbitrary unitary operation $U$ to both states. In the first case, in which the initial state is $\vert \psi \rangle,$ the state becomes

and in the second case, in which the initial state is $\vert \phi\rangle,$ it becomes

That is, the two resulting states still differ by the same global phase $\alpha.$

Consequently, the two quantum states $\vert\psi\rangle$ and $\vert\phi\rangle$ that differ by a global phase are completely indistinguishable: no matter what operation, or sequence of operations, we apply to the two states, they will always differ by a global phase, and performing a standard basis measurement will produce outcomes with precisely the same probabilities as the other. For this reason, two quantum state vectors that differ by a global phase are considered to be equivalent, and are effectively viewed as being the same state.

For example, the quantum states

differ by a global phase (which is $-1$ in this example), and are therefore considered to be the same state.

On the other hand, the quantum states

do not differ by a global phase. Although the only difference between the two states is that a plus sign turns into a minus sign, this is not a global phase difference, it is a relative phase difference because it does not affect every vector entry, but only a proper subset of the entries. This is consistent with what we have already observed in Lesson 1, which is that the states $\vert + \rangle$ and $\vert - \rangle$ can be discriminated perfectly — performing a Hadamard operation and then measuring yields outcome probabilities as follows:

As an aside, here we find another advantage of the general description of quantum information based on density matrices over the simplified description based on quantum state vectors. In the general description of quantum information, the degeneracy in which two quantum state vectors can differ by a global phase, and hence effectively represent the same quantum state, disappears. That is, two density matrices that differ necessarily represent two distinct quantum states that can be discriminated in a statistical sense.

No-cloning theorem

The no-cloning theorem shows it is impossible to create a perfect copy of an unknown quantum state.

Theorem (No-cloning theorem). Let $\mathsf{X}$ and $\mathsf{Y}$ be systems sharing the same classical state set $\Sigma$ having at least two elements. There does not exist a quantum state $\vert \phi\rangle$ of $\mathsf{Y}$ and a unitary operation $U$ on the pair $(\mathsf{X},\mathsf{Y})$ such that $U \bigl( \vert \psi \rangle \otimes \vert\phi\rangle\bigr) = \vert \psi \rangle \otimes \vert\psi\rangle$ for every state $\vert \psi \rangle$ of $\mathsf{X}.$

That is, there is no way to initialize the system $\mathsf{Y}$ (to any state $\vert\phi\rangle$ whatsoever) and perform a unitary operation $U$ on the joint system $(\mathsf{X},\mathsf{Y})$ so that the effect is for the state $\vert\psi\rangle$ of $\mathsf{X}$ to be cloned — resulting in $(\mathsf{X},\mathsf{Y})$ being in the state $\vert \psi \rangle \otimes \vert\psi\rangle.$

The proof of this theorem is actually quite simple: it boils down to the observation that the mapping

is not linear in $\vert\psi\rangle.$

In particular, because $\Sigma$ has at least two elements, we may choose $a,b\in\Sigma$ with $a\neq b.$ If there did exist a quantum state $\vert \phi\rangle$ of $\mathsf{Y}$ and a unitary operation $U$ on the pair $(\mathsf{X},\mathsf{Y})$ for which $U \bigl( \vert \psi \rangle \otimes \vert\phi\rangle\bigr) = \vert \psi \rangle \otimes \vert\psi\rangle$ for every quantum state $\vert\psi\rangle$ of $\mathsf{X},$ then it would be the case that