# Operations on multiple qubits

This article reviews the rules used to build multi-qubit states out of single-qubit states and discusses the gate operations needed to include in a gate set to form a universal many-qubit quantum computer. These tools are absolutely necessary to understand the gate sets that are commonly used in Q# code, and also to gain intuition about why quantum effects such as entanglement or interference render quantum computing more powerful than classical computing.

## Single-qubit vs. multi-qubit gates

The true power of quantum computing only becomes evident as you increase the number of qubits. Single-qubit gates possess some counter-intuitive features, such as the ability to be in more than one state at a given time. However, if all you had in a quantum computer were single-qubit gates, then a calculator and certainly a classical supercomputer would dwarf its computational power.

Quantum computing power arises, in part, because the dimension of the vector space of quantum state vectors grows exponentially with the number of qubits. This means that while a single qubit can be trivially modeled, simulating a fifty-qubit quantum computation would arguably push the limits of existing supercomputers. Increasing the size of the computation by only one additional qubit *doubles* the memory required to store the state and roughly *doubles* the computational time. This rapid doubling of computational power is why a quantum computer with a relatively small number of qubits can far surpass the most powerful supercomputers of today, tomorrow, and beyond for some computational tasks.

## Two-qubit states

If you are given two separate qubits, one in the state $\psi=\begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ and the other in the state $\phi=\begin{bmatrix} \gamma \\ \delta \end{bmatrix}$, the corresponding two-qubit state is given by the tensor product (or Kronecker product of vectors, which is defined as follows

$$ \psi \otimes \phi = \begin{bmatrix} \alpha \\ \beta \end{bmatrix} \otimes \begin{bmatrix} \gamma \\ \delta \end{bmatrix} =\begin{bmatrix} \alpha \begin{bmatrix} \gamma \\ \delta \end{bmatrix} \\ \beta \begin{bmatrix}\gamma \\ \delta \end{bmatrix} \end{bmatrix} = \begin{bmatrix} \alpha\gamma \\ \alpha\delta \\ \beta\gamma \\ \beta\delta \end{bmatrix}. $$

Therefore, given two single-qubit states $\psi$ and $\phi$, each of dimension 2, the corresponding two-qubit state $\psi \otimes \phi$ is 4-dimensional. The vector

$$ \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} $$

represents a quantum state on two qubits if $|\alpha_{00}|^2+|\alpha_{01}|^2+|\alpha_{10}|^2+|\alpha_{11}|^2=1$. More generally, you can see that the quantum state of $n$ qubits is represented by a unit vector $v_1 \otimes v_2 \otimes \cdots \otimes v_n$ of dimension $2 \cdot 2 \cdot 2 \cdots = 2^n$ using this construction. Just as with single qubits, the quantum state vector of multiple qubits holds all the information needed to describe the system's behavior. For more information about vectors and tensor products, see Vectors and Matrices in Quantum Computing.

The computational basis for two-qubit states is formed by the tensor products of one-qubit states. For example, you have

\begin{align} 00 \equiv \begin{bmatrix}1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix}1 \\ 0 \end{bmatrix} &= \begin{bmatrix}1 \\ 0\\ 0\\ 0 \end{bmatrix},\qquad 01 \equiv \begin{bmatrix}1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix}0 \\ 1 \end{bmatrix} = \begin{bmatrix}0 \\ 1\\ 0\\ 0 \end{bmatrix},\\ 10 \equiv \begin{bmatrix}0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix}1 \\ 0 \end{bmatrix} &= \begin{bmatrix}0 \\ 0\\ 1\\ 0 \end{bmatrix},\qquad 11 \equiv \begin{bmatrix}0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix}0 \\ 1 \end{bmatrix} = \begin{bmatrix}0 \\ 0\\ 0\\ 1 \end{bmatrix}. \end{align}

Note that while you can always take the tensor product of two single-qubit states to form a two-qubit state, not all two-qubit quantum states can be written as the tensor product of two single-qubit states. For example, there are no states $\psi=\begin{bmatrix} \alpha \\ \beta \end{bmatrix}$ and $\phi=\begin{bmatrix} \gamma \\ \delta \end{bmatrix}$ such that their tensor product is the state

$$\psi\otimes \phi = \begin{bmatrix} 1/\sqrt{2} \\ 0 \\ 0 \\ 1/\sqrt{2} \end{bmatrix}.$$

Such a two-qubit state, which cannot be written as the tensor product of single-qubit states, is called an "entangled state"; the two qubits are said to be *entangled*. Loosely speaking, because the quantum state cannot be thought of as a tensor product of single qubit states, the information that the state holds is not confined to either of the qubits individually. Rather, the information is stored non-locally in the correlations between the two states. This non-locality of information is one of the major distinguishing features of quantum computing over classical computing and is essential for a number of quantum protocols including quantum teleportation and quantum error correction.

### Measuring two-qubit states

Measuring two-qubit states is very similar to single-qubit measurements. Measuring the state

$$ \begin{bmatrix} \alpha_{00} \\ \alpha_{01} \\ \alpha_{10} \\ \alpha_{11} \end{bmatrix} $$

yields $00$ with probability $|\alpha_{00}|^2$, $01$ with probability $|\alpha_{01}|^2$, $10$ with probability $|\alpha_{10}|^2$, and $11$ with probability $|\alpha_{11}|^2$. The variables $\alpha_{00}, \alpha_{01}, \alpha_{10},$ and $\alpha_{11}$ were deliberately named to make this connection clear. After the measurement, if the outcome is $00$ then the quantum state of the two-qubit system has collapsed and is now

$$ 00 \equiv \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}. $$

It is also possible to measure just one qubit of a two-qubit quantum state. In cases where you measure only one of the qubits, the impact of measurement is subtly different because the entire state is not collapsed to a computational basis state, rather it is collapsed to only one sub-system. In other words, in such cases measuring only one qubit only collapses one of the subsystems but not all of them.

To see this consider measuring the first qubit of the following state, which is formed by applying the Hadamard transform $H$ on two qubits initially set to the "0" state: $$ H^{\otimes 2} \left( \begin{bmatrix}1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix}1 \\ 0 \end{bmatrix} \right) = \frac{1}{2}\begin{bmatrix}1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}\begin{bmatrix}1\\ 0\\ 0\\ 0\end{bmatrix} = \frac{1}{2}\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix} \mapsto \begin{cases}\text{outcome }=0 & \frac{1}{\sqrt{2}}\begin{bmatrix}1\\ 1\\ 0\\ 0 \end{bmatrix}\\ \text{outcome }=1 & \frac{1}{\sqrt{2}}\begin{bmatrix}0\\ 0\\ 1\\ 1 \end{bmatrix}\\ \end{cases}. $$ Both outcomes have 50% probability of occurring. The outcome being 50% probability for both can be intuited from the fact that the initial quantum state vector is invariant under swapping $0$ with $1$ on the first qubit.

The mathematical rule for measuring the first or second qubit is simple. If you let $e_k$ be the $k^{\rm th}$ computational basis vector and let $S$ be the set of all $e_k$ such that the qubit in question takes the value $1$ for that value of $k$. For example, if you are interested in measuring the first qubit then $S$ would consist of $e_1\equiv 10$ and $e_3\equiv 11$. Similarly, if you are interested in the second qubit $S$ would consist of $e_2\equiv 01$ and $e_3 \equiv 11$. Then the probability of measuring the chosen qubit to be $1$ is for state vector $\psi$

$$ P(\text{outcome}=1)= \sum_{e_k \text{ in the set } S}\psi^\dagger e_k e_k^\dagger \psi. $$

Note

This article uses the *little-endian* format to label the computational basis. In little endian format, the least significant bits come first. For example, the number four in little-endian format is represented by the string of bits 001.

Since each qubit measurement can only yield $0$ or $1$, the probability of measuring $0$ is simply $1-P(\text{outcome}=1)$. This is why you only explicitly give a formula for the probability of measuring $1$.

The action that such a measurement has on the state can be expressed mathematically as

$$ \psi \mapsto \frac{\sum_{e_k \text{ in the set } S} e_k e_k^\dagger \psi}{\sqrt{P(\text{outcome}=1)}}. $$

The cautious reader may worry about what happens when the probability of the measurement is zero. While the resultant state is technically undefined in this case, you never need to worry about such eventualities because the probability is zero!

If you take $\psi$ to be the uniform state vector given above and are interested in measuring the first qubit then

$$ P(\text{measurement of first qubit}=1) = (\psi^\dagger e_1)(e_1^\dagger \psi)+(\psi^\dagger e_3)(e_3^\dagger \psi)=|e_1^\dagger \psi|^2+|e_3^\dagger \psi|^2. $$

Note that this is just the sum of the two probabilities that would be expected for measuring the results $10$ and $11$ were all the qubits to be measured. For our example, this evaluates to

$$ \frac{1}{4}\left|\begin{bmatrix}0&0&1&0\end{bmatrix}\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix} \right|^2+\frac{1}{4}\left|\begin{bmatrix}0&0&0&1\end{bmatrix}\begin{bmatrix}1\\ 1\\ 1\\ 1\end{bmatrix} \right|^2=\frac{1}{2}. $$

which perfectly matches what our intuition tells us the probability should be. Similarly, the state can be written as

$$ \frac{\frac{e_1}{2}+\frac{e_3}{2}}{\sqrt{\frac{1}{2}}}=\frac{1}{\sqrt{2}}\begin{bmatrix} 0\\ 0\\ 1\\ 1\end{bmatrix} $$

again in accordance with our intuition.

### Two-qubit operations

As in the single-qubit case, any unitary transformation is a valid operation on qubits. In general, a unitary transformation on $n$ qubits is a matrix $U$ of size $2^n \times 2^n$ (so that it acts on vectors of size $2^n$), such that $U^{-1} = U^\dagger$. For example, the CNOT (controlled-NOT) gate is a commonly used two-qubit gate and is represented by the following unitary matrix:

$$ \operatorname{CNOT} = \begin{bmatrix} 1\ 0\ 0\ 0 \\ 0\ 1\ 0\ 0 \\ 0\ 0\ 0\ 1 \\ 0\ 0\ 1\ 0 \end{bmatrix} $$

We can also form two-qubit gates by applying single-qubit gates on both qubits. For example, if you apply the gates

$$ \begin{bmatrix} a\ b\\ c\ d \end{bmatrix} $$

and

$$\begin{bmatrix} e\ f\\ g\ h \end{bmatrix} $$

to the first and second qubits, respectively, this is equivalent to applying the two-qubit unitary given by their tensor product: $$\begin{bmatrix} a\ b\\ c\ d \end{bmatrix} \otimes \begin{bmatrix} e\ f\\ g\ h \end{bmatrix}= \begin{bmatrix} ae\ af\ be\ bf \\ ag\ ah\ bg\ bh \\ ce\ cf\ de\ df \\ cg\ ch\ dg\ dh \end{bmatrix}.$$

Thus, you can form two-qubit gates by taking the tensor product of some known single-qubit gates. Some examples of two-qubit gates include $H \otimes H$, $X \otimes \boldone$, and $X \otimes Z$.

Note that while any two single-qubit gates define a two-qubit gate by taking their tensor product, the converse is not true. Not all two-qubit gates can be written as the tensor product of single-qubit gates. Such a gate is called an *entangling* gate. One example of an entangling gate is the CNOT gate.

The intuition behind a controlled-not gate can be generalized to arbitrary gates. A controlled gate in general is a gate that acts as identity (for example, it has no action) unless a specific qubit is $1$. You denote a controlled unitary, controlled in this case on the qubit labeled $x$, with a $\Lambda_x(U)$. As an example $\Lambda_0(U) e_{1}\otimes {\psi}=e_{1}\otimes U{\psi}$ and $\Lambda_0(U) e_{0}\otimes {\psi}=e_{0}\otimes{\psi}$, where $e_0$ and $e_1$ are the computational basis vectors for a single qubit corresponding to the values $0$ and $1$. For example, consider the following controlled-$Z$ gate then you can express this as $$ \Lambda_0(Z)= \begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1 \end{bmatrix}=(\boldone\otimes H)\operatorname{CNOT}(\boldone\otimes H). $$

Building controlled unitaries in an efficient manner is a major challenge. The simplest way to implement this requires forming a database of controlled versions of fundamental gates and replacing every fundamental gate in the original unitary operation with its controlled counterpart. This is often quite wasteful and clever insight often can be used to just replace a few gates with controlled versions to achieve the same impact. For this reason, the framework provides the ability to perform either the naive method of controlling or allow the user to define a controlled version of the unitary if an optimized hand-tuned version is known.

Gates can also be controlled using classical information. A classically controlled not-gate, for example, is just an ordinary not-gate but it is only applied if a classical bit is $1$ as opposed to a quantum bit. In this sense, a classically controlled gate can be thought of as an if statement in the quantum code wherein the gate is applied only in one branch of the code.

As in the single-qubit case, a two-qubit gate set is universal if any $4\times 4$ unitary matrix can be approximated by a product of gates from this set to arbitrary precision. One example of a universal gate set is the Hadamard gate, the T gate, and the CNOT gate. By taking products of these gates, you can approximate any unitary matrix on two qubits.

## Quantum entanglement

Consider two qubits $A$ and $B$ in superpositions such that the state of the global system is

$$\ket{\psi}_{AB}=\frac1{\sqrt2}\ket{00}\frac1{\sqrt2}\ket{11}$$

In such a state, only two outcomes are possible when you measure the state of both qubits in the standard basis: **$|00\rangle$** and **$|11\rangle$**. Notice that each outcome has the same probability of $\frac{1}{2}$. There's zero probability of obtaining **$|01\rangle$** and **$|10\rangle$**. If you measure the first qubit and you get that it is in **$|0\rangle$** state, then you can be positive that the second qubit is also in **$|0\rangle$** state, even without measuring it. The measurement outcomes are correlated, and the qubits are *entangled*.

Note

This examples uses two qubits, but quantum entanglement is not limited to two qubits. In general it's possible that multiple-qubit systems share entanglement.

Entangled qubits are correlated such that they cannot be described independently from each other. That is, whatever operation happens to the state of one qubit in an entangled pair, also affects to the state of the other qubit.

For a practical implementation, see the tutorial exploring quantum entanglement with Q# and Azure Quantum.

### Entanglement in pure states

Pure quantum states are those that are characterized by a single ket vector or wavefunction, and cannot be written as a statistical mixture (or *convex combination*) of other quantum states. On the Bloch sphere, pure states are represented by a point on the surface of the sphere, whereas mixed states are represented by an interior point.

A pure state $\ket{\phi}*{AB}$ is entangled if it cannot be written as a combination of product states of the subsystems, that is $\ket{\phi}*{AB} = \ket{a}_A \otimes \ket{b}_B$.

For example, consider the state $$ \ket{\psi}_{AB} = \frac{1}{2} (\ket{00} + \ket{10} +\ket{01} +\ket{11})$$

At first, the state $\ket{\psi}_{AB}$ doesn't look like a product state, but if we rewrite the state as

$$ \ket{\psi}_{AB} = \frac{1}{\sqrt{2}} (\ket{0}_A +\ket{1}_A) \otimes \frac{1}{\sqrt{2}} (\ket{0}_B +\ket{1}_B)= \ket{+}_A \ket{+}_B$$

the state $\ket{\psi}_{AB}$ is a product state, therefore it's not entangled.

### Entanglement in mixed states

Mixed quantum states are a statistical ensemble of pure states. A mixed state $\rho$ has neither quantum nor classical correlations if it can be written as a product state $\rho = \rho^{A} \otimes \rho^{B}$ for some density matrices $\rho^{A} \qeq 0 , \rho^{B} \geq 0$.

A mixed state $\rho$ is separable if it can be written as a convex combination of product states of the subsystems, such as $\rho = \sum_j p_j \rho^{A}*{j} \otimes \rho^{B}*{j}$ where $p_j \geq 0, \sum p_j = 1$ and $\rho^{A}*{j} \geq 0, \rho^{B}*{j} \geq 0$.

A mixed state $\rho$ is entangled if it's not separable, that is, it cannot be written as a convex combination of product states.

Tip

A separable state contains only classical correlations.

### Understanding classical correlations

Classical correlations are due to our lack of knowledge of the state of the system. That is, there's some randomness associated to classical correlation, but it can be eliminated by gaining knowledge.

For example, consider two boxes, each containing one ball. We know that both balls are the same color, either blue or red. If we open one box and find out that the ball inside is blue, then we know that the other ball is blue too. Therefore, they are correlated. However, the uncertainty that we have when opening the box is due to our lack of knowledge, it's not fundamental. The ball was blue before we opened the box. Thus, this is a classical correlation, not a quantum correlation.

The mixed quantum state of the system formed by the two boxes $\rho_{boxes}$ can be written as

$$ \rho_{boxes} = \frac{1}{2} (\ket{red}\bra{red}_A \otimes \ket{red}\bra{red}_B) +\frac{1}{2} (\ket{blue}\bra{blue}_A \otimes \ket{blue}\bra{blue}_B)

Notice that the state $\rho_{boxes}$ is separable, where $p_1 = p_2 = \frac{1}{2}$ then it contains only classical correlations. Another example of a mixed separable state is

$$ \rho = \frac{1}{2} (\ket{0}\bra{0}_A \otimes \ket{0}\bra{0}_B) +\frac{1}{2} (\ket{1}\bra{1}_A \otimes \ket{1}\bra{1}_B)

Now, consider the following state:

$$ \rho = \frac{1}{4} (\ket{00}\bra{00} + \ket{00}\bra{11} + \ket{11}\bra{00} + \ket{11}\bra{11}) = \ket{\phi^+}\bra{\phi^+}$$

In this case, our knowledge of the state is perfect, we know with maximal certainty that the system $AB$ is in the Bell state $\ket{\phi^+}$ and $\rho$ is a pure state. Therefore, there aren't classical correlations. But if we measure an observable on subsystem $A$, we obtain a random result which gives us information about the state of the subsystem $B$. This randomness is fundamental, namely these are quantum correlations.

An example of a quantum state that contains both classical and quantum correlations is

$$ \rho = \frac{1}{2} (\ket{\phi^+}\bra{\phi^+} + \ket{\phi^-}\bra{\phi^-})$$

Tip

- If an entangled state $\rho$ is pure, then it contains only quantum correlations.
- If an entangled state $\rho$ is mixed, then it contains both classical and quantum correlations.

## Many-qubit systems

We follow exactly the same patterns explored in the two-qubit case to build many-qubit quantum states from smaller systems. Such states are built by forming tensor products of smaller states. For example, consider encoding the bit string $1011001$ in a quantum computer. You can encode this as

$$ 1011001 \equiv \begin{bmatrix} 0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix}. $$

Quantum gates work in exactly the same way. For example, if you wish to apply the $X$ gate to the first qubit and then perform a CNOT between the second and third qubits you may express this transformation as

\begin{align} &(X \otimes \operatorname{CNOT}_{12}\otimes \boldone\otimes \boldone \otimes \boldone \otimes \boldone) \begin{bmatrix} 0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix} \otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 1 \\ 0 \end{bmatrix}\otimes \begin{bmatrix} 0 \\ 1 \end{bmatrix}\\ &\qquad\qquad\equiv 0011001. \end{align}

In many qubit systems, there is often a need to allocate and deallocate qubits that serve as temporary memory for the quantum computer. Such a qubit is said to be *auxiliary*. By default, you can assume the qubit state is initialized to $e_0$ upon allocation. You can further assume that it is returned again to $e_0$ before deallocation. This assumption is important because if an auxiliary qubit becomes entangled with another qubit register when it becomes deallocated then the process of deallocation will damage the auxiliary qubit. For this reason, you always assume that such qubits are reverted to their initial state before being released.

Finally, although new gates needed to be added to our gate set to achieve universal quantum computing for two qubit quantum computers, no new gates need to be introduced in the multi-qubit case. The gates $H$, $T$ and CNOT form a universal gate set on many qubits because any general unitary transformation can be broken into a series of two qubit rotations. You then can leverage the theory developed for the two-qubit case and use it again here when you have many qubits.

Note

While the linear algebraic notation that has been used thus far can certainly be used to describe multi-qubit states, it becomes increasingly cumbersome as you increase the size of the states. The resulting column-vector for a length 7 bit string, for example, is $128$ dimensional, which makes expressing it using the notation described previously very cumbersome. Instead, *Dirac notation*, a symbolic shorthand that simplifies the representation of quantum states, is used. For more information, see Dirac notation.

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