13.6.1: Kets, Bras, Brackets, and Operators

Kets, bras, brackets and operators are the building bricks of bracket notation, which is the most commonly used notation for quantum mechanical systems. They can be thought of as column vectors, row vectors, dot products and matrices respectively.

\(| Ket \rangle\)

A ket is just a column vector composed of complex numbers. It is represented as:

\(|k\rangle=\left(\begin{array}{l}
k_{1} \\
k_{2}
\end{array}\right)=\overrightarrow {k}. \tag{13.5}\)

The symbol \(k\) inside the ket \(| k \rangle\) is the label by which we identify this vector. The two kets \(| 0\rangle\) and \(| 1\rangle\) are used to represent the two logical states of qubits, and have a standard vector representation

\[
|0\rangle=\left(\begin{array}{l}
1 \\
0
\end{array}\right) \quad|1\rangle=\left(\begin{array}{l}
0 \\
1
\end{array}\right). \tag{13.6}
\nonumber \]

Recall from Equation 13.1 that the superposition of two quantum states \(\psi_0\) and \(\psi_1\) is

\(\psi = \alpha \psi_0 + \beta \psi_1 \tag{13.7}\)

where \(\alpha\) and \(\beta\) are complex numbers. In bracket notation, this superposition of \(| 0\rangle\) and \(| 1\rangle\) can be written as

\[
\begin{align*}
|\psi\rangle &=\alpha|0\rangle+\beta|1\rangle \tag{13.8}\\
&=\alpha\left(\begin{array}{l}
1 \\
0
\end{array}\right)+\beta\left(\begin{array}{l}
0 \\
1
\end{array}\right) \tag{13.9}\\
&=\left(\begin{array}{c}
\alpha \\
\beta
\end{array}\right) \tag{13.10}
\end{align*}
\nonumber \]

\(\langle Bra |\)

Bras are the Hermitian conjugates of kets. That is, for a given ket, the corresponding bra is a row vector (the transpose of a ket), where the elements have been complex conjugated. For example, the qubit from 13.10 has a corresponding bra \(\langle \psi |\) that results from taking the Hermitian conjugate of equation 13.10

\begin{align*}
(|\psi\rangle)^{\dagger} &=(\alpha|0\rangle+\beta|1\rangle)^{\dagger} \tag{13.11}\\
&=\alpha^{*}(|0\rangle)^{\dagger}+\beta^{*}(|1\rangle)^{\dagger} \tag{13.12}\\
&=\alpha^{*}\left(\begin{array}{l}
1 \\
0
\end{array}\right)^{\dagger}+\beta^{*}\left(\begin{array}{l}
0 \\
1
\end{array}\right)^{\dagger} \tag{13.13}\\
&=\alpha^{*}\left(\begin{array}{ll}
1 & 0
\end{array}\right)+\beta^{*}\left(\begin{array}{ll}
0 & 1
\end{array}\right) \tag{13.14}\\
&=\left(\begin{array}{ll}
\alpha^{*} & \beta^{*} \tag{13.15}
\end{array}\right)
\end{align*}

The symbol \(\dagger\) is used to represent the operation of hermitian conjugation of a vector or a matrix.\(^2\) The star (*) is the conventional notation for the conjugate of a complex number: \((a + i b)* = a − i b\) if \(a\) and \(b\) are real numbers.

\(\langle Bra | Ket \rangle\)

The dot product is the product of a bra (row vector) \(\langle q |\), by a ket (column vector) \(| k\rangle\), it is called bracket and denoted \(\langle q | k\rangle\), and is just what you would expect from linear algebra

\[\langle q \mid k\rangle=\left(\begin{array}{ll}
q_{1}^{*} & q_{2}^{*}
\end{array}\right) \times\left(\begin{array}{l}
k_{1} \\
k_{2}
\end{array}\right)=\sum_{j} q_{j}^{*} k_{j}. \tag{13.16} \]

Note that the result of \(\langle q | k\rangle \) is a complex number.

Brackets allow us to introduce a very important property of kets. Kets are always assumed to be normalized, which means that the dot product of a ket by itself is equal to 1. This implies that at least one of the elements in the column vector of the ket must be nonzero. For example, the dot product of an arbitrary qubit \((| \psi \rangle\)) by itself, \(\rangle \psi | \psi \rangle\) = 1, so

\begin{align*}
\langle\psi \mid \psi\rangle &=\left(\alpha ^ { * } \left\langle0\left|+\beta^{*}\langle 1|\right) \cdot(\alpha|0\rangle+\beta|1\rangle)\right.\right.\\
&=\alpha^{*} \alpha\langle 0 \mid 0\rangle+\beta^{*} \alpha\langle 1 \mid 0\rangle \alpha^{*} \beta\langle 0 \mid 1\rangle+\beta^{*} \beta\langle 1 \mid 1\rangle \\
&=\alpha^{*} \alpha+\beta^{*} \beta \\
&=|\alpha|^{2}+|\beta|^{2}=1 \tag{13.17}
\end{align*}

This is precisely the result we postulated when we introduced the qubit as a superposition of wave functions. In Chapter 10, we saw that the product of a wavefunction by its complex conjugate is a probability distribution and must integrate to one. This requirement is completely analogous to requiring that a ket be normalized.\(^3\)

The dot product can be used to compute the probability of a qubit of being in either one of the possible states \(| 0\rangle\) and \(| 1\rangle\). For example, if we wish to compute the probability that the outcome of a measurement on the qubit \(| \psi \rangle\) is state \(| 0\rangle\), we just take the dot product of \(| 0\rangle\) and \(| \psi\rangle\) and square the result

\begin{align*}
\operatorname{Pr}(|0\rangle) &=|\langle 0 \mid \psi\rangle|^{2} \\
&=|\alpha\langle 0 \mid 0\rangle+\beta\langle 0 \mid 1\rangle|^{2} \\
&=\left|\alpha_{1}+\beta_{0}\right|^{2} \\
&=|\alpha|^{2} \tag{13.18}
\end{align*}

\(\widehat{Operators}\)

Operators are objects that transform one ket \(| k\rangle\) into another ket \(| q\rangle\). Operators are represented with hats: \(\hat{O}\). It follows from our definition of ket that an operator is just a matrix,

\begin{align*}
\widehat {O}|k\rangle &=\left(\begin{array}{ll}
o_{11} & o_{12} \\
o_{21} & o_{22}
\end{array}\right) \times\left(\begin{array}{l}
k_{1} \\
k_{2}
\end{array}\right) \\
&=\left(\begin{array}{l}
q_{1} \\
q_{2}
\end{array}\right) \\
&=|q\rangle \tag{13.19}
\end{align*}

Operators act on bras in a similar manner

\(\langle k| \widehat{O}^\dagger = \langle q | \tag{13.20}\)

Equations 13.19 and 13.20 and the requirement that kets be normalized allow us to derive an important property of operators. Multiplying both equations we obtain

\begin{align*}
\langle k|\widehat{O}^{\dagger} \widehat{O}| k\rangle &=\langle q \mid q\rangle \tag{13.21}\\
\langle k|\widehat{O}^{\dagger} \widehat{O}| k\rangle &=1, \tag{13.22}
\end{align*}

the second line follows from assuming that \(\widehat{O}\) preserves the normalization of the ket, and since \(\langle k | k\rangle\) = 1, it implies that \(\widehat{O}^{\dagger} \widehat{O} = \mathbb{I}\). Operators that have this property are said to be unitary, and their inverse is equal to their adjoint. All quantum mechanical operators must be unitary, or else, the normalization of the probability distribution would not be preserved by the transformations of the ket. Note that this is the exact same reasoning we employed to require that time evolution be unitary back in Chapter 10. From the physical standpoint unitarity means that doing and then undoing the operation defined by \(\widehat{O}\) should leave us with the same we had in origin (Note the similarity with the definition of reversibility).

There is an easy way to construct an operator if we know the input and output kets. We can use the exterior product, that is, the product of a column vector by a row vector (the dot product is often also called inner or interior product, hence the name of exterior product). We can construct the operator \(\widehat{O}\) using the exterior product of a ket by a bra

\[\widehat{O}|k\rangle=(|q\rangle\langle k|)\;|k\rangle=|q\rangle\langle k \mid k\rangle=|q\rangle \tag{13.23} \]

note that this would not be possible if the kets were not normalized to 1; another way to put it is that the normalization of the ket inforces the fact that operators built in this way are unitary.

For example, to transform a qubit in the state \(| 0\rangle\) into the qubit in the state \(| \psi \rangle\) defined above, we construct the operator

\[\widehat{O}_{2}=\alpha \;|\;0\rangle\langle 0\;|+\beta| \;1\rangle\langle 0\;| \tag{13.24} \]

we can verify that this operator produces the expected result

\begin{align*}
\widehat{O}_{2}|\;0\rangle &=\alpha\;|\;0\rangle\langle 0 \mid 0\rangle+\beta\;|\;1\rangle\langle 0 \mid 0\rangle \\
&=\alpha\;|\;0\rangle 1+\beta\;|\;1\rangle 1 \\
&=\alpha\;|\;0\rangle+\beta\;|\;1\rangle \\
&=|\;\psi\rangle \tag{13.25}
\end{align*}

we have just performed our first quantum computation!

In quantum mechanics books it is custommary to drop the hat from the operators \((\widehat{O} → O)\) to “simplify notation.” Often at an introductory level (and an advanced level as well), this simplification causes confusion between operators and scalars; in these notes we will try to avoid doing so.

---

\(^2\)This operation is known by several different names, including “complex transpose” and “adjoint.”

\({ }^{3}\) It may not be immediately obvious that \(\int \Psi^{*} \Psi d x\) is a dot product. To see that it is, discretize the integral \(\int \Psi^{*} \Psi d x \rightarrow\) \(\sum_{i} \Psi_{i}^{*} \Psi_{i}\) and compare to the definition of dot product. You may argue that in doing so we have transformed a function \(\Psi\) into a vector with elements \(\Psi_{i} ;\) but we defined a ket as a vector to relate it to linear algebra. If the ket were to represent a function, then the appropriate definition of the dot product would be \(\langle\Phi \mid \Psi\rangle=\int \Phi^{*} \Psi d x\).