1 Ket
A ket is simply a vector of complex numbers:
\[\begin{equation}|a\rangle = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \end{pmatrix}\end{equation}\]
Note that the argument of the ket is simply a label for the vector and doesn’t carry any intrinsic meaning. Writting
\(|\mathrm{boom!}\rangle\) or
\(|+\rangle\) is just as valid as writing
\(|\psi\rangle\). Sometimes the label will be used as part of the calculation so that you may find expressions like
\[\begin{equation*}\mathbf{A}|\alpha \psi\rangle = \alpha \mathbf{A}|\psi\rangle\end{equation*}\]
but bear in mind this is just assuming the additional notation
\(|\alpha \psi\rangle\equiv\alpha|\psi\rangle\) and is not something inherent to the braket notation. Often the label will be an index into a set of vectors like a basis.
2 Bra
The bra \(\langle a|\) is the transpose complex-conjugate of the ket vector, an operation that is usually denoted by a dagger symbol:
\[\begin{equation}\langle a| = \begin{pmatrix} a_1^* & a_2^* & \cdots \end{pmatrix} =
|a\rangle^\dagger\end{equation}\]
The complex-conjugation operation simply replaces each
\(i\) with a
\(-i\):
\((a+ib)^*=a-ib\), and the transpose makes each row a column.
Note that \(\langle a|\) is in a sense a different object from \(|a\rangle\). Whereas \(|a\rangle\) is a vector, \(\langle a|\) is sometimes called a co-vector, covariant vector, or a one-form. It lives in a dual space to the vector space that \(|a\rangle\) lives in.
3 Inner-product
Multiplying a bra and ket yields the usual inner product, and the resulting braket is the source of the whimsical names for the notation:
\[\begin{equation}\label{eq:braket}
\langle a | b\rangle = \begin{pmatrix} a_1^* & a_2^* & \cdots \end{pmatrix}\begin{pmatrix} b_1 \\ b_2 \\ \vdots \end{pmatrix} =
\sum_j a_j^* b_j = c\end{equation}\]
It is usual to drop the extra bar and have
\(\langle a | b\rangle\) instead of
\(\langle a||b\rangle\) and there is actually a subtle “syntatic-sugar” at work here.
\(\langle a | b\rangle\) is a complex number, whereas
\(\langle a||b\rangle\) is the dot product of two vectors—implicit in the notation is that the dot-product will produce a number. It’s these small tricks of perception that make the braket notation a very elegant representation of vectors. We’ll see some more below.
First let’s confirm that the braket really is an inner-product. There are only three requirements to check other that the inner product takes two vectors and produces a number, which is demonstrated in \eqref{eq:braket},
- Complex conjugation swaps the order:
\[\begin{equation*}\langle a | b\rangle^*=\langle a|^*|b\rangle^*=(|b\rangle^{*T}\langle a|^{*T})^T=\langle b | a\rangle^T=\langle b | a\rangle.\end{equation*}\]
- Linear in the second argument: say \(|c\rangle=\sum_j \gamma_j|c_j\rangle\)
\[\begin{equation*}\langle a | c\rangle = \langle a| \sum_j \gamma_j|c_j\rangle = \sum_j \gamma_j\langle a | c_j\rangle.\end{equation*}\]
(Some conventions have an inner product \((\cdot,\cdot)\) which is linear in its first argument, in which case \((b,a)\equiv\langle a | b\rangle\)). - Self-inner product is a non-negative real number:
\[\begin{equation*}\langle a | a\rangle = (a_1^* \ldots a_n^*)\cdot \begin{pmatrix} a_1\\ \vdots \\ a_n\end{pmatrix}
= |a_1|^2 + \ldots + |a_n|^2 \ge 0.\end{equation*}\]
Now that we have an inner product, we can describe vectors as being orthogonal if \(\langle a | b\rangle=0\), and normalised if \(\langle a | a\rangle=1\), and so we can define an orthonormal basis \(\{|a_j\rangle\}\) with the property that
\[\begin{equation}\label{eq:orthonormal}
\langle a_j | a_k\rangle = \delta_{jk}.\end{equation}\]
Any vector from that vector space can then be expanded in the basis
\[\begin{equation*}|b\rangle = \sum_j a_j |a_j\rangle.\end{equation*}\]
Note that we have used the symbol
\(a_j\) to denote two things in the above expression. The bare
\(a_j\) is the complex coefficient in the expansion, and the
\(a_j\) in the ket is a label for the basis vector. In practice, the context is more than sufficient to avoid ambiguity.
Since the \(\{|a_j\rangle\}\) basis we have introduced is orthonormal, we have
\[\begin{equation*}\langle a_k | b\rangle = a_k|a_k\rangle,\end{equation*}\]
so we can write the expansion in basis vectors as
\[\begin{equation}\label{eq:basisexpansion}
|b\rangle = \sum_j \langle a_j | b\rangle|a_j\rangle.\end{equation}\]
4 Tensor-product
We can define the product of two kets to be the tensor-product, with the result that it forms a larger ket. If the dimensions of \(|a\rangle\) and \(|b\rangle\) are \(m\) and \(n\) respectively, the dimension of the product will be \(mn\):
\[\begin{equation*}|a\rangle|b\rangle \equiv |a\rangle\otimes|b\rangle=(a_1b_1, a_1b_2, \dots, a_1b_n, a_2b_1, a_2b_2, \ldots, a_2b_n, \ldots, a_mb_n)^T\end{equation*}\]
A ket multiplying a bra forms a matrix and again you can imagine there is an implicit tensor-product between the two:
\[\begin{equation}|a \rangle\langle b| \equiv
\begin{pmatrix} b_1 \\ b_2 \\ \vdots \end{pmatrix}\otimes\begin{pmatrix} a_1^* & a_2^* & \cdots \end{pmatrix} =
\begin{pmatrix} b_1 a_1^* & b_1 a_2^* & \cdots \\ b_2 a_1^* & b_2 a_2^* & \cdots \\ \vdots & \vdots & \ddots \end{pmatrix}\end{equation}\]
Every matrix can be written as a sum of outer-products of this form. Its easiest to see this with the standard basis composed of vectors that have a single `1’ in the \(j^\mathrm{th}\) row: \(|e_j\rangle=(0,\ldots,1,\ldots,0)^T\). The matrix formed by \(|e_j \rangle\langle e_k|\) will have a single `1’ on the \(j^\mathrm{th}\) row and \(k^\mathrm{th}\) column. So
\[\begin{equation*}\mathbf{M} = \sum_{jk} m_{jk}|e_j \rangle\langle e_k|\end{equation*}\]
where
\(m_{jk}\) is the element on the
\(j^\mathrm{th}\) row and
\(k^\mathrm{th}\) column.
5 In the eye of the beholder
Notice that by judicious rearrangement of the elements we can split up an expression between matrices, vectors and numbers:
\[\begin{equation}|a \rangle\langle b||c\rangle = |a\rangle\langle b | c\rangle = \langle b | c\rangle |a\rangle.\end{equation}\]
This is where the notation comes into its own.
Have another look at the expansion of a vector in an orthonormal basis, Eq.\eqref{eq:basisexpansion}. With a trivial rearrangement we get:
\[\begin{equation*}|b\rangle = \sum_j \langle a_j | b\rangle|a_j\rangle = \sum_j|a_j \rangle\langle a_j||b\rangle.\end{equation*}\]
Since this is true for any
\(|b\rangle\), this is a proof that \begin{equation*} \sum_j|a_j \rangle\langle a_j| = I \end{equation*} for an orthonormal basis. A result that may not have been obvious but is very useful.