So far we have only seen vectors as a way to mathematically describe a quantum state. This can get quite inconvenient if the vector get bigger and also often contains not that much useful information (e.g. a lot of \(0\) entries). We therefore introduce a new form of writing quantum states called the ket notation.
The idea works as follows: We can rewrite a quantum state \(\psi\) in the following way \[
\psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \\ \psi_N \end{pmatrix} = \psi_1 \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} + \psi_2 \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix} + \dots + \psi_N \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} = \sum_{i=1}^{N} \psi_i \cdot e_i.
\] The vector \(e_i\) denotes the vector with 0 entries at every position except the \(i\)-th position, where the entry is \(1\).
From this notation we already get an advantage, since we can drop out all \(0\)-entries. But we still have no intuitive mapping from the vector \(e_i\) to the classical possibility represented by \(e_i\). For example, \(e_{123}\) can represent the classical possibility “red,4,top”. For this we use a \(\ket{}\) symbol. More precise this means for a classical possibility \(x\), which is the \(i\)-th possibility and is represented by \(e_i\), we write \[
\ket{x}:= e_i = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \leftarrow \text{1 at the $i$-th position}.
\] In the example, we would therefore write \(\ket{\text{red,4,top}}\) for \(e_{123}\).
Given a quantum system with the classical possibilities \(00\), \(01\), \(10\) and \(11\), the quantum state \(\psi = \begin{pmatrix} \frac{1}{\sqrt{2}} & 0 & 0 & \frac{1}{\sqrt{2}} \end{pmatrix}^T\) can be written as \[
\psi = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ 0 \\ \frac{1}{\sqrt{2}} \end{pmatrix} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{\sqrt{2}} \ket{11}.
\]
Written like this, we can see at first glance that this is a superposition of the classical possibilities \(00\) and \(11\). Writing \(\psi = \frac{1}{\sqrt{2}} e_1 + \frac{1}{\sqrt{2}} e_4\) would be less obvious.
Note that the ket notation can also be used in a few other ways. We can use it as described above to the state \(\ket{x}\) corresponding to the classical possibility \(x\), but we also use it to emphasize that \(\psi\) is a quantum state by writing \(\ket{\psi}\) (here \(\psi\) is not a classical possibility). We also have two special cases \(\ket{+}\) and \(\ket{-}\) which are defined as follows: \[
\begin{aligned}
\ket{+} &:= \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1},\\
\ket{-} &:= \frac{1}{\sqrt{2}} \ket{0} - \frac{1}{\sqrt{2}} \ket{1}.
\end{aligned}
\] Which of the meanings of the symbol \(\ket{}\) is meant has to be deduced from the context.
Teleportation
We take another look at the example from the last chapter with ket notation.
Once again, Alice has the qubit \(\psi\) and Alice and Bob have shared the state \(\beta_{00}\).
Alice has the state \(\ket{\psi} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \alpha\ket{0} + \beta\ket{1}\) and the shared state is \(\ket{\beta_{00}} = \begin{pmatrix} \tfrac{1}{\sqrt{2}} & 0 & 0 & \tfrac{1}{\sqrt{2}} \end{pmatrix}^T = \tfrac{1}{\sqrt{2}} \ket{00} + \tfrac{1}{\sqrt{2}} \ket{11}\). This means that the entire state is \[
\begin{aligned}
\ket{\phi_1} &= \ket{\psi} \otimes \ket{\beta_{00}} \\
&= (\alpha\ket{0} + \beta\ket{0}) \otimes \Bigl(\tfrac{1}{\sqrt{2}}\ket{00} + \tfrac{1}{\sqrt{2}} \ket{11}\Bigr) \\
&= \tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{100} + \tfrac{\beta}{\sqrt{2}} \ket{111}.
\end{aligned}
\]
We can now translate each ket notation individually and get the result much simpler: \[
\begin{aligned}
\ket{\phi_2} &= (\operatorname{CNOT} \otimes \operatorname{I}_2) \ket{\phi_1} t \\
&= (\operatorname{CNOT} \otimes \operatorname{I}_2) \Bigl(\tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{100} + \tfrac{\beta}{\sqrt{2}} \ket{111}\Bigr) \\
&= \phantom{(\operatorname{CNOT} \otimes \operatorname{I}_2) \bigl(} \tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{110} + \tfrac{\beta}{\sqrt{2}} \ket{101}.
\end{aligned}
\]
Identical to step 2, we can look at each ket notation individually: \[
\begin{aligned}
\ket{\phi_3} &= (\operatorname{H} \otimes \operatorname{I}_4) \ket{\phi_2} \\
&= (\operatorname{H} \otimes \operatorname{I}_4) \Bigl(\tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{101} + \tfrac{\beta}{\sqrt{2}} \ket{110}\Bigr) \\
&= \tfrac{\alpha}{\sqrt{2}} (H\ket{0} \otimes \ket{00}) + \tfrac{\alpha}{\sqrt{2}} (H\ket{0} \otimes \ket{11}) + \tfrac{\beta}{\sqrt{2}} (H\ket{1} \otimes \ket{01}) + \tfrac{\beta}{\sqrt{2}} (H\ket{1} \otimes \ket{10}) \\
&= \tfrac{\alpha}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} + \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{00}\biggr) + \tfrac{\alpha}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} + \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{11}\biggr) + \\
& \quad\:\, \tfrac{\beta}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} - \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{01}\biggr) + \tfrac{\beta}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} - \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{10}\biggr) \\
&= \tfrac{\alpha}{2} \ket{000} + \tfrac{\alpha}{2} \ket{100} + \tfrac{\alpha}{2} \ket{011} + \tfrac{\alpha}{2} \ket{111} + \tfrac{\beta}{2} \ket{001} - \tfrac{\beta}{2} \ket{101} + \tfrac{\beta}{2} \ket{010} - \tfrac{\beta}{2} \ket{110}.
\end{aligned}
\]
We again assume that \(a=0\) and \(b=1\) and therefore only \(\frac{\alpha}{2} \ket{011}\) and \(\frac{\beta}{2} \ket{010}\) are relevant. It applies \(|\alpha|^2 + |\beta|^2 = 1\) because \(\psi\) is a quantum state. Therefore the probability for this is \[
\left|\frac{\beta}{2}\right|^2 + \left|\frac{\alpha}{2}\right|^2 = \frac{|\alpha|^2 + |\beta|^2}{4} = \frac{1}{4}
\] and the post-measurement-state \[
\ket{\phi_4}
= \frac{\frac{\alpha}{2} \ket{011}}{\sqrt{\tfrac{1}{4}}} + \frac{\frac{\beta}{2} \ket{011}}{\sqrt{\tfrac{1}{4}}}
= \alpha \ket{011} + \beta \ket{011}
= \ket{01} \otimes (\alpha \ket{1} + \beta \ket{0}).
\]
Since \(b = 1\), the Pauli-matrix \(\operatorname{X}\) is used: \[
\begin{aligned}
\ket{\phi_5} &= (\operatorname{I}_4 \otimes \operatorname{X}) \ket{\phi_4} \\
&= (\operatorname{I}_4 \otimes \operatorname{X}) \Bigl(\ket{01} \otimes (\alpha \ket{1} + \beta \ket{0})\Bigr) \\
&= \operatorname{I}_4\ket{01} \otimes \Bigl(\operatorname{X}(\alpha \ket{1} + \beta \ket{0})\Bigr) \\
&= \ket{01} \otimes (\alpha\operatorname{X}\ket{1} + \beta\operatorname{X}\ket{0}) \\
&= \ket{01} \otimes (\alpha\ket{0} + \beta\ket{1}).
\end{aligned}
\]
Since \(a = 0\), nothing happens in this step: \[
\ket{\phi_6}
= \ket{\phi_5}
= \ket{01} \otimes (\alpha\ket{0} + \beta\ket{1})
= \ket{01} \otimes \ket{\psi}.
\]
As expected, we get the same result as in the previous chapter.
The ket notation can save a lot of work and sources of error. (Keep in mind that the example here got a bit lengthy because we wrote out a lot of intermediate steps.)