8  Ket Notation

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 \[ \ket{x} \coloneqq e_i = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix} \leftarrow \text{1 at the $i$-th position}. \] In the example, we would therefore write \(\ket{\text{red,4,top}}\) for \(e_{123}\).

Example: Ket notation

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: \[ \begin{aligned} \ket{+} &\coloneqq \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt{2}} \ket{1},\\ \ket{-} &\coloneqq \frac{1}{\sqrt{2}} \ket{0} - \frac{1}{\sqrt{2}} \ket{1}. \end{aligned} \] Which of the meanings of the symbol \(\ket{}\) is meant has to be deduced from the context.

8.1 Teleportation

We take another look at the example from the last chapter with ket notation.

Example: Teleportation

Once again, Alice has the qubit \(\psi\) and Alice and Bob have shared the state \(\beta_{00}\).

Circuit for qubit teleportation

  1. 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 \[ \begin{aligned} \ket{\phi_1} &= \ket{\psi} \otimes \ket{\beta_{00}} \\ &= (\alpha\ket{0} + \beta\ket{0}) \otimes \Bigl(\tfrac{1}{\sqrt{2}}\ket{00} + \tfrac{1}{\sqrt{2}} \ket{11}\Bigr) \\ &= \tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{100} + \tfrac{\beta}{\sqrt{2}} \ket{111}. \end{aligned} \]

  2. We can now translate each ket notation individually and get the result much simpler: \[ \begin{aligned} \ket{\phi_2} &= (\operatorname{CNOT} \otimes \operatorname{I}_2) \ket{\phi_1} t \\ &= (\operatorname{CNOT} \otimes \operatorname{I}_2) \Bigl(\tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{100} + \tfrac{\beta}{\sqrt{2}} \ket{111}\Bigr) \\ &= \phantom{(\operatorname{CNOT} \otimes \operatorname{I}_2) \bigl(} \tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{110} + \tfrac{\beta}{\sqrt{2}} \ket{101}. \end{aligned} \]

  3. Identical to step 2, we can look at each ket notation individually: \[ \begin{aligned} \ket{\phi_3} &= (\operatorname{H} \otimes \operatorname{I}_4) \ket{\phi_2} \\ &= (\operatorname{H} \otimes \operatorname{I}_4) \Bigl(\tfrac{\alpha}{\sqrt{2}} \ket{000} + \tfrac{\alpha}{\sqrt{2}} \ket{011} + \tfrac{\beta}{\sqrt{2}} \ket{101} + \tfrac{\beta}{\sqrt{2}} \ket{110}\Bigr) \\ &= \tfrac{\alpha}{\sqrt{2}} (H\ket{0} \otimes \ket{00}) + \tfrac{\alpha}{\sqrt{2}} (H\ket{0} \otimes \ket{11}) + \tfrac{\beta}{\sqrt{2}} (H\ket{1} \otimes \ket{01}) + \tfrac{\beta}{\sqrt{2}} (H\ket{1} \otimes \ket{10}) \\ &= \tfrac{\alpha}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} + \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{00}\biggr) + \tfrac{\alpha}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} + \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{11}\biggr) + \\ & \quad\:\, \tfrac{\beta}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} - \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{01}\biggr) + \tfrac{\beta}{\sqrt{2}} \biggl(\Bigl(\tfrac{1}{\sqrt{2}}\ket{0} - \tfrac{1}{\sqrt{2}} \ket{1}\Bigr) \otimes \ket{10}\biggr) \\ &= \tfrac{\alpha}{2} \ket{000} + \tfrac{\alpha}{2} \ket{100} + \tfrac{\alpha}{2} \ket{011} + \tfrac{\alpha}{2} \ket{111} + \tfrac{\beta}{2} \ket{001} - \tfrac{\beta}{2} \ket{101} + \tfrac{\beta}{2} \ket{010} - \tfrac{\beta}{2} \ket{110}. \end{aligned} \]

  4. 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 \[ \ket{\phi_4} = \frac{\frac{\alpha}{2} \ket{011}}{\sqrt{\tfrac{1}{4}}} + \frac{\frac{\beta}{2} \ket{011}}{\sqrt{\tfrac{1}{4}}} = \alpha \ket{011} + \beta \ket{011} = \ket{01} \otimes (\alpha \ket{1} + \beta \ket{0}). \]

  5. Since \(b = 1\), the Pauli-matrix \(\operatorname{X}\) is used: \[ \begin{aligned} \ket{\phi_5} &= (\operatorname{I}_4 \otimes \operatorname{X}) \ket{\phi_4} \\ &= (\operatorname{I}_4 \otimes \operatorname{X}) \Bigl(\ket{01} \otimes (\alpha \ket{1} + \beta \ket{0})\Bigr) \\ &= \operatorname{I}_4\ket{01} \otimes \Bigl(\operatorname{X}(\alpha \ket{1} + \beta \ket{0})\Bigr) \\ &= \ket{01} \otimes (\alpha\operatorname{X}\ket{1} + \beta\operatorname{X}\ket{0}) \\ &= \ket{01} \otimes (\alpha\ket{0} + \beta\ket{1}). \end{aligned} \]

  6. 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.)