Section 1.7 Qubit States
Subsection 1.7.1 Qubits
As a reminder, a qubit exists in a superposition between the states \(\ket{0}\) and \(\ket{1}\text{,}\) which means that they can be expressed as a linear combination of the vectors \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\begin{pmatrix} 0 \\ 1 \end{pmatrix} \text{.}\) This means a qubit \(\psi \) can be expressed
\begin{equation*}
\ket{\psi} = \alpha \ket{0} + \beta \ket{1}
\end{equation*}
Where \(\alpha \) and \(\beta \) are complex coefficiants that relate to the probability that a qubit is in the states \(\ket{0}\) and \(\ket{1} \) respectively. This means that qubits exist within a complex vector space. The exact space that qubits exist within is known as a hilbert Space.
Subsection 1.7.2 Hilbert Space
A hilbert space is a complex vector space in which the inner product is defined as an operation. A single qubit exists within the hilbert space \(\mathbb{C}^2\text{,}\) which is the set of all two dimensional vectors with complex entries.
Subsection 1.7.3 Bra Vectors and the Inner Product for Quantum Computation
We now introduce the bra in Bra-Ket Notation. Each vector representation of a qubit has a corresponding bra vector (sometimes also referred to as the "Dual Vector") that is equal to the complex conjugate of the transpose of the ket. This operation of taking the transpose and complex conjugating is known as the adjoint and is represented by the \(\dagger\) symbol. Bra vectors are represented by the symbol \(\bra{}\text{.}\) An example of finding the bra vector for the ket \(\psi\) is shown below (where the \(*\) symbol represents complex conjugating).
\begin{equation*}
\ket{\psi}^{\dagger} = (\alpha \ket{0} + \beta \ket{1})^{\dagger} = (\alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ 1 \end{pmatrix})^{\dagger} = \alpha^{*} \begin{pmatrix} 1 & 0 \end{pmatrix} + \beta^{*} \begin{pmatrix} 0 & 1 \end{pmatrix} = \bra{\psi}
\end{equation*}
Bra vectors allow us to use a new notation for the inner product. We write the inner product between two vectors, \(\ket{\psi}\) and \(\ket{\phi}\text{,}\) as \(\braket{\psi | \phi}\) and perform the calculation as matrix multiplication, as described in Subsection 1.5.4
Find the inner product between the following two qubits.
\begin{equation*}
\ket{\psi} = \frac{1}{2} \ket{0} + \frac{\sqrt{3}}{2} \ket{1} = \begin{pmatrix} \frac{1}{2} \\ \frac{\sqrt{3}}{2} \end{pmatrix}
\end{equation*}
\begin{equation*}
\ket{\phi} = \frac{1}{\sqrt{2}} \ket{0} + \frac{1}{\sqrt2} \ket{1} = \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix}
\end{equation*}
Solution.
\begin{equation*}
\ket{\psi} \cdot \ket{\phi} = \braket{\psi|\phi} = \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{pmatrix} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1 + \sqrt3}{2\sqrt{2}}
\end{equation*}
Subsection 1.7.4 Qubit Measurements
When a qubit is measured, it collapses from being a superposition of the states \(\ket{0}\) and \(\ket{1}\) to being in one state or the other. The probability that the qubit will be in either state is related to the coefficient on that state. Consider a qubit \(\psi \text{,}\)
\begin{equation*}
\ket{\psi} = \alpha \ket{0} + \beta \ket{1} = \alpha \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}
\end{equation*}
Since all qubits are normalized, \(\psi\) has a length of \(1\text{,}\) so \(\sqrt{\alpha^2 + \beta^2}=1\text{,}\) which means \(\alpha^2 + \beta^2=1\text{.}\) Since the combined probability that the qubit will be in the state \(\ket{0}\) or \(\ket{1}\) is equal to \(1\text{,}\) we use the normal property of qubits as a method to determine the probability that the qubit will be in either state. For the qubit \(\psi\) the probability that it will be in the state \(\ket{0}\) after it is measured is \(\alpha^2\) and the probability that it will be in the state \(\ket{1}\) is \(\beta^2\text{.}\)
Born Rule: The Born Rule tells us that if we express a qubit as a linear combination of basis states, then upon measurement, the probability that the qubit collapses into any given basis state is equal to the square of the coefficient for that basis state. A qubit \(\ket{\psi}\) in a vector space defined by the basis \(\{ \ket{\beta_1},\ket{\beta_2},\ldots,\ket{\beta_n} \} \)
\begin{equation*}
\ket{\psi} = \alpha_1 \ket{\beta_1} + \alpha_2 \ket{\beta_2} + \ldots + \alpha_n \ket{\beta_n}
\end{equation*}
The probability that \(\ket{\psi}\) will collapse into the basis state \(\ket{\beta_i}\) is given by \(\alpha_i^2\text{.}\) Furthermore, since qubits are normalized
\begin{equation*}
\sum_{i=1}^n P(\beta_i) = \sum_{i=1}^n \alpha_i^2 = 1
\end{equation*}
Checkpoint 1.7.2.
For the following qubit, find the probability that it will be in the state \(\ket{0}\) and the probability that it will be in the state \(\ket{1}\) after measurement
\begin{equation*}
\ket{\psi} = \frac{1}{2} \ket{0} + \frac{\sqrt{3}}{2} \ket{1}
\end{equation*}
Solution.
The probability that \(\ket{\psi}\) will be in the state \(\ket{0}\) after measurement is 25% because \((\frac{1}{2})^2 = 1/4 = 0.25\text{.}\) The probability that \(\ket{\psi}\) will be in the state \(\ket{1}\) after measurement is 75% because \((\frac{\sqrt{3}}{2})^2 = 3/4 = 0.75\text{.}\)
Another way to find the probability that a qubit is in any given state is to take the square of the inner product of the outcome state and the qubit. For a qubit \(\ket{\psi}\text{,}\) the probability that it is in the state \(\ket{0}\) after measurement would be \((\braket{0|\psi})^2\text{,}\) the probability that it is in the state \(\ket{1}\) would be \((\braket{1|\psi})^2\text{,}\) and the probability that it is in some mixed state \(\ket{x}\) would be \((\braket{x|\psi})^2\text{.}\)
Checkpoint 1.7.3.
Using an inner product, calculate the probability that the following system is in the state \(\ket{0}\) and \(\ket{1}\text{.}\)
\begin{equation*}
\ket{\psi} = \frac{\sqrt{7}}{\sqrt{8}} \ket{0} + \frac{1}{\sqrt{8}} \ket{1} = \begin{pmatrix} \frac{\sqrt{7}}{\sqrt{8}} \\ \frac{1}{\sqrt{8}} \end{pmatrix}
\end{equation*}
Solution.
\begin{equation*}
P(0) = (\braket{0|\psi})^2 = (\begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{7}}{\sqrt{8}} \\ \frac{1}{\sqrt{8}} \end{pmatrix}+0)^2 = (\frac{\sqrt{7}}{\sqrt{8}})^2 = 7/8 = 0.875
\end{equation*}
The probability that \(\ket{\psi}\) will collapse into the state \(\ket{0}\) is 87.5%.
\begin{equation*}
P(1) = (\braket{1|\psi})^2 = (\begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{\sqrt{7}}{\sqrt{8}} \\ \frac{1}{\sqrt{8}} \end{pmatrix})^2 = (0+\frac{1}{\sqrt{8}})^2 = 1/8 = 0.125
\end{equation*}
The probability that \(\ket{\psi}\) will collapse into the state \(\ket{1}\) is 12.5%.
