14  From Quantum Physics to a Quantum Computer

In the previous chapter, we have introduced the fundamentals of quantum physics. We now relate this to quantum computers.

As an example, we consider the construction of a simple hypothetical single-qubit quantum computer. We encode the qubit using the excitation of an electron. For this purpose, we assume that the electron has only the two basis states \(\ket{0}\) and \(\ket{1}\), where \(\ket{0}\) represents the ground state. The states have the energy levels \(\pi\) and \(2 \pi\).

To fully construct our one qubit quantum computer, we need to be able to perform three basic operations:

  1. We need to initialize our qubit with \(\ket{0}\). This is also called cooling.
  2. We need to be able to measure the qubit.
  3. We need to be able to apply a unitary on the qubit.

We look into how to construct a unitary. The cooling and measuring operations are out of scope of this chapter.

The Hamiltonian \(H\) of our single-qubit quantum computer is given by: \[ H = \begin{pmatrix} \pi & 0 \\ 0 & 2\pi \end{pmatrix}. \]

According to Section 13.2 (with our assumption \(\hbar = 1\)), the time evolution of our quantum computer is: \[ \begin{aligned} \ket{0} \mapsto e^{-i \pi t} \ket{0}, \\ \ket{1} \mapsto e^{-i 2\pi t} \ket{1}. \end{aligned} \] In other words, after time \(t\) the quantum state is multiplied with the unitary \[ U_t = \begin{pmatrix} e^{-i \pi t} & 0 \\ 0 & e^{-i 2\pi t} \end{pmatrix} = e^{-i H t}. \] Thus, waiting for a duration \(t\) is equivalent to applying the unitary \(U_t\).

As an example, waiting for one second corresponds to: \[ U_1 = \begin{pmatrix} e^{-i \pi} & 0 \\ 0 & e^{-i 2\pi} \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} = -Z, \] where \(Z\) is the Pauli-\(Z\)-gate.

This means that the unitary \(-Z\) is applied every second “automatically”. The factor \(-1\) does not make a physical difference as it is a global phase factor, so the \(-Z\) is physically equal to the Pauli-\(Z\)-gate.

How can we obtain new unitaries beyond just \(Z\)?

One solution is to apply external energy to the system, for example shining a laser on the atom, applying a magnetic field, \(\dots\). This modifies the system’s Hamiltonian from \(H\) to \(H + \delta H\).

As an example, we consider \[ \delta H = \begin{pmatrix} 0 & 1000 \\ 1000 & 0 \end{pmatrix}. \] \(\delta H\) is much bigger than \(H\), so \(H + \delta H \approx \delta H\).

Then after time \(t\), the time evolution unitary becomes \[ U_t' = e^{-i \frac{H + \delta H}{\hbar} t} \approx e^{-i \frac{\delta H}{\hbar} t} \eqcolon U_t^{\text{approx}}. \]

For \(t = \frac{\pi}{2000}\), this gives \[ U_\frac{\pi}{2000}^{\text{approx}} = \begin{pmatrix} 0 & -i \\ -i & 0 \end{pmatrix} = -i \cdot X \] where \(X\) is the Pauli-\(X\)-gate. Again the factor \(-i\) does not make a physical difference as it is a global phase factor, so the \(-i X\) is physically equal to the Pauli-\(X\)-gate.

In reality, the exact time evolution unitary would be (up to a position of three digits) \[ U_\frac{\pi}{2000}' = \begin{pmatrix} 0.002i & -0.007 - 0.999 i \\ -0.007 - 0.999 i & -0.002 \end{pmatrix} \] which approximates the unitary \(-i X\) with only an error up to about \(0,2\%\).