192.036 Introduction to Quantum Computing

1. Mathematical and Quantum Mechanical Preliminaries:
   • Complex Number
     • Imaginary Unit
     • Complex Modulus
     • Complex-Valued Function
   • Hilbert Space
   • L2 Space
     • Square-Integrable Function
   • Norm
   • Cauchy Sequence
   • Convergent Sequence
   • Limit
   • Banach Space
   • Inner Product
   • Inner Product Space
   • Orthonormal Basis
   • Linear Operator
   • Adjoint Operator
   • Hermitian Operator
   • Unitary Operator
   • Orthogonal Projection
   • Idempotent Operator
   • Eigenvector
   • Tensor Product
   • Bra-Ket Notation
   • Spectral Representation
   • Operator
   • Expected Value

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Exercise Sheet 1

Exercise 1

Show that the Pauli matrices

$$X:=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),Y:=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),Z:=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right),$$

are Hermitian, unitary, and that they square to the identity.

Solution:

$$\begin{array}{rl}{X}^{\ast }& =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\\ Y^{\ast }& =\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\\ Z^{\ast }& =\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\end{array}$$

Given that $\forall A\in \{X,Y,Z\}:A=A^{\ast }$, all three Pauli matrices are hermitian.

$$\begin{array}{rl}X{X}^{\ast }& =X^{\ast }X={\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)}^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2\\ Y{Y}^{\ast }& =Y^{\ast }Y={\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)}^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2\\ Z{Z}^{\ast }& =Z^{\ast }Z={\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)}^2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2\end{array}$$

Given that all matrices are Hermitian and $\forall A\in \{X,Y,Z\}:X^2=I_2$, all three Pauli matrices are unitary.

Exercise 2

Show:

1. The eigenvalues of a self-adjoined operator are always real.
2. Projection operators have only $0$ and $1$ as eigenvalues.

Solution:

Let $T$ be a self-adjoined operator, i.e. $T=T^{\ast }$. Assume there exists an eigenvalue $\lambda$ of $T$ such that $\lambda$ is not equal to its complex conjugate $\overline{\lambda }$:

$$\lambda \ne \overline{\lambda }$$

By definition, there is a corresponding eigenvector for each $\lambda$ such that:

$$Tx=\lambda x$$

Given that $T$ is self-adjoined, the following holds:

$$⟨Ty,z⟩=⟨T^{\ast }y,z⟩=⟨y,T^{\ast }z⟩=⟨y,Tz⟩\forall y,z\in \mathcal{H}$$

Choose $y=z=x$ and substituting $Tx=\lambda x$ yields:

$$⟨\lambda x,x⟩=⟨x,\lambda x⟩$$

Rewriting it into Bra-Ket notation:

$$⟨\lambda x,x⟩=⟨\overline{\lambda }x|x⟩$$