Quantum Fundamentals: NoTerm-2022 Matrix Practice : Due Day 8 W 2/16 Mathbits
Matrix Refresher
S0 4365S
Calculate the following quantities for the matrices:
\[A\doteq
\begin{pmatrix}
1&0&0\\ 0&0&1\\ 0&-1&0\\
\end{pmatrix}
\hspace{2em}
B\doteq
\begin{pmatrix}
a&b&c\\ d&e&f\\ g&h&j\\
\end{pmatrix}
\hspace{2em}
C\doteq
\begin{pmatrix}
\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\
\end{pmatrix}
\]
and the vector:
\[\left|D\right\rangle\doteq
\begin{pmatrix}
1\\ i\\ -1\\
\end{pmatrix}
\hspace{2em}
\]
\(AB\)
\({\rm tr} (B)\)
\(A\vert D\rangle\)
\(\det(\lambda{\cal I}-A)\) where \(\lambda\) is a scalar.
\(C^{-1}\) (Hint: Geometrically, what is the \(C\) transformation? What transformation undoes what \(C\) does?)
Pauli Practice
S0 4365S
The Pauli spin matrices \(\sigma_x\), \(\sigma_y\), and \(\sigma_z\) are
defined by:
\[\sigma_x=
\begin{pmatrix}
0&1\\ 1&0\\
\end{pmatrix}
\hspace{2em}
\sigma_y=
\begin{pmatrix}
0&-i\\ i&0\\
\end{pmatrix}
\hspace{2em}
\sigma_z=
\begin{pmatrix}
1&0\\ 0&-1\\
\end{pmatrix}
\]
These matrices are related to angular momentum in
quantum mechanics. Prove, and become familiar with, the identities
listed below.
Show that each of the Pauli matrices is hermitian. (A matrix is
hermitian if it is equal to its hermitian adjoint.)
Show that the determinant of each of the Pauli matrices is \(-1\).
Show that \(\sigma_i^2={\cal I}\) for each of the Pauli matrices,
i.e. for \(i\in\left\{x,y,z\right\}\).
Hermitian Adjoints
S0 4365S
Calculate the following quantities for the matrices:
\[A\doteq
\begin{pmatrix}
1&0&0\\ 0&0&1\\ 0&-1&0\\
\end{pmatrix}
\hspace{2em}
B\doteq
\begin{pmatrix}
a&b&c\\ d&e&f\\ g&h&j\\
\end{pmatrix}
\hspace{2em}
C\doteq
\begin{pmatrix}
\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\\
\end{pmatrix}
\]
and the vectors:
\[\left|D\right\rangle\doteq
\begin{pmatrix}
1\\ i\\ -1\\
\end{pmatrix}
\hspace{2em}
\left|E\right\rangle\doteq
\begin{pmatrix}
1\\ i\\
\end{pmatrix}
\hspace{2em}
\left|F\right\rangle\doteq
\begin{pmatrix}
1\\ -1\\
\end{pmatrix}
\]
\(A^{\dagger}\)
\(\vert E\rangle^{\dagger}\equiv\langle E\vert\)
\(\langle D\vert A\vert D\rangle\)
\(\left(A\vert D\rangle\right)^{\dagger}\)
Using explicit matrix multiplication (without using a theorem)
verify that \(\left(A\vert D\rangle\right)^{\dagger}
=\langle D\vert
A^{\dagger}\)