Question #146817
Are there values of (a) belongs to C for which the matrix [
1 0 0
0 -1/√2 1/√2
0 1/√2. a
]
is unitary? Justify your answer
1
Expert's answer
2020-12-01T15:52:46-0500

Yes.LetA=(10001212012a)=(1+0i0+0i0+0i0+0i12+0i12+0i0+0i12+0ia+0i)Conjugate ofA=Ac=(10001212012a)=(10i00i00i00i120i120i00i120ia0i)=(10001212012a)Conjugate transpose ofA=A=(10001212012a)AA=A2=(10001a2120a212a2+12)By definitionAA=(100010001)Comparing both elementsa2+12=1a2=12a=±12=±22a212=0a2=12a=22The positive value ofasatisfiesthe definition of a unitary matrix.The positive value satisfies theidentity matrix because onsubstituting the negative value ofa,we do not have an identity matrix,the1son substitution ofthe positive valueabecomes1on the substitution of the negative valueofa.Also,22C.\displaystyle\textbf{\textsf{Yes.}}\\ \textsf{Let}\, A = \begin{pmatrix} 1 & 0 & 0\\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & a \end{pmatrix} = \begin{pmatrix} 1 + 0i & 0 + 0i & 0 + 0i\\ 0 + 0i & -\frac{1}{\sqrt{2}} + 0i & \frac{1}{\sqrt{2}} + 0i\\ 0 + 0i & \frac{1}{\sqrt{2}} + 0i & a + 0i \end{pmatrix}\\ \textsf{Conjugate of}\, A = A^{c} = \begin{pmatrix} 1 & 0 & 0\\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & a \end{pmatrix} =\begin{pmatrix} 1 - 0i & 0 - 0i & 0 - 0i\\ 0 - 0i & -\frac{1}{\sqrt{2}} - 0i & \frac{1}{\sqrt{2}} - 0i\\ 0 - 0i & \frac{1}{\sqrt{2}} - 0i & a - 0i \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ 0 & \frac{1}{\sqrt{2}} & a \end{pmatrix}\\ \textsf{Conjugate transpose of}\, A = A^{\ast} = \begin{pmatrix} 1 & 0 & 0\\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ 0 & \frac{1}{\sqrt{2}}& a \end{pmatrix}\\ AA^{\ast} = A^2 = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & \frac{a}{\sqrt{2}} - \frac{1}{2}\\ 0 & \frac{a}{\sqrt{2}} - \frac{1}{2}& a^2 + \frac{1}{2} \end{pmatrix}\\ \textsf{By definition}\, AA^{\ast} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}\\ \textsf{Comparing both elements}\\ a^2 + \frac{1}{2} = 1 \\ a^2 = \frac{1}{2} \\ a = \pm\frac{1}{\sqrt{2}} = \pm\frac{\sqrt{2}}{2}\\ \frac{a}{\sqrt{2}} - \frac{1}{2} = 0\\ \frac{a}{\sqrt{2}} = \frac{1}{2} \\ a = \frac{\sqrt{2}}{2}\\ \textsf{The positive value of}\, a \,\,\textsf{satisfies}\\ \textsf{the definition of a unitary matrix}.\\ \textsf{The positive value satisfies the}\\ \textsf{identity matrix because on}\\ \textsf{substituting the negative value of}\, a, \\ \textsf{we do not have an identity matrix,}\\ \textsf{the}\, 1's \, \textsf{on substitution of}\\ \textsf{the positive value}\,a\, \textsf{becomes}\, -1\\ \textsf{on the substitution of the negative value}\\ \textsf{of}\,a.\\ \textsf{Also,}\,\,\frac{\sqrt{2}}{2}\in \mathbb{C}.


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