Chapter 26: Problem 28
For this problem, we return to the Dirac bra and ket notation. Let \(\mathbf{T}\) be an isometry in the real vector space \(\mathcal{V}\). Then \(|y\rangle=(\mathbf{T}-\mathbf{1})|x\rangle\) is the vector, which, in three- dimensions, connects the tip of \(|x\rangle\) to its isometric image. (a) Show that \(\langle y \mid y\rangle=2\langle x|(\mathbf{1}-\mathbf{T})| x\rangle\). (b) Show that $$ \mathbf{P}_{y}=(\mathbf{T}-\mathbf{1}) \frac{|x\rangle\langle x|}{2\langle x|(\mathbf{1}-\mathbf{T})| x\rangle}\left(\mathbf{T}^{t}-\mathbf{1}\right) $$ and $$ \mathbf{R}_{y}=\mathbf{1}-(\mathbf{T}-\mathbf{1}) \frac{|x\rangle\langle x|}{\langle x|(\mathbf{1}-\mathbf{T})| x\rangle}\left(\mathbf{T}^{t}-\mathbf{1}\right) . $$ (c) Verify that \(\mathbf{R}_{y}|x\rangle=\mathbf{T}|x\rangle\), as we expect.
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.