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I only know of these: http://www.youtube.com/watch?v=LYNOGk3ZjFM (PERIMETER INSTITUTE RECORDED SEMINAR ARCHIVE) http://www.youtube.com/watch?v=b5VUnapu-qs&list=PLiUVvsKxTUr66oLF6Pzirc1EgSstMbRZR (Indian University of Technology Madras) The first two I recommend because they are simply the same courses as the one that you are attending. The following ...

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Background You already seem to know this stuff but it's worth going over again. So, the adjoint of an operator is the equivalent effect of the operator on the other side of the wavefunction inner product: $$\langle \Phi | \hat A | \Psi\rangle = \int_{-\infty}^{\infty} dx~\Phi^*(x) ~ A[\Psi](x) = \int_{-\infty}^{\infty} dx~{A^\dagger[\Phi]}^*(x) ~ \Psi(x) ... 0 Intuitively, shifting then reflecting is not the same as reflecting then shifting. Consider the case of first shifting 1 unit to the right from 0, then reflecting: you end up at x=-1. If you reflect first, it does nothing, and then shifting to the right by 1 means you end up at x=1. The problem is that$$\hat{T}(-a)\hat{R}\psi(x) =\hat{T}(-a)\psi(-x) ...

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Let $\{|e_i\rangle: 1\leq i\leq n\}$ be a basis. Since $|e_1\rangle$ is a vector, and $\hat{T}$ is a linear transformation then $\hat{T}|e_1\rangle$ is a vector. Since $\hat{T}|e_1\rangle$ is a vector and $\{|e_i\rangle: 1\leq i\leq n\}$ is a basis then $\hat{T}|e_1\rangle$ is uniquely expressed as a linear combination of the vectors in $\{|e_i\rangle: ... 1 First get the idea of kets as some component in$\mathbb{R}^n$out of your mind. kets are elements of a complex vector space, in this case a finite dimensional one. Yes the space will be isomorphic to$\mathbb{R}^n$, or rather$\mathbb{C}^n$, but by assuming they are$\mathbb{C}^n$, you may imbue them with properties of$\mathbb{C}^n\$ that are not true of a ...

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