1 Ket-bra operators
1.1 Kets and bras
A ket operator \(\left\lvert \cdot \right\rangle \) on a Hilbert space \(\mathcal{H}\) is defined as the linear map \(\mathcal{B}(\mathcal{H},\mathcal{B}(\mathbb {C},\mathcal{H}))\) and is given by \(x\mapsto (\alpha \mapsto \alpha x)\).
A bra operator \(\left\langle \cdot \right\rvert \) on a Hilbert space \(\mathcal{H}\) is defined as the anti-linear map \(\mathcal{H}\to \mathcal{B}(\mathcal{H},\mathbb {C})\) and is given by \(x\mapsto (y\mapsto \left\langle x \mid y\right\rangle )\).
Given \(x\in \mathcal{H}\), we get \(\left\langle x\right\rvert ^*=\left\lvert x\right\rangle \).
Easy computation.
Given \(x,y\in \mathcal{H}\), we get \(\left\langle x\right\rvert \circ \left\lvert y\right\rangle (1)=\left\langle x \mid y\right\rangle \).
Straightforward computation.
Let \(\mathcal{A}\) be an algebra and a Hilbert space. Then the unit map \(\eta \colon \mathbb {C}\to \mathcal{A}\) (which is given by \(\alpha \mapsto \alpha 1\)) is exactly \(\left\lvert 1\right\rangle \).
True by definition.
The adjoint of the unit map \(\eta \colon \mathbb {C}\to \mathcal{A}\) in an algebra and Hilbert space \(\mathcal{A}\) is \(\left\langle 1\right\rvert \).
Let \(f\in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\) and \(x\in \mathcal{H}_1\). Then \(f\circ \left\lvert x\right\rangle =\left\lvert f(x)\right\rangle \).
Simple computation.
Let \(f\in \mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\) and \(x\in \mathcal{H}_2\). Then \(\left\langle x\right\rvert \circ f=\left\langle f^*(x)\right\rvert \).
Simple computation.
1.2 Ket-bras
A ket-bra operator \(\left\lvert \cdot \right\rangle \! \! \left\langle \cdot \right\rvert \) is defined as the linear map from \(E_2\) to the anti-linear map \(E_1\to \mathcal{B}(E_1,E_2)\) and is given by
This is exactly \(\left\lvert \cdot \right\rangle \! \! \left\langle \cdot \right\rvert =\left\lvert \cdot \right\rangle \, \circ \, \left\langle \cdot \right\rvert \).
Let \(E_1,E_2,E_3\) be inner product spaces over \(\mathbb {C}\). So given \(x\in {E_2}\) and \(y\in {E_1}\), we write \(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert \) to mean the map \(u\mapsto \left\langle y \mid u\right\rangle x\).
Given \(x\) and \(y\) in a finite-dimensional Hilbert space \(A\), we get \(\operatorname {Tr}(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert )=\left\langle y \mid x\right\rangle \).
Obvious.
Given a linear map \(T_1\in \mathcal{B}(E_2,E_3)\) and elements \(x\in E_2\), \(y\in {E_1}\), we get, \(T_1\circ \left\lvert x\right\rangle \! \! \left\langle y\right\rvert =\left\lvert T_1(x)\right\rangle \! \! \left\langle y\right\rvert \)
Straightforward computation.
Given a linear map \(T_2\in \mathcal{B}(E_3,E_1)\) and elements \(x\in E_2\), \(y\in E_1\), we get \(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert \circ T_2=\left\lvert x\right\rangle \! \! \left\langle T_2^*(y)\right\rvert \).
Straightforward computation.
Given \(x\in E_2\) and \(y\in E_1\), we get \(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert ^*=\left\lvert y\right\rangle \! \! \left\langle x\right\rvert \).
Using Lemma 1.1.3.
Given an orthonormal basis \((u_i)\) of a \(\mathbb {C}\)-inner product space \(E\), we get \(\sum _i\left\lvert u_i\right\rangle \! \! \left\langle u_i\right\rvert =\operatorname {id}\).
Straightforward computation.
Given a Hilbert space \(\mathcal{H}\), we have \({\mathcal{B}(\mathcal{H})}^\prime =\{ \alpha \operatorname {id}:\alpha \in \mathbb {C}\} \).
In other words, \(x\in \mathcal{B}(\mathcal{H})\) commutes with all operators \(y\in \mathcal{B}(\mathcal{H})\) if and only if \(x=\alpha \operatorname {id}\) for some \(\alpha \in \mathbb {C}\).
Let \(x\in \mathcal{B}(\mathcal{H})\). Obviously, if \(x=\alpha \operatorname {id}\) for some \(\alpha \in \mathbb {C}\), then it commutes with every other operator. Now suppose \(x\) commutes with every operator in \(\mathcal{B}(\mathcal{H})\). So this means \(\left\lvert a\right\rangle \! \! \left\langle x^*(b)\right\rvert =\left\lvert a\right\rangle \! \! \left\langle b\right\rvert x=x\left\lvert a\right\rangle \! \! \left\langle b\right\rvert =\left\lvert x(a)\right\rangle \! \! \left\langle b\right\rvert \) for all \(a,b\in \mathcal{H}\). Suppose there exists some non-zero \(a\in \mathcal{H}\), otherwise this is trivial. Then, for any \(b\in \mathcal{H}\), we have
Thus \(x=\alpha \operatorname {id}\) where \(\alpha =\left\langle x^*(a) \mid a\right\rangle /\left\| a\right\| ^2\).
Let \(\mathcal{H}\) be a Hilbert space, and let \(x,y\in \mathcal{H}\). Then if \(\left\lvert x\right\rangle \! \! \left\langle x\right\rvert =\left\lvert y\right\rangle \! \! \left\langle y\right\rvert \), then there exists some \(0\neq \alpha \in \mathbb {C}\) such that \(x=\alpha {y}\) (i.e., they are co-linear).
Suppose \(\left\lvert x\right\rangle \! \! \left\langle x\right\rvert =\left\lvert y\right\rangle \! \! \left\langle y\right\rvert \). Then it is clear that we get \(x=0\) if and only if \(y=0\). So we assume \(x\neq 0\) (and so \(y\neq 0\)), otherwise this is trivial. Then we have
And as \(x\neq 0\), we get \(x=\dfrac {\left\langle y \mid x\right\rangle }{\left\| x\right\| ^2}y\). We have \(\left\langle y \mid x\right\rangle \neq 0\) (otherwise, this would mean \(x=0\) which is a contradiction). Thus we can let \(\alpha =\left\langle y \mid x\right\rangle /\left\| x\right\| ^2\neq 0\) such that \(x=\alpha {y}\).
Let \(\mathcal{H}_1,\mathcal{H}_2\) be Hilbert spaces, and let \(a,c\in \mathcal{H}_1\setminus \{ 0\} \) and \(b,d\in \mathcal{H}_2\setminus \{ 0\} \). Then if \(\left\lvert a\right\rangle \! \! \left\langle b\right\rvert =\left\lvert c\right\rangle \! \! \left\langle d\right\rvert \), then there exists some \(0\neq \alpha ,\beta \in \mathbb {C}\) such that \(a = \alpha c\) and \(b=\alpha \beta d\).
As \(\left\lvert a\right\rangle \! \! \left\langle b\right\rvert =\left\lvert c\right\rangle \! \! \left\langle d\right\rvert \), we get \(\left\langle b \mid b\right\rangle a = \left\langle d \mid b\right\rangle c\) (so \(a = \dfrac {\left\langle d \mid b\right\rangle }{\left\| b\right\| ^2}c\)). Taking adjoints of the hypothesis, we get \(\left\lvert b\right\rangle \! \! \left\langle a\right\rvert =\left\lvert d\right\rangle \! \! \left\langle c\right\rvert \), and so
Clearly, \(\left\langle d \mid b\right\rangle \neq 0\) (otherwise, we get \(a=0\)). So then we can let \(\alpha =\left\langle d \mid b\right\rangle /\left\| b\right\| ^2\) and \(\beta =\left\| c\right\| ^2/\left\| a\right\| ^2\).
Given a finite-dimensional inner product space \(E\) over \(\mathbb {C}\) and \(T\in \mathcal{B}(E)\), we get
\(T\) is positive semi-definite \(\Leftrightarrow \) \(T=\sum _i\left\lvert v_i\right\rangle \! \! \left\langle v_i\right\rvert \) for some tuple \((v_i)\) in \(E\).
- \((\Rightarrow )\)
Suppose \(0\leq {T}\). We use the spectral theorem and let \((v_i)\) be the eigenbasis of \(T\) in \(E\) with corresponding eigenvalues \((\lambda _i)\). Note that, as \(0\leq {T}\), we also get each \(0\leq \lambda _i\). So then let each \(x_i=\sqrt{\lambda _i}u_i\). Then we have \(\sum _i\left\lvert x_i\right\rangle \! \! \left\langle x_i\right\rvert =\sum _i\sqrt{\lambda _i}\overline{\sqrt{\lambda _i}}\left\lvert u_i\right\rangle \! \! \left\langle u_i\right\rvert =\sum _i\lambda _i\left\lvert u_i\right\rangle \! \! \left\langle u_i\right\rvert =T\), where the last equality comes from Corollary 1.2.6.
- \((\Leftarrow )\)
Suppose we have some tuple \((v_i)\) in \(E\) such that \(T=\sum _i\left\lvert v_i\right\rangle \! \! \left\langle v_i\right\rvert \). Then, for any \(x\in {E}\), we get \(\left\langle x \mid T(x)\right\rangle =\sum _i\left\langle x \mid v_i\right\rangle \left\langle v_i \mid x\right\rangle =\sum _i\left\lvert {\left\langle x \mid v_i\right\rangle }\right\rvert ^2\geq 0\). Thus \(T\) is positive semi-definite.
Given an orthonormal basis \(b=(b_i)\) of a finite-dimensional Hilbert space \(\mathcal{H}\), we define \(R_b\) to be the linear isomorphism \(\mathcal{H}\cong \mathbb {C}^{\dim \mathcal{H}}\) given by \(R_b(x)_i=\left\langle b_i \mid x\right\rangle \) with its inverse given by \(x\mapsto \sum _ix_ib_i\).
Let \(e=(e_i)\) be an orthonormal basis of a finite-dimensional Hilbert space \(\mathcal{H}\). Then \(R_e^*=R_e^{-1}\).
Let \(x\in \mathcal{H}\) and \(y\in \mathbb {C}^{\dim \mathcal{H}}\). Then we compute,
Thus \(R_e^*=R_e^{-1}\).
Given orthonormal bases \(b=(b_i)\) and \(c=(c_j)\) of Hilbert spaces \(\mathcal{H}_1,\mathcal{H}_2\), then we let \(\mathcal{M}\) denote the identification from \(\mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)\) to \(M_{\dim \mathcal{H}_2,\dim \mathcal{H}_1}\), which is given by \(\mathcal{M}_{b,c}(T)_{kp}=\left\langle c_k \mid T(b_p)\right\rangle \).
Given orthonormal bases \(b=(b_i)\) and \(c=(c_j)\) of finite-dimensional Hilbert spaces \(\mathcal{H}_1,\mathcal{H}_2\), and elements \(x\in \mathcal{H}_1\) and \(y\in \mathcal{H}_2\), we have \(\mathcal{M}_{c,b}(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert )=R_b(x){R_c(y)}^*\).
For any \(i\in [\dim \mathcal{H}_1],j\in [\dim \mathcal{H}_2]\), we compute,
Thus \(\mathcal{M}_{c,b}(\left\lvert x\right\rangle \! \! \left\langle y\right\rvert )=R_b(x){R_c(y)}^*\).