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Question Number 118651    Answers: 1   Comments: 2

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Question Number 118634    Answers: 0   Comments: 3

Prove that the equation of the circle passing through the points of intersection of these two curves: y=1+(c/x) ; y=x^2 (c < (2/(3(√3))) ) is (x−(c/2))^2 +(y−1)^2 =1+(c^2 /4) .

$${Prove}\:{that}\:{the}\:{equation}\:{of}\:{the}\:{circle} \\ $$$${passing}\:{through}\:{the}\:{points}\:{of} \\ $$$${intersection}\:{of}\:{these}\:{two}\:{curves}: \\ $$$$\:\:{y}=\mathrm{1}+\frac{{c}}{{x}}\:;\:\:{y}={x}^{\mathrm{2}} \:\:\:\:\:\left({c}\:<\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\right)\: \\ $$$${is}\:\:\:\left({x}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\:\:. \\ $$

Question Number 118633    Answers: 0   Comments: 2

(4.1) ψ_μ (x)≡⟨x,μ∣ψ⟩ (4.2) ψ_μ (x−a)=[1−a•(∂/∂x)+(1/(2!))(a•(∂/∂x))^2 −…]ψ_μ (x) =exp(−a•(∂/∂x))ψ_μ (x)=⟨x,μ∣exp(−i((a•p)/h^ ))∣ψ⟩ (4.3) ∣ψ^′ ⟩≡U(a)∣ψ⟩ ; U(a)≡exp(−ia•p/h^ ) translation operator (4.4) ψ_μ (x−a)=⟨x,μ∣U(a)∣ψ⟩=⟨x,μ∣ψ^′ ⟩=ψ_μ ^′ (x) (4.5) ih^ ((∂∣ψ⟩)/∂a_x )=−ih^ ((∂∣ψ⟩)/∂x)=p_x ∣ψ⟩−− Passive transformations (1) ψ_μ (x;a+y)=exp(a•(∂/∂y))ψ_μ (x;y) (2) ⟨x,μ;0∣U(−a)∣ψ⟩=ψ_μ (x+a;0)=ψ_μ (x;a) (4.6) ∣x_0 ,μ⟩=∫d^3 p∣p,μ⟩⟨p,μ∣x_0 ,μ⟩=(1/h^(3/2) )∫d^3 pe^(−ix_0 •p/h^ ) ∣p,μ⟩ (4.7) U(a)∣x_0 ,μ⟩=(1/h^(3/2) )∫d^3 pe^(−ix_0 •p/h^ ) U(a)∣p,μ⟩ =(1/h^(3/2) )∫d^3 pe^(−i(x_0 +a)•p/h^ ) ∣p,μ⟩ =∣x_0 ,a⟩,μ⟩ Operators from expectation values (1) ⟨ψ∣A∣ψ⟩=⟨ψ∣B∣ψ⟩ (2) λ(⟨∅∣A∣χ⟩−⟨∅∣B∣χ⟩)=λ^∗ (⟨χ∣B∣∅⟩−⟨χ∣A∣∅⟩) (4.8) 1=⟨ψ^′ ∣ψ^′ ⟩=⟨ψ∣U^+ U∣ψ⟩ unitiry operator U^+ U=I ; U^+ =U^(−1) (4.9) U(δθ)=I−iδθτ+O(δθ)^2 (4.10) I=U^+ (δθ)U(δθ)=I+iδθ((τ^+ −τ)+O(δθ)^2 (4.11) i((∂∣ψ^′ ⟩)/∂θ)=τ∣ψ^′ ⟩ (4.12) U(θ)≡lim_(N→∞) (1−i(θ/N)τ)^N =e^(−iθτ) τ (hermition) = generator of U & the transformation (4.13) U(𝛂)=exp(−i𝛂•J) ; J_i :angular-momentum operators (4.14) i((∂∣ψ⟩)/∂α)=𝛂^ •J∣ψ⟩q (4.15) parity transformation P≡ (((−1),0,0),(0,(−1),0),(0,0,(−1)) ) ; Px=−x (4.16) quantum parity operator P: P ψ_μ ^′ (x)≡⟨x,μ∣P∣ψ⟩≡ψ_μ (Px)=ψ_μ (−x)=⟨−x,μ∣ψ⟩ (4.17) ψ_μ ^(′′) (x)=⟨x,μ∣P∣ψ^′ ⟩=⟨−x,μ∣ψ^′ ⟩ =⟨−x,μ∣P∣ψ⟩=⟨x,μ∣ψ⟩=ψ_μ (x) (4.18) ⟨∅∣P∣ψ⟩^∗ =∫d^3 xΣ_μ (⟨∅⇂x,μ⟩⟨x,μ⇂P⇂ψ⟩)^∗ =∫d^3 xΣ_μ (⟨∅⇂x,μ⟩⟨−x,μ⇂ψ⟩)^∗ =∫d^3 xΣ_μ (⟨ψ⇂−x,μ⟩⟨x,μ⇂P^2 ⇂∅⟩) =∫d^3 xΣ_μ (⟨ψ⇂−x,μ⟩⟨−x,μ⇂P⇂∅⟩)=⟨ψ∣P∣∅⟩ (4.19) Mirror operators ⟨x,y∣M∣ψ⟩=⟨y,x∣ψ⟩ (4.20) U^+ (a)xU(a)=x+a (4.21) x+δa⋍(1+i((δa•p)/h^ ))x(1−i((δa•p)/h^ )) =x−(i/h^ )[x,δa•p]+O(δa)^2 (4.22) [x_i ,p_j ]=ih^ δ_(ij) (4.23) U^+ (a)xU(a)=U^+ (a)U(a)x+U^+ (a)[x,U(a)]=x+U^+ (a)[x,U(a)] (4.24) U^+ (a)xU(a)=x−(i/h^ )U^+ (a)[x,a•p]U(a)=x+a Rotations in ordinary space R^T =R^(−1) ; det(R)=+1 ; R(𝛂)𝛂^ =𝛂^ TrR(𝛂)=1+2cos∣𝛂∣ ; v^′ =v+𝛂×v (4.25) R(𝛂)⟨ψ∣x∣ψ⟩=⟨ψ^′ ∣x∣ψ^′ ⟩=⟨ψ∣U^+ (𝛂)xU(𝛂)∣ψ⟩ (4.26) R(𝛂)x=U^+ (𝛂)xU(𝛂) (4.27) x+δ𝛂×x⋍(1+iδ𝛂•J)x(1−iδ𝛂•J) =x+i[δ𝛂•J,x]+O(δ𝛂)^2 (4.28) (δ𝛂×x)_i =Σ_(ij) ε_(ijk) δα_j x_k (4.29) Σ_(ij) ε_(ijk) δα_j x_k =iΣ_j δα_j [J_j ,x_i ] (4.30) [J_i ,x_j ]=iΣ_k ε_(ijk) x_k (4.31) [J_i ,v_j ]=iΣ_k ε_(ijk) v_k (4.32) [J_i ,p_j ]=iΣ_k ε_(ijk) p_k (4.33) [J_i ,J_j ]=iΣ_k ε_(ijk) J_k (4.34) ⟨ψ^′ ∣S∣ψ^′ ⟩=⟨ψ∣U^+ (𝛂)SU(𝛂)∣ψ⟩=⟨ψ∣S∣ψ⟩ (4.35) S⋍(1+iδ𝛂•J)S(1−iδ𝛂•J) =S+iδ𝛂•[J,S]+O(δ𝛂)^2 (4.36) [J,S]=0 (4.37) [J,J^2 ]=0 (4.38) The parity operator: x→Px=−x −⟨ψ∣x∣ψ⟩=P⟨ψ∣x∣ψ⟩=⟨ψ^′ ∣x∣ψ^′ ⟩=⟨ψ∣P^+ xP∣ψ⟩ (4.39) {x,P}≡xP+Px=0 (4.40) {v,P}≡vP+Pv=0 (4.41) v⇂𝛚^′ ⟩=v(P⇂𝛚⟩)=−Pv⇂𝛚⟩=−𝛚P⇂𝛚⟩=−𝛚⇂𝛚^′ ⟩ (4.42) −⟨±∣v∣±⟩=P⟨±∣v∣±⟩=⟨±∣P^+ vP∣±⟩=(±)^2 ⟨±∣v∣±⟩ (4.43a) ⟨x∣PV∣ψ⟩=⟨−x∣V∣ψ⟩=V(−x)⟨−x∣ψ⟩=V(x)⟨−x∣ψ⟩ (4.43b) ⟨x∣VP∣ψ⟩=V(x)⟨x∣P∣ψ⟩=V(x)⟨−x∣ψ⟩ (4.44) p^2 P=Σ_k p_k p_k P=−Σ_k p_k Pp_k =Σ_k Pp_k p_k =Pp^2 ⇒[p^2 ,P]=0 (4.45) {P,[v_i ,J_j ]}=iΣ_k ε_(ijk) {P,v_k }=0 (4.46) 0={P,[v_i ,J_j ]}=[{P,v_i },J_j ]−{[P,J_j ],v_i }=−{[P,J_j ],v_i } (4.47) [P,J_j ]=λP (4.48) ⟨ψ^′ ∣J∣ψ^′ ⟩=⟨ψ∣P^+ JP∣ψ⟩=⟨ψ∣J∣ψ⟩ (4.48) ⟨ψ∣M^+ xM∣ψ⟩=⟨ψ∣y∣ψ⟩ Mirror operators (4.50) M^+ xM=y ⇒ xM=My (4.51) ∣ψ,t⟩=e^(−iHt/h^ ) ∣ψ,0⟩ (4.52) U(t)=e^(−iHt/h^ ) time-evolution operator (4.53) U(θ)U(t)∣ψ⟩=U(t)U(θ)∣ψ⟩ (4.54a) ⟨x∣VU(𝛂)∣ψ⟩=V(x)⟨x∣U(𝛂)∣ψ⟩=V(x)⟨R(𝛂)x∣ψ⟩ (4.54b) ⟨x∣U(𝛂)V∣ψ⟩=⟨R(𝛂)x∣V∣ψ⟩=V(R(𝛂)x)⟨R(𝛂)x∣ψ⟩ (4.55) H=Σ_(i=1) ^n (p_i ^2 /(2m_i ))+Σ_(i<j) V(x_i −x_j ) (4.56) ∣ψ,t⟩=U(t)∣ψ,0⟩ (4.57) Q_t ^∼ ≡U^+ (t)QU(t) (4.58) ⟨Q⟩_t =⟨ψ,t∣Q∣ψ,t⟩=⟨ψ,0∣U^+ (t)QU(t)∣ψ,0⟩=⟨ψ,0∣Q_t ^∼ ∣ψ,0⟩ (4.59) ⟨∅,t⇂ψ,t⟩=⟨∅,0⇂ψ,0⟩ ; ∣∅,t⟩≡U(t)∣∅,0⟩ (4.60) (dQ_t ^∼ /dt)=(dU^+ /dt)QU+U^+ Q(dU/dt) (4.61) (dU/dt)=−((iH)/h^ )U⇒(dU^+ /dt)=((iH)/h^ )U^+ (4.62) ih^ (dQ_t ^∽ /dt)=−HU^+ QU+U^+ QUH=[Q_t ^∽ ,H] (4.63) exp(−i𝛂•J)≡R(𝛂) (4.64) I=R^T (𝛂)R(𝛂)=exp(−i𝛂•J)^T exp(−i𝛂•J) =exp(−i𝛂•J^T )exp(−i𝛂•J) (4.65) 0=−in•J^T exp(−iθn•J^T )exp(−iθn•J) +exp(−iθn•J^T )exp(−iθn•J)(−in•J) −in•{J^T +J} (4.66) {R^T (𝛂)R(𝛃)R(𝛂)}𝛃^′ =R^T (𝛂)R(𝛃)𝛃=R^T (𝛂)𝛃^′ (4.67) R^T (𝛂)R(𝛃)R(𝛂)=R(𝛃^′ )=R(R(−𝛂)𝛃) (4.68) (1+i𝛂•J)(1+i𝛃•J)(1−i𝛂•J)⋍1−i(𝛃−𝛂×𝛃)•J (4.69) α_i β_j [J_i ,J_j ]=iα_i β_j Σ_k ε_(ijk) J_k (4.70) [J_i ,J_j ]=iΣ_k ε_(ijk) J_k (4.71) Prob(at x⇂ψ)=Σ_μ ∣⟨x,μ⇂ψ⟩∣^2 (4.72) R(∅)= (((cos ∅),(−sin ∅),0),((sin ∅),(cos ∅),0),(0,0,1) ) (4.73) J_z ^′ =≡M•J_z •M^+ (4.74) S_x =(1/( (√2))) ((0,1,0),(1,0,1),(0,1,0) ) ; S_y =(1/( (√2))) ((0,(−i),0),(i,0,(−i)),(0,i,0) ) ; S_z = ((1,0,0),(0,0,0),(0,0,(−1)) ) (4.75) ⟨x∣p⟩=e^(ip•x/h^ ) (4.76) [{A,B},C]={A,[B,C]}+{[A,C],B} (4.77) G≡(1/2)(1−P) (4.78) S⟨ψ∣x∣ψ⟩=⟨ψ∣S^+ xS∣ψ⟩ (4.79) S_(ij) =δ_(ij) −2n_i n_j (4.80) V(x)=f(R)+λxy ; R=(√(x^2 +y^2 ))

$$ \\ $$$$\left(\mathrm{4}.\mathrm{1}\right)\:\:\:\psi_{\mu} \left(\boldsymbol{{x}}\right)\equiv\langle\boldsymbol{{x}},\mu\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{2}\right)\:\:\:\psi_{\mu} \left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)=\left[\mathrm{1}−\boldsymbol{{a}}\bullet\frac{\partial}{\partial\boldsymbol{{x}}}+\frac{\mathrm{1}}{\mathrm{2}!}\left(\boldsymbol{{a}}\bullet\frac{\partial}{\partial\boldsymbol{{x}}}\right)^{\mathrm{2}} −\ldots\right]\psi_{\mu} \left(\boldsymbol{{x}}\right) \\ $$$$\:\:\:\:\:\:={exp}\left(−\boldsymbol{{a}}\bullet\frac{\partial}{\partial\boldsymbol{{x}}}\right)\psi_{\mu} \left(\boldsymbol{{x}}\right)=\langle\boldsymbol{{x}},\mu\mid{exp}\left(−{i}\frac{\boldsymbol{{a}}\bullet\boldsymbol{{p}}}{\bar {{h}}}\right)\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{3}\right)\:\:\:\mid\psi^{'} \rangle\equiv{U}\left(\boldsymbol{{a}}\right)\mid\psi\rangle\:\:\:;\:{U}\left(\boldsymbol{{a}}\right)\equiv{exp}\left(−{i}\boldsymbol{{a}}\bullet\boldsymbol{{p}}/\bar {{h}}\right)\:\boldsymbol{{translation}}\:\boldsymbol{{operator}} \\ $$$$\left(\mathrm{4}.\mathrm{4}\right)\:\:\:\psi_{\mu} \left(\boldsymbol{{x}}−\boldsymbol{{a}}\right)=\langle\boldsymbol{{x}},\mu\mid{U}\left(\boldsymbol{{a}}\right)\mid\psi\rangle=\langle\boldsymbol{{x}},\mu\mid\psi^{'} \rangle=\psi_{\mu} ^{'} \left(\boldsymbol{{x}}\right) \\ $$$$\left(\mathrm{4}.\mathrm{5}\right)\:\:\:{i}\bar {{h}}\frac{\partial\mid\psi\rangle}{\partial{a}_{{x}} }=−{i}\bar {{h}}\frac{\partial\mid\psi\rangle}{\partial{x}}={p}_{{x}} \mid\psi\rangle−− \\ $$$$\boldsymbol{{Passive}}\:\boldsymbol{{transformations}} \\ $$$$\left(\mathrm{1}\right)\:\:\:\psi_{\mu} \left(\boldsymbol{{x}};\boldsymbol{{a}}+\boldsymbol{{y}}\right)={exp}\left(\boldsymbol{{a}}\bullet\frac{\partial}{\partial\boldsymbol{{y}}}\right)\psi_{\mu} \left(\boldsymbol{{x}};\boldsymbol{{y}}\right) \\ $$$$\left(\mathrm{2}\right)\:\:\:\langle\boldsymbol{{x}},\mu;\mathrm{0}\mid{U}\left(−\boldsymbol{{a}}\right)\mid\psi\rangle=\psi_{\mu} \left(\boldsymbol{{x}}+\boldsymbol{{a}};\mathrm{0}\right)=\psi_{\mu} \left(\boldsymbol{{x}};\boldsymbol{{a}}\right) \\ $$$$\left(\mathrm{4}.\mathrm{6}\right)\:\:\:\mid\boldsymbol{{x}}_{\mathrm{0}} ,\mu\rangle=\int{d}^{\mathrm{3}} \boldsymbol{{p}}\mid\boldsymbol{{p}},\mu\rangle\langle\boldsymbol{{p}},\mu\mid\boldsymbol{{x}}_{\mathrm{0}} ,\mu\rangle=\frac{\mathrm{1}}{{h}^{\mathrm{3}/\mathrm{2}} }\int{d}^{\mathrm{3}} \boldsymbol{{p}}{e}^{−{i}\boldsymbol{{x}}_{\mathrm{0}} \bullet\boldsymbol{{p}}/\bar {\boldsymbol{{h}}}} \mid\boldsymbol{{p}},\mu\rangle \\ $$$$\left(\mathrm{4}.\mathrm{7}\right)\:\:\:{U}\left(\boldsymbol{{a}}\right)\mid\boldsymbol{{x}}_{\mathrm{0}} ,\mu\rangle=\frac{\mathrm{1}}{{h}^{\mathrm{3}/\mathrm{2}} }\int{d}^{\mathrm{3}} \boldsymbol{{p}}{e}^{−{i}\boldsymbol{{x}}_{\mathrm{0}} \bullet\boldsymbol{{p}}/\bar {\boldsymbol{{h}}}} {U}\left(\boldsymbol{{a}}\right)\mid\boldsymbol{{p}},\mu\rangle \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{{h}^{\mathrm{3}/\mathrm{2}} }\int{d}^{\mathrm{3}} \boldsymbol{{p}}{e}^{−{i}\left(\boldsymbol{{x}}_{\mathrm{0}} +\boldsymbol{\mathrm{a}}\right)\bullet\boldsymbol{{p}}/\bar {\boldsymbol{{h}}}} \mid\boldsymbol{{p}},\mu\rangle \\ $$$$\:\:\:\:\:\:=\mid\boldsymbol{{x}}_{\mathrm{0}} ,\boldsymbol{{a}}\rangle,\mu\rangle \\ $$$$\boldsymbol{{Operators}}\:\boldsymbol{{from}}\:\boldsymbol{{expectation}}\:\boldsymbol{{values}} \\ $$$$\left(\mathrm{1}\right)\:\:\:\langle\psi\mid{A}\mid\psi\rangle=\langle\psi\mid{B}\mid\psi\rangle \\ $$$$\left(\mathrm{2}\right)\:\:\:\lambda\left(\langle\emptyset\mid{A}\mid\chi\rangle−\langle\emptyset\mid{B}\mid\chi\rangle\right)=\lambda^{\ast} \left(\langle\chi\mid{B}\mid\emptyset\rangle−\langle\chi\mid{A}\mid\emptyset\rangle\right) \\ $$$$\left(\mathrm{4}.\mathrm{8}\right)\:\:\:\mathrm{1}=\langle\psi^{'} \mid\psi^{'} \rangle=\langle\psi\mid{U}^{+} {U}\mid\psi\rangle \\ $$$$\boldsymbol{{unitiry}}\:\boldsymbol{{operator}}\:{U}^{+} {U}={I}\:\:\:;\:{U}^{+} ={U}^{−\mathrm{1}} \\ $$$$\left(\mathrm{4}.\mathrm{9}\right)\:\:\:{U}\left(\delta\theta\right)={I}−{i}\delta\theta\tau+{O}\left(\delta\theta\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{4}.\mathrm{10}\right)\:\:{I}={U}^{+} \left(\delta\theta\right){U}\left(\delta\theta\right)={I}+{i}\delta\theta\left(\left(\tau^{+} −\tau\right)+{O}\left(\delta\theta\right)^{\mathrm{2}} \right. \\ $$$$\left(\mathrm{4}.\mathrm{11}\right)\:\:{i}\frac{\partial\mid\psi^{'} \rangle}{\partial\theta}=\tau\mid\psi^{'} \rangle \\ $$$$\left(\mathrm{4}.\mathrm{12}\right)\:\:{U}\left(\theta\right)\equiv\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}−{i}\frac{\theta}{{N}}\tau\right)^{{N}} ={e}^{−{i}\theta\tau} \\ $$$$\:\:\:\:\:\:\tau\:\left({hermition}\right)\:=\:\boldsymbol{{generator}}\:\boldsymbol{{of}}\:{U}\:\&\:{the}\:\boldsymbol{\mathrm{transformation}} \\ $$$$\left(\mathrm{4}.\mathrm{13}\right)\:\:{U}\left(\boldsymbol{\alpha}\right)={exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{{J}}\right)\:\:\:;\:{J}_{{i}} :\boldsymbol{{angular}}-\boldsymbol{{momentum}}\:\boldsymbol{{operators}} \\ $$$$\left(\mathrm{4}.\mathrm{14}\right)\:\:{i}\frac{\partial\mid\psi\rangle}{\partial\alpha}=\hat {\boldsymbol{\alpha}}\bullet\boldsymbol{{J}}\mid\psi\rangle{q} \\ $$$$\left(\mathrm{4}.\mathrm{15}\right)\:\:\boldsymbol{{parity}}\:\boldsymbol{{transformation}} \\ $$$$\:\:\:\:\:\:\mathcal{P}\equiv\begin{pmatrix}{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{1}}\end{pmatrix}\:\:\:\:;\:\mathcal{P}\boldsymbol{{x}}=−\boldsymbol{{x}} \\ $$$$\left(\mathrm{4}.\mathrm{16}\right)\:\:\boldsymbol{{quantum}}\:\boldsymbol{{parity}}\:\boldsymbol{{operator}}\:{P}: \\ $$$$\:\:\:\:\:\:{P}\:\psi_{\mu} ^{'} \left(\boldsymbol{{x}}\right)\equiv\langle\boldsymbol{{x}},\mu\mid{P}\mid\psi\rangle\equiv\psi_{\mu} \left(\mathcal{P}\boldsymbol{{x}}\right)=\psi_{\mu} \left(−\boldsymbol{{x}}\right)=\langle−\boldsymbol{{x}},\mu\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{17}\right)\:\:\psi_{\mu} ^{''} \left(\boldsymbol{{x}}\right)=\langle\boldsymbol{{x}},\mu\mid{P}\mid\psi^{'} \rangle=\langle−\boldsymbol{{x}},\mu\mid\psi^{'} \rangle \\ $$$$\:\:\:\:\:\:=\langle−\boldsymbol{{x}},\mu\mid{P}\mid\psi\rangle=\langle\boldsymbol{{x}},\mu\mid\psi\rangle=\psi_{\mu} \left(\boldsymbol{{x}}\right) \\ $$$$\:\left(\mathrm{4}.\mathrm{18}\right)\:\:\langle\emptyset\mid{P}\mid\psi\rangle^{\ast} =\int{d}^{\mathrm{3}} \boldsymbol{{x}}\underset{\mu} {\sum}\left(\langle\emptyset\downharpoonright\boldsymbol{{x}},\mu\rangle\langle\boldsymbol{{x}},\mu\downharpoonright{P}\downharpoonright\psi\rangle\right)^{\ast} \\ $$$$\:\:\:\:\:\:=\int{d}^{\mathrm{3}} \boldsymbol{{x}}\underset{\mu} {\sum}\left(\langle\emptyset\downharpoonright\boldsymbol{{x}},\mu\rangle\langle−\boldsymbol{{x}},\mu\downharpoonright\psi\rangle\right)^{\ast} \\ $$$$\:\:\:\:\:\:=\int{d}^{\mathrm{3}} \boldsymbol{{x}}\underset{\mu} {\sum}\left(\langle\psi\downharpoonright−\boldsymbol{{x}},\mu\rangle\langle\boldsymbol{{x}},\mu\downharpoonright{P}^{\mathrm{2}} \downharpoonright\emptyset\rangle\right) \\ $$$$\:\:\:\:\:\:=\int{d}^{\mathrm{3}} \boldsymbol{{x}}\underset{\mu} {\sum}\left(\langle\psi\downharpoonright−\boldsymbol{{x}},\mu\rangle\langle−\boldsymbol{{x}},\mu\downharpoonright{P}\downharpoonright\emptyset\rangle\right)=\langle\psi\mid{P}\mid\emptyset\rangle \\ $$$$\left(\mathrm{4}.\mathrm{19}\right)\:\:\boldsymbol{{Mirror}}\:\boldsymbol{{operators}} \\ $$$$\:\:\:\:\:\:\langle{x},{y}\mid{M}\mid\psi\rangle=\langle{y},{x}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{20}\right)\:\:{U}^{+} \left(\boldsymbol{{a}}\right)\boldsymbol{{x}}{U}\left(\boldsymbol{{a}}\right)=\boldsymbol{{x}}+\boldsymbol{{a}} \\ $$$$\left(\mathrm{4}.\mathrm{21}\right)\:\:\boldsymbol{{x}}+\delta\boldsymbol{{a}}\backsimeq\left(\mathrm{1}+{i}\frac{\delta\boldsymbol{{a}}\bullet\boldsymbol{{p}}}{\bar {{h}}}\right)\boldsymbol{{x}}\left(\mathrm{1}−{i}\frac{\delta\boldsymbol{{a}}\bullet\boldsymbol{{p}}}{\bar {{h}}}\right) \\ $$$$\:\:\:\:\:\:=\boldsymbol{{x}}−\frac{{i}}{\bar {{h}}}\left[\boldsymbol{{x}},\delta\boldsymbol{{a}}\bullet\boldsymbol{{p}}\right]+{O}\left(\delta\boldsymbol{{a}}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{4}.\mathrm{22}\right)\:\:\left[{x}_{{i}} ,{p}_{{j}} \right]={i}\bar {{h}}\delta_{{ij}} \\ $$$$\left(\mathrm{4}.\mathrm{23}\right)\:\:{U}^{+} \left(\boldsymbol{{a}}\right)\boldsymbol{{x}}{U}\left(\boldsymbol{{a}}\right)={U}^{+} \left(\boldsymbol{{a}}\right){U}\left(\boldsymbol{{a}}\right)\boldsymbol{{x}}+{U}^{+} \left(\boldsymbol{{a}}\right)\left[\boldsymbol{{x}},{U}\left(\boldsymbol{{a}}\right)\right]=\boldsymbol{{x}}+{U}^{+} \left(\boldsymbol{{a}}\right)\left[\boldsymbol{{x}},{U}\left(\boldsymbol{{a}}\right)\right] \\ $$$$\left(\mathrm{4}.\mathrm{24}\right)\:\:{U}^{+} \left(\boldsymbol{{a}}\right)\boldsymbol{{x}}{U}\left(\boldsymbol{{a}}\right)=\boldsymbol{{x}}−\frac{{i}}{\bar {{h}}}{U}^{+} \left(\boldsymbol{{a}}\right)\left[\boldsymbol{{x}},\boldsymbol{{a}}\bullet\boldsymbol{{p}}\right]{U}\left(\boldsymbol{{a}}\right)=\boldsymbol{{x}}+\boldsymbol{{a}} \\ $$$$\boldsymbol{{Rotations}}\:\boldsymbol{{in}}\:\boldsymbol{{ordinary}}\:\boldsymbol{{space}} \\ $$$$\:\:\:\:\:\:\boldsymbol{{R}}^{{T}} =\boldsymbol{{R}}^{−\mathrm{1}} \:\:\:;\:{det}\left(\boldsymbol{{R}}\right)=+\mathrm{1}\:\:\:;\:\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\hat {\boldsymbol{\alpha}}=\hat {\boldsymbol{\alpha}} \\ $$$$\:\:\:\:\:\:{Tr}\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)=\mathrm{1}+\mathrm{2}{cos}\mid\boldsymbol{\alpha}\mid\:\:\:;\:\boldsymbol{{v}}^{'} =\boldsymbol{{v}}+\boldsymbol{\alpha}×\boldsymbol{{v}} \\ $$$$\left(\mathrm{4}.\mathrm{25}\right)\:\:\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\langle\psi\mid\boldsymbol{{x}}\mid\psi\rangle=\langle\psi^{'} \mid\boldsymbol{{x}}\mid\psi^{'} \rangle=\langle\psi\mid{U}^{+} \left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}{U}\left(\boldsymbol{\alpha}\right)\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{26}\right)\:\:\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}={U}^{+} \left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}{U}\left(\boldsymbol{\alpha}\right) \\ $$$$\left(\mathrm{4}.\mathrm{27}\right)\:\:\boldsymbol{{x}}+\delta\boldsymbol{\alpha}×\boldsymbol{{x}}\backsimeq\left(\mathrm{1}+{i}\delta\boldsymbol{\alpha}\bullet\boldsymbol{{J}}\right)\boldsymbol{{x}}\left(\mathrm{1}−{i}\delta\boldsymbol{\alpha}\bullet\boldsymbol{{J}}\right) \\ $$$$\:\:\:\:\:\:=\boldsymbol{{x}}+{i}\left[\delta\boldsymbol{\alpha}\bullet\boldsymbol{{J}},\boldsymbol{{x}}\right]+{O}\left(\delta\boldsymbol{\alpha}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{4}.\mathrm{28}\right)\:\:\left(\delta\boldsymbol{\alpha}×\boldsymbol{{x}}\right)_{{i}} =\underset{{ij}} {\sum}\epsilon_{{ijk}} \delta\alpha_{{j}} {x}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{29}\right)\:\:\underset{{ij}} {\sum}\epsilon_{{ijk}} \delta\alpha_{{j}} {x}_{{k}} ={i}\underset{{j}} {\sum}\delta\alpha_{{j}} \left[{J}_{{j}} ,{x}_{{i}} \right] \\ $$$$\left(\mathrm{4}.\mathrm{30}\right)\:\:\left[{J}_{{i}} ,{x}_{{j}} \right]={i}\underset{{k}} {\sum}\epsilon_{{ijk}} {x}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{31}\right)\:\:\left[{J}_{{i}} ,{v}_{{j}} \right]={i}\underset{{k}} {\sum}\epsilon_{{ijk}} {v}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{32}\right)\:\:\left[{J}_{{i}} ,{p}_{{j}} \right]={i}\underset{{k}} {\sum}\epsilon_{{ijk}} {p}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{33}\right)\:\:\left[{J}_{{i}} ,{J}_{{j}} \right]={i}\underset{{k}} {\sum}\epsilon_{{ijk}} {J}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{34}\right)\:\:\langle\psi^{'} \mid{S}\mid\psi^{'} \rangle=\langle\psi\mid{U}^{+} \left(\boldsymbol{\alpha}\right){SU}\left(\boldsymbol{\alpha}\right)\mid\psi\rangle=\langle\psi\mid{S}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{35}\right)\:\:{S}\backsimeq\left(\mathrm{1}+{i}\delta\boldsymbol{\alpha}\bullet\boldsymbol{{J}}\right){S}\left(\mathrm{1}−{i}\delta\boldsymbol{\alpha}\bullet\boldsymbol{{J}}\right) \\ $$$$\:\:\:\:\:\:={S}+{i}\delta\boldsymbol{\alpha}\bullet\left[\boldsymbol{{J}},{S}\right]+{O}\left(\delta\boldsymbol{\alpha}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{4}.\mathrm{36}\right)\:\:\left[\boldsymbol{{J}},{S}\right]=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{37}\right)\:\:\left[\boldsymbol{{J}},{J}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{38}\right)\:\:\boldsymbol{{The}}\:\boldsymbol{{parity}}\:\boldsymbol{{operator}}:\:\boldsymbol{{x}}\rightarrow\mathcal{P}\boldsymbol{{x}}=−\boldsymbol{{x}} \\ $$$$\:\:\:\:\:\:−\langle\psi\mid\boldsymbol{{x}}\mid\psi\rangle=\mathcal{P}\langle\psi\mid\boldsymbol{{x}}\mid\psi\rangle=\langle\psi^{'} \mid\boldsymbol{{x}}\mid\psi^{'} \rangle=\langle\psi\mid{P}^{+} \boldsymbol{{x}}{P}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{39}\right)\:\:\left\{\boldsymbol{{x}},{P}\right\}\equiv\boldsymbol{{x}}{P}+{P}\boldsymbol{{x}}=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{40}\right)\:\:\left\{\boldsymbol{{v}},{P}\right\}\equiv\boldsymbol{{v}}{P}+{P}\boldsymbol{{v}}=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{41}\right)\:\:\boldsymbol{{v}}\downharpoonright\boldsymbol{\omega}^{'} \rangle=\boldsymbol{{v}}\left({P}\downharpoonright\boldsymbol{\omega}\rangle\right)=−{P}\boldsymbol{{v}}\downharpoonright\boldsymbol{\omega}\rangle=−\boldsymbol{\omega}{P}\downharpoonright\boldsymbol{\omega}\rangle=−\boldsymbol{\omega}\downharpoonright\boldsymbol{\omega}^{'} \rangle \\ $$$$\left(\mathrm{4}.\mathrm{42}\right)\:\:−\langle\pm\mid\boldsymbol{{v}}\mid\pm\rangle=\mathcal{P}\langle\pm\mid\boldsymbol{{v}}\mid\pm\rangle=\langle\pm\mid{P}^{+} \boldsymbol{{v}}{P}\mid\pm\rangle=\left(\pm\right)^{\mathrm{2}} \langle\pm\mid\boldsymbol{{v}}\mid\pm\rangle \\ $$$$\left(\mathrm{4}.\mathrm{43}{a}\right)\:\langle\boldsymbol{{x}}\mid{PV}\mid\psi\rangle=\langle−\boldsymbol{{x}}\mid{V}\mid\psi\rangle={V}\left(−\boldsymbol{{x}}\right)\langle−\boldsymbol{{x}}\mid\psi\rangle={V}\left(\boldsymbol{{x}}\right)\langle−\boldsymbol{{x}}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{43}{b}\right)\:\langle\boldsymbol{{x}}\mid{VP}\mid\psi\rangle={V}\left(\boldsymbol{{x}}\right)\langle\boldsymbol{{x}}\mid{P}\mid\psi\rangle={V}\left(\boldsymbol{{x}}\right)\langle−\boldsymbol{{x}}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{44}\right)\:\:{p}^{\mathrm{2}} {P}=\underset{{k}} {\sum}{p}_{{k}} {p}_{{k}} {P}=−\underset{{k}} {\sum}{p}_{{k}} {Pp}_{{k}} =\underset{{k}} {\sum}{Pp}_{{k}} {p}_{{k}} ={Pp}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\Rightarrow\left[{p}^{\mathrm{2}} ,{P}\right]=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{45}\right)\:\:\left\{{P},\left[{v}_{{i}} ,{J}_{{j}} \right]\right\}={i}\underset{{k}} {\sum}\epsilon_{{ijk}} \left\{{P},{v}_{{k}} \right\}=\mathrm{0} \\ $$$$\left(\mathrm{4}.\mathrm{46}\right)\:\:\mathrm{0}=\left\{{P},\left[{v}_{{i}} ,{J}_{{j}} \right]\right\}=\left[\left\{{P},{v}_{{i}} \right\},{J}_{{j}} \right]−\left\{\left[{P},{J}_{{j}} \right],{v}_{{i}} \right\}=−\left\{\left[{P},{J}_{{j}} \right],{v}_{{i}} \right\} \\ $$$$\left(\mathrm{4}.\mathrm{47}\right)\:\:\left[{P},{J}_{{j}} \right]=\lambda{P} \\ $$$$\left(\mathrm{4}.\mathrm{48}\right)\:\:\langle\psi^{'} \mid\boldsymbol{{J}}\mid\psi^{'} \rangle=\langle\psi\mid{P}^{+} \boldsymbol{{J}}{P}\mid\psi\rangle=\langle\psi\mid\boldsymbol{{J}}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{48}\right)\:\:\langle\psi\mid{M}^{+} {xM}\mid\psi\rangle=\langle\psi\mid{y}\mid\psi\rangle\:\:\:\boldsymbol{{Mirror}}\:\boldsymbol{{operators}} \\ $$$$\left(\mathrm{4}.\mathrm{50}\right)\:\:{M}^{+} {xM}={y}\:\Rightarrow\:{xM}={My} \\ $$$$\left(\mathrm{4}.\mathrm{51}\right)\:\:\mid\psi,{t}\rangle={e}^{−{iHt}/\bar {{h}}} \mid\psi,\mathrm{0}\rangle \\ $$$$\left(\mathrm{4}.\mathrm{52}\right)\:\:{U}\left({t}\right)={e}^{−{iHt}/\bar {{h}}} \:\:\:\boldsymbol{{time}}-\boldsymbol{{evolution}}\:\boldsymbol{{operator}} \\ $$$$\left(\mathrm{4}.\mathrm{53}\right)\:\:{U}\left(\theta\right){U}\left({t}\right)\mid\psi\rangle={U}\left({t}\right){U}\left(\theta\right)\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{54}{a}\right)\:\langle\boldsymbol{{x}}\mid{VU}\left(\boldsymbol{\alpha}\right)\mid\psi\rangle={V}\left(\boldsymbol{{x}}\right)\langle\boldsymbol{{x}}\mid{U}\left(\boldsymbol{\alpha}\right)\mid\psi\rangle={V}\left(\boldsymbol{{x}}\right)\langle\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{54}{b}\right)\:\langle\boldsymbol{{x}}\mid{U}\left(\boldsymbol{\alpha}\right){V}\mid\psi\rangle=\langle\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}\mid{V}\mid\psi\rangle={V}\left(\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}\right)\langle\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\boldsymbol{{x}}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{55}\right)\:\:{H}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\boldsymbol{{p}}_{{i}} ^{\mathrm{2}} }{\mathrm{2}{m}_{{i}} }+\underset{{i}<{j}} {\sum}{V}\left(\boldsymbol{{x}}_{{i}} −\boldsymbol{{x}}_{{j}} \right) \\ $$$$\left(\mathrm{4}.\mathrm{56}\right)\:\:\mid\psi,{t}\rangle={U}\left({t}\right)\mid\psi,\mathrm{0}\rangle \\ $$$$\left(\mathrm{4}.\mathrm{57}\right)\:\:\overset{\sim} {{Q}}_{{t}} \equiv{U}^{+} \left({t}\right){QU}\left({t}\right) \\ $$$$\left(\mathrm{4}.\mathrm{58}\right)\:\:\langle{Q}\rangle_{{t}} =\langle\psi,{t}\mid{Q}\mid\psi,{t}\rangle=\langle\psi,\mathrm{0}\mid{U}^{+} \left({t}\right){QU}\left({t}\right)\mid\psi,\mathrm{0}\rangle=\langle\psi,\mathrm{0}\mid\overset{\sim} {{Q}}_{{t}} \mid\psi,\mathrm{0}\rangle \\ $$$$\left(\mathrm{4}.\mathrm{59}\right)\:\:\langle\emptyset,{t}\downharpoonright\psi,{t}\rangle=\langle\emptyset,\mathrm{0}\downharpoonright\psi,\mathrm{0}\rangle\:\:\:;\:\mid\emptyset,{t}\rangle\equiv{U}\left({t}\right)\mid\emptyset,\mathrm{0}\rangle \\ $$$$\left(\mathrm{4}.\mathrm{60}\right)\:\:\frac{{d}\overset{\sim} {{Q}}_{{t}} }{{dt}}=\frac{{dU}^{+} }{{dt}}{QU}+{U}^{+} {Q}\frac{{dU}}{{dt}} \\ $$$$\left(\mathrm{4}.\mathrm{61}\right)\:\:\frac{{dU}}{{dt}}=−\frac{{iH}}{\bar {{h}}}{U}\Rightarrow\frac{{dU}^{+} }{{dt}}=\frac{{iH}}{\bar {{h}}}{U}^{+} \\ $$$$\left(\mathrm{4}.\mathrm{62}\right)\:\:{i}\bar {{h}}\frac{{d}\overset{\backsim} {{Q}}_{{t}} }{{dt}}=−{HU}^{+} {QU}+{U}^{+} {QUH}=\left[\overset{\backsim} {{Q}}_{{t}} ,{H}\right] \\ $$$$\left(\mathrm{4}.\mathrm{63}\right)\:\:{exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right)\equiv\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right) \\ $$$$\left(\mathrm{4}.\mathrm{64}\right)\:\:\boldsymbol{{I}}=\boldsymbol{{R}}^{{T}} \left(\boldsymbol{\alpha}\right)\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)={exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right)^{{T}} {exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right) \\ $$$$\:\:\:\:\:\:={exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}^{{T}} \right){exp}\left(−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right) \\ $$$$\left(\mathrm{4}.\mathrm{65}\right)\:\:\mathrm{0}=−{i}\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}^{{T}} {exp}\left(−{i}\theta\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}^{{T}} \right){exp}\left(−{i}\theta\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}\right) \\ $$$$\:\:\:\:\:\:+{exp}\left(−{i}\theta\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}^{{T}} \right){exp}\left(−{i}\theta\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}\right)\left(−{i}\boldsymbol{{n}}\bullet\boldsymbol{\mathcal{J}}\right) \\ $$$$\:\:\:\:\:\:−{i}\boldsymbol{{n}}\bullet\left\{\boldsymbol{\mathcal{J}}^{{T}} +\boldsymbol{\mathcal{J}}\right\} \\ $$$$\left(\mathrm{4}.\mathrm{66}\right)\:\:\left\{\boldsymbol{{R}}^{{T}} \left(\boldsymbol{\alpha}\right)\boldsymbol{{R}}\left(\boldsymbol{\beta}\right)\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)\right\}\boldsymbol{\beta}^{'} =\boldsymbol{{R}}^{{T}} \left(\boldsymbol{\alpha}\right)\boldsymbol{{R}}\left(\boldsymbol{\beta}\right)\boldsymbol{\beta}=\boldsymbol{{R}}^{{T}} \left(\boldsymbol{\alpha}\right)\boldsymbol{\beta}^{'} \\ $$$$\left(\mathrm{4}.\mathrm{67}\right)\:\:\boldsymbol{{R}}^{{T}} \left(\boldsymbol{\alpha}\right)\boldsymbol{{R}}\left(\boldsymbol{\beta}\right)\boldsymbol{{R}}\left(\boldsymbol{\alpha}\right)=\boldsymbol{{R}}\left(\boldsymbol{\beta}^{'} \right)=\boldsymbol{{R}}\left(\boldsymbol{{R}}\left(−\boldsymbol{\alpha}\right)\boldsymbol{\beta}\right) \\ $$$$\left(\mathrm{4}.\mathrm{68}\right)\:\:\left(\mathrm{1}+{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right)\left(\mathrm{1}+{i}\boldsymbol{\beta}\bullet\boldsymbol{\mathcal{J}}\right)\left(\mathrm{1}−{i}\boldsymbol{\alpha}\bullet\boldsymbol{\mathcal{J}}\right)\backsimeq\mathrm{1}−{i}\left(\boldsymbol{\beta}−\boldsymbol{\alpha}×\boldsymbol{\beta}\right)\bullet\boldsymbol{\mathcal{J}} \\ $$$$\left(\mathrm{4}.\mathrm{69}\right)\:\:\alpha_{{i}} \beta_{{j}} \left[\mathcal{J}_{{i}} ,\mathcal{J}_{{j}} \right]={i}\alpha_{{i}} \beta_{{j}} \underset{{k}} {\sum}\epsilon_{{ijk}} \mathcal{J}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{70}\right)\:\:\left[\mathcal{J}_{{i}} ,\mathcal{J}_{{j}} \right]={i}\underset{{k}} {\sum}\epsilon_{{ijk}} \mathcal{J}_{{k}} \\ $$$$\left(\mathrm{4}.\mathrm{71}\right)\:\:{Prob}\left({at}\:\boldsymbol{{x}}\downharpoonright\psi\right)=\underset{\mu} {\sum}\mid\langle\boldsymbol{{x}},\mu\downharpoonright\psi\rangle\mid^{\mathrm{2}} \\ $$$$\left(\mathrm{4}.\mathrm{72}\right)\:\:\boldsymbol{{R}}\left(\emptyset\right)=\begin{pmatrix}{\mathrm{cos}\:\emptyset}&{−\mathrm{sin}\:\emptyset}&{\mathrm{0}}\\{\mathrm{sin}\:\emptyset}&{\mathrm{cos}\:\emptyset}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$$$\left(\mathrm{4}.\mathrm{73}\right)\:\:\boldsymbol{\mathcal{J}}_{{z}} ^{'} =\equiv\boldsymbol{{M}}\bullet\boldsymbol{\mathcal{J}}_{{z}} \bullet\boldsymbol{{M}}^{+} \\ $$$$\left(\mathrm{4}.\mathrm{74}\right)\:\:{S}_{{x}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\begin{pmatrix}{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\end{pmatrix}\:\:;\:{S}_{{y}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\begin{pmatrix}{\mathrm{0}}&{−{i}}&{\mathrm{0}}\\{{i}}&{\mathrm{0}}&{−{i}}\\{\mathrm{0}}&{{i}}&{\mathrm{0}}\end{pmatrix}\:\:;\:{S}_{{z}} =\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{−\mathrm{1}}\end{pmatrix}\: \\ $$$$\left(\mathrm{4}.\mathrm{75}\right)\:\:\langle\boldsymbol{{x}}\mid\boldsymbol{{p}}\rangle={e}^{{i}\boldsymbol{{p}}\bullet\boldsymbol{{x}}/\bar {{h}}} \\ $$$$\left(\mathrm{4}.\mathrm{76}\right)\:\:\left[\left\{{A},{B}\right\},{C}\right]=\left\{{A},\left[{B},{C}\right]\right\}+\left\{\left[{A},{C}\right],{B}\right\} \\ $$$$\left(\mathrm{4}.\mathrm{77}\right)\:\:{G}\equiv\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{P}\right) \\ $$$$\left(\mathrm{4}.\mathrm{78}\right)\:\:{S}\langle\psi\mid\boldsymbol{{x}}\mid\psi\rangle=\langle\psi\mid{S}^{+} \boldsymbol{{x}}{S}\mid\psi\rangle \\ $$$$\left(\mathrm{4}.\mathrm{79}\right)\:\:{S}_{{ij}} =\delta_{{ij}} −\mathrm{2}{n}_{{i}} {n}_{{j}} \\ $$$$\left(\mathrm{4}.\mathrm{80}\right)\:\:{V}\left(\boldsymbol{{x}}\right)={f}\left({R}\right)+\lambda{xy}\:\:\:;\:{R}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 118650    Answers: 1   Comments: 0

What is the (explicit) formula of this sequence ? 19, 151, 1021, 50389

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\left(\mathrm{explicit}\right)\:\mathrm{formula}\:\mathrm{of}\:\mathrm{this}\:\mathrm{sequence}\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{19},\:\mathrm{151},\:\mathrm{1021},\:\mathrm{50389} \\ $$

Question Number 118627    Answers: 1   Comments: 0

find the infmum(s) if S=Σ_(n=1) ^∞ (((−1)^n )/(n )) help me sir

$${find}\:{the}\:{infmum}\left({s}\right)\:{if}\:{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\:}\: \\ $$$${help}\:{me}\:{sir} \\ $$

Question Number 118612    Answers: 1   Comments: 0

Question Number 118606    Answers: 0   Comments: 0

Question Number 118594    Answers: 0   Comments: 0

Question Number 118588    Answers: 1   Comments: 1

x(x+y) dy −x^2 dx = 0

$${x}\left({x}+{y}\right)\:{dy}\:−{x}^{\mathrm{2}} \:{dx}\:=\:\mathrm{0}\: \\ $$

Question Number 118585    Answers: 0   Comments: 4

v2.227 reverts app first screen to two parts, saved and forum. To start in forum by default change settings and Enable start in forum. Non English users were having difficulty in locating offline editor. Work is in progress to translate in-app menu to different languages.

$${v}\mathrm{2}.\mathrm{227}\:\mathrm{reverts}\:\mathrm{app}\:\mathrm{first}\:\mathrm{screen}\:\mathrm{to} \\ $$$$\mathrm{two}\:\mathrm{parts},\:\mathrm{saved}\:\mathrm{and}\:\mathrm{forum}. \\ $$$$\mathrm{To}\:\mathrm{start}\:\mathrm{in}\:\mathrm{forum}\:\mathrm{by}\:\mathrm{default}\:\mathrm{change} \\ $$$$\mathrm{settings}\:\mathrm{and}\:\mathrm{Enable}\:\mathrm{start}\:\mathrm{in}\:\mathrm{forum}. \\ $$$$\mathrm{Non}\:\mathrm{English}\:\mathrm{users}\:\mathrm{were}\:\mathrm{having} \\ $$$$\mathrm{difficulty}\:\mathrm{in}\:\mathrm{locating}\:\mathrm{offline}\:\mathrm{editor}. \\ $$$$\mathrm{Work}\:\mathrm{is}\:\mathrm{in}\:\mathrm{progress}\:\mathrm{to}\:\mathrm{translate} \\ $$$$\mathrm{in}-\mathrm{app}\:\mathrm{menu}\:\mathrm{to}\:\mathrm{different}\:\mathrm{languages}. \\ $$

Question Number 118575    Answers: 1   Comments: 0

∫ cos (2cot^(−1) (√((1−x)/(1+x))) ) dx ?

$$\:\:\int\:\mathrm{cos}\:\left(\mathrm{2cot}^{−\mathrm{1}} \sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\:\right)\:{dx}\:? \\ $$

Question Number 118574    Answers: 1   Comments: 0

find particular solution of (D^3 +4D) y = sin 2x by using inverse operation ?

$${find}\:{particular}\:{solution}\:{of}\: \\ $$$$\left({D}^{\mathrm{3}} +\mathrm{4}{D}\right)\:{y}\:=\:\mathrm{sin}\:\mathrm{2}{x}\:{by}\:{using} \\ $$$${inverse}\:{operation}\:? \\ $$

Question Number 118570    Answers: 2   Comments: 0

Find the dimensions of the largest rectangular garden that can be enclosed by 80m of fencing

$${Find}\:{the}\:{dimensions}\:{of}\:{the} \\ $$$${largest}\:{rectangular}\:{garden}\:{that} \\ $$$${can}\:{be}\:{enclosed}\:{by}\:\mathrm{80}{m}\:{of} \\ $$$${fencing} \\ $$

Question Number 118566    Answers: 6   Comments: 2

solve ∫_0 ^1 (dx/( (√x) (√(1−x)) )) .

$$\:\:{solve}\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{dx}}{\:\sqrt{{x}}\:\sqrt{\mathrm{1}−{x}}\:}\:. \\ $$

Question Number 118555    Answers: 2   Comments: 0

Question Number 118545    Answers: 1   Comments: 1

Question Number 118541    Answers: 3   Comments: 1

∫ tan (x).tan (2x).tan (3x) dx = ?

$$\:\:\:\int\:\mathrm{tan}\:\left({x}\right).\mathrm{tan}\:\left(\mathrm{2}{x}\right).\mathrm{tan}\:\left(\mathrm{3}{x}\right)\:{dx}\:=\:? \\ $$

Question Number 118546    Answers: 3   Comments: 7

nilai maksimum?fungsi y= 1+ sin 2x +cos 2x

$${nilai}\:{maksimum}?{fungsi}\:{y}=\:\mathrm{1}+\:{sin}\:\mathrm{2}{x}\:+{cos}\:\mathrm{2}{x} \\ $$

Question Number 118513    Answers: 2   Comments: 0

If f(x) + f(((x − 1)/x)) = 1 + x, find f(2).

$$\mathrm{If}\:\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:\:+\:\:\mathrm{f}\left(\frac{\mathrm{x}\:\:−\:\:\mathrm{1}}{\mathrm{x}}\right)\:\:\:=\:\:\:\mathrm{1}\:\:+\:\:\mathrm{x},\:\:\:\:\:\:\:\mathrm{find}\:\:\:\mathrm{f}\left(\mathrm{2}\right). \\ $$

Question Number 121164    Answers: 0   Comments: 0

...ADVANCED CALCULUS... If ∫_0 ^( ∞) ln(x)sin(x^2 )dx =λ∫_0 ^( ∞) sin(x^2 )dx then find the value of ′′λ′′ . ...m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\mathrm{ADVANCED}\:\:\mathrm{CALCULUS}... \\ $$$$\:\:\:\:\mathrm{If}\:\:\:\int_{\mathrm{0}} ^{\:\infty} {ln}\left({x}\right){sin}\left({x}^{\mathrm{2}} \right){dx}\:=\lambda\int_{\mathrm{0}} ^{\:\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}\: \\ $$$$\:\:\:\:\:\:\:\:{then}\:\:{find}\:\:{the}\:\:{value}\:{of}\:''\lambda''\:. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\mathrm{m}.\mathrm{n}.\mathrm{july}.\mathrm{1970}... \\ $$

Question Number 118506    Answers: 2   Comments: 0

Question Number 118511    Answers: 3   Comments: 0

How many positive integers x satisfy log_(x/8) (x^2 /4)<7+log_2 (8/x)

$$\mathrm{How}\:\mathrm{many}\:\mathrm{positive}\:\mathrm{integers}\:{x}\:\mathrm{satisfy} \\ $$$$\mathrm{log}_{\frac{{x}}{\mathrm{8}}} \frac{{x}^{\mathrm{2}} }{\mathrm{4}}<\mathrm{7}+\mathrm{log}_{\mathrm{2}} \frac{\mathrm{8}}{{x}} \\ $$

Question Number 118491    Answers: 1   Comments: 0

... ⧫Advanced Calculus⧫... Evaluate:: Ω = ∫_0 ^( ∞) ((secθ)/( (√(4tan^2 θ+5))))dθ ...♠L𝛗rD ∅sE♠... ...♣GooD LucK♣

$$ \\ $$$$...\:\blacklozenge\mathrm{Advanced}\:\mathrm{Calculus}\blacklozenge... \\ $$$$ \\ $$$$\mathrm{Evaluate}:: \\ $$$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sec}\theta}{\:\sqrt{\mathrm{4tan}^{\mathrm{2}} \theta+\mathrm{5}}}\mathrm{d}\theta \\ $$$$ \\ $$$$...\spadesuit\boldsymbol{\mathrm{L}\phi\mathrm{rD}}\:\boldsymbol{\varnothing\mathrm{sE}}\spadesuit... \\ $$$$ \\ $$$$...\clubsuit\boldsymbol{\mathrm{GooD}}\:\boldsymbol{\mathrm{LucK}}\clubsuit \\ $$

Question Number 118489    Answers: 2   Comments: 0

(1) solve the equation ((x−49)/(50)) + ((x−50)/(49)) = ((49)/(x−50)) + ((50)/(x−49)) (2) How many numbers from 12 to 12345 inclusive have digits which are consecutive an in increasing order, reading from left to right ?

$$\left(\mathrm{1}\right)\:{solve}\:{the}\:{equation}\:\frac{{x}−\mathrm{49}}{\mathrm{50}}\:+\:\frac{{x}−\mathrm{50}}{\mathrm{49}}\:=\:\frac{\mathrm{49}}{{x}−\mathrm{50}}\:+\:\frac{\mathrm{50}}{{x}−\mathrm{49}} \\ $$$$\left(\mathrm{2}\right)\:{How}\:{many}\:{numbers}\:{from}\:\mathrm{12}\:{to}\:\mathrm{12345}\: \\ $$$${inclusive}\:{have}\:{digits}\:{which}\:{are}\: \\ $$$${consecutive}\:{an}\:{in}\:{increasing}\:{order}, \\ $$$${reading}\:{from}\:{left}\:{to}\:{right}\:?\: \\ $$

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