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Question Number 201502    Answers: 0   Comments: 3

A generation is about one-third of a lifetime.Approximately about how many generations have passed since the year 0AD?

$${A}\:{generation}\:{is}\:{about}\:{one}-{third}\:{of}\:{a} \\ $$$${lifetime}.{Approximately}\:{about}\:{how} \\ $$$${many}\:{generations}\:{have}\:{passed}\:{since} \\ $$$${the}\:{year}\:\mathrm{0}{AD}? \\ $$

Question Number 192257    Answers: 0   Comments: 3

A bullet with a velocity of 30 ms^(−1) after pentrating a 6 cm whole tree the velocity is reduced by one−third and then the bullet travels for 1s more. Will the bullet penetratee th tree? Analyze mathematically.

$$ \\ $$$$\mathrm{A}\:\mathrm{bullet}\:\mathrm{with}\:\mathrm{a}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{30}\:\mathrm{ms}^{−\mathrm{1}} \:\mathrm{after} \\ $$$$\mathrm{pentrating}\:\mathrm{a}\:\mathrm{6}\:{cm}\:\mathrm{whole}\:\mathrm{tree}\:\mathrm{the}\:\mathrm{velocity}\:\mathrm{is}\: \\ $$$$\mathrm{reduced}\:\mathrm{by}\:\mathrm{one}−\mathrm{third}\:\mathrm{and}\:\mathrm{then}\:\mathrm{the}\:\mathrm{bullet} \\ $$$$\mathrm{travel}{s}\:\mathrm{for}\:\mathrm{1s}\:\mathrm{more}.\: \\ $$$$ \\ $$$$ \\ $$$${Will}\:\mathrm{the}\:\mathrm{bullet}\:\mathrm{penetratee} \\ $$$$\mathrm{th}\:\mathrm{tree}?\:\mathrm{Analyze}\:\mathrm{mathematically}. \\ $$

Question Number 192248    Answers: 0   Comments: 1

Question Number 189932    Answers: 0   Comments: 0

Question Number 181253    Answers: 0   Comments: 5

define microscopic and macroscopic with one one example.

$${define}\:{microscopic}\:{and}\:{macroscopic} \\ $$$${with}\:{one}\:{one}\:{example}. \\ $$

Question Number 169972    Answers: 0   Comments: 0

Question Number 168164    Answers: 0   Comments: 1

Find the total energy (in Joules) of a particle of mass 4.0×10^(−11) kg moving at 80% the speed of light.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{total}\:\mathrm{energy}\:\left({in}\:{Joules}\right) \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{4}.\mathrm{0}×\mathrm{10}^{−\mathrm{11}} \mathrm{kg} \\ $$$$\mathrm{moving}\:\mathrm{at}\:\mathrm{80\%}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{light}. \\ $$

Question Number 163988    Answers: 0   Comments: 1

if here are 20 apple in a box, how many are the significant figures in it?

$${if}\:{here}\:{are}\:\mathrm{20}\:{apple}\:{in}\:{a}\:{box},\:{how}\:{many}\: \\ $$$${are}\:{the}\:{significant}\:{figures}\:{in}\:{it}? \\ $$

Question Number 163858    Answers: 0   Comments: 2

which object or substance has the 2rd number velocity after light?

$${which}\:{object}\:{or}\:{substance}\:{has}\:\:{the}\: \\ $$$$\mathrm{2}{rd}\:{number}\:{velocity}\:{after}\:{light}? \\ $$

Question Number 158878    Answers: 0   Comments: 0

Question Number 158694    Answers: 0   Comments: 0

Help me sir in phase sppace d^3 p=dp_x dp_(y ) dp_z then find ∫_0 ^∞ P^(2 ) e^(p^2 /(2mKT )) d^3 p =....

$${Help}\:{me}\:{sir} \\ $$$$\: \\ $$$${in}\:{phase}\:{sppace}\:\:{d}^{\mathrm{3}} {p}={dp}_{{x}} {dp}_{{y}\:} {dp}_{{z}} \:\:{then} \\ $$$${find} \\ $$$$\:\int_{\mathrm{0}} ^{\infty} \:{P}^{\mathrm{2}\:} {e}^{\frac{{p}^{\mathrm{2}} }{\mathrm{2}{mKT}\:}} \:\:{d}^{\mathrm{3}} {p}\:\:=.... \\ $$$$ \\ $$$$ \\ $$

Question Number 156968    Answers: 0   Comments: 0

Question Number 156660    Answers: 0   Comments: 0

Question Number 154554    Answers: 2   Comments: 1

Question Number 152652    Answers: 0   Comments: 1

Question Number 141700    Answers: 0   Comments: 2

Question Number 140938    Answers: 0   Comments: 0

Question Number 139002    Answers: 0   Comments: 2

Question Number 137429    Answers: 0   Comments: 4

A nuclide _(81)^(210) X decays to another nuclide _(80)^A Y in four successive radioactive decays. Each decay involves the emmision of either an alpha particle or a beta particle. The value of A is: A. 120 B. 206 C. 208 D. 212

$$\mathrm{A}\:\mathrm{nuclide}\:_{\mathrm{81}} ^{\mathrm{210}} {X}\:\mathrm{decays}\:\mathrm{to}\:\mathrm{another}\:\mathrm{nuclide}\:_{\mathrm{80}} ^{{A}} {Y}\:\mathrm{in}\: \\ $$$$\mathrm{four}\:\mathrm{successive}\:\mathrm{radioactive}\:\mathrm{decays}.\:\mathrm{Each}\:\mathrm{decay} \\ $$$$\mathrm{involves}\:\mathrm{the}\:\mathrm{emmision}\:\mathrm{of}\:\mathrm{either}\:\mathrm{an}\:\mathrm{alpha}\:\mathrm{particle} \\ $$$$\mathrm{or}\:\mathrm{a}\:\mathrm{beta}\:\mathrm{particle}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{A}\:\mathrm{is}: \\ $$$$\mathrm{A}.\:\mathrm{120}\:\:\:\:\:\:\:\:\:\:\:\mathrm{B}.\:\mathrm{206} \\ $$$$\mathrm{C}.\:\mathrm{208}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{D}.\:\mathrm{212} \\ $$

Question Number 130180    Answers: 0   Comments: 0

Question Number 121092    Answers: 0   Comments: 0

Question Number 121082    Answers: 0   Comments: 0

Question Number 120132    Answers: 0   Comments: 0

Suppose you are in a imagnary train which travels at the half of the speed of light. Suppose You have a brother who is 7 year smaller than you. He stands on the platform which you had left. After 1 hour of travelling on the train you come back on the platform. Then you observe something strange. You can see your brother looks older . So what is his age?(He was ten years old)

$${Suppose}\:{you}\:{are}\:{in}\:{a}\:{imagnary}\:{train}\:{which}\:{travels}\:{at}\:{the}\:{half}\: \\ $$$${of}\:{the}\:{speed}\:{of}\:{light}.\:{Suppose}\:{You}\:{have}\:{a}\:{brother}\:{who}\:{is}\:\mathrm{7}\:{year} \\ $$$${smaller}\:{than}\:{you}.\:{He}\:{stands}\:{on}\:{the}\:{platform}\:{which}\:{you}\:{had}\:{left}. \\ $$$${After}\:\mathrm{1}\:{hour}\:{of}\:{travelling}\:{on}\:{the}\:{train}\:{you}\:{come}\:{back}\:{on}\:{the} \\ $$$${platform}.\:{Then}\:{you}\:{observe}\:{something}\:{strange}.\:{You}\:{can}\:{see} \\ $$$${your}\:{brother}\:{looks}\:{older}\:.\:{So}\:{what}\:{is}\:{his}\:{age}?\left({He}\:{was}\:{ten}\:{years}\right. \\ $$$$\left.{old}\right) \\ $$

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 116079    Answers: 1   Comments: 0

prove that Re=((ρ∙v∙d)/μ) renulds number

$${prove}\:{that}\:{Re}=\frac{\rho\centerdot{v}\centerdot{d}}{\mu}\:\:\:\:\:{renulds}\:{number} \\ $$

Question Number 116078    Answers: 0   Comments: 0

prove that Fr=(v^2 /(gh)) froude numer

$${prove}\:{that}\:\:\:{Fr}=\frac{{v}^{\mathrm{2}} }{{gh}}\:\:\:\:{froude}\:{numer} \\ $$

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