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CRC Press eBooks

Design of quantum gates using electrons, positrons and photons, quantum information theory and quantum stochastic filtering

Harish Parthasarathy

Published Feb 3, 2021• 0 citations
Publisher

Abstract

Assume that system 1 evolves according to the dynamics ρ ′ ( t ) ⁢ = i [ H 1 , ρ ( t ) ] + ⁢ θ 1 ( ρ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math407.tif"/> 78 and system 2 evolves according to σ ′ ( t ) = − i [ H 2 , σ ( t ) ] + ⁢ θ 2 ( σ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math408.tif"/> where θ 1 , θ 2 are Lindblad maps: θ 1 ( ρ ) ⁢ = ⁢ ( - 1 / 2 ) Σ k = 1 N ( L k * L k ρ + ρ L k * L k - 2 L k ρ L k * ) θ 2 ( σ ) ⁢ = ⁢ ( - 1 / 2 ) ⁢ Σ k = 1 M ⁢ ( P k * P k σ ⁢ + ⁢ σ P k * P k - 2 P k σ P k * ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math409.tif"/> Then calculate the rate of change of the relative entropy between the two states at time t : H ( ρ ( t ) , σ ( t ) ) ⁢ = T r ( ρ ( t ) ( l o g ( ρ ( t ) ) ⁢ − l o g ( σ ( t ) ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math410.tif"/> hint: Use the formula for the differential of the exponential map in the form: d d t e x p ( Z ( t ) ) ⁢ = e x p ( Z ( t ) ) g ( a d ( Z ( t ) ) − 1 ( Z ′ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math411.tif"/> where g ( z ) = z 1 − exp ⁡ ( − z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math412.tif"/> Equivalently, Z ′ ( t ) ⁢ = g ( α d ( Z ( t ) ) ( e x p ( − Z ( t ) ) . d d t e x p ( Z ( t ) ) ) f https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math413.tif"/> Take ρ ( t )= exp ( Z ( t )) and deduce that d d t l o g ( ρ ( t ) ) ⁢ = g ( a d ( l o g ( ρ ( t ) ) ) ) ( ρ ( t ) − 1 ρ ′ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math414.tif"/> Now note that g ( z ) ⁢ = 1 1 − z / 2 ! + z 2 / 3 ! + … = - 1 + Σ k ≥ 1 c [ k ] z k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math415.tif"/> Note that ρ(t) has eigenvalues in the range [0,1]. So log(ρ(t)) has all eigenvalues in the range (−∞,0]. Hence, ad(log(ρ(t)) has eigenvalues in the range ℝ = (−∞,∞). So in order that we be able to substitute ad(log(ρ(t))) for z in the above equation, we require that the Taylor series for g(z) be convergent for all z ∈ ℝ which may not be the case. Note that Tr(ρ ' (t)) = 0 and that for k ≥ 1, T r ( ρ ( t ) a d ( l o g ( ρ ( t ) ) k ( ρ ( t ) − 1 ρ ′ ( t ) ) ) ⁢ - T r ( a d ( l o g ( ρ ( t ) ) ) k ( ρ ′ ( t ) ) ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math416.tif"/> since the commutator of two operators has trace zero (provided that the operators are bounded). Thus, we get T r ( ρ ( t ) d d t l o g ( ρ ( t ) ) ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math417.tif"/> 79 and so d d t T r ( ρ ( t ) . ⁢ l o g ( ρ ( t ) ) ) ⁢ = T r ( ρ ′ ( t ) . ⁢ l o g ( ρ ( t ) ) ) - T r ( T ( ρ ( t ) ) . ⁢ l o g ( ρ ( t ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math418.tif"/> where T ( X ) ⁢ = - i [ H 1 , X ] + θ 1 ( X ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math419.tif"/> Now, T r ( [ H 1 , ρ ] . ⁢ l o g ( ρ ) ) ⁢ = T r ( [ H 1 ⁢ . ⁢ l o g ( ρ ) , ρ ] ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math420.tif"/> so d d t ⁢ T r ( ρ ( t ) ⁢ . ⁢ l o g ( ρ ( t ) ) ) ⁢ = T r ( θ 1 ( ρ ( t ) ) ⁢ . ⁢ l o g ( ρ ( t ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math421.tif"/>

DOI
10.1201/9781003163626-5

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CRC Press eBooks•Feb 3, 2021•0 citations

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Q36(1) The spherical harmonics Y 1,1 ( θ , φ ), Y 1,0 ( θ , φ ), Y 1, −1( θ , φ ) are given by Eqs. (16.65) to (16.67). In Cartesian coordinates, these functions are given by 1 Y 1 , − 1 ( x , y , z ) = 3 8 π x − i y r , Y 1 , 0 ( x , y , z ) = 3 4 π z r , Y 1 , 1 ( x , y , z ) = − 3 8 π x + i y r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0667.tif"/> (a) Using the expression in Cartesian coordinates for the quantised orbital angular momentum operator L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1694.tif"/> in Eq. (27.86), verify by explicit calculations that Y 1,1 ( x , y , z ), Y 1,0 ( x , y , z )and Y 1,-1 ( x , y , z ) are the eigenfunctions of L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1695.tif"/> corresponding to eigenvalues ħ , 0 and – ħ . (b) Consider the following coordinate transformations: x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0668.tif"/> (1) 202 Show that the component of the orbital angular momentum operator along the z ′-direction is the same as that along the x -direction, i.e., show that L ^ c z ′ = L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1696.tif"/> . (2) Show that the simultaneous eigenfunctions of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1697.tif"/> and L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1698.tif"/> corresponding to eigenvalues of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1699.tif"/> equal to 2 ħ 2 are given in Cartesian coordinates by X 1 , − 1 ( x , y , z ) = 3 8 π y − i z r , X 1 , 0 ( x , y , z ) = 3 4 π x r , X 1 , 1 ( x , y , z ) = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0669.tif"/> SQ36(1)(a) In Cartesian coordinates we have 2 : p ^ x Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ x x − i y r = − i ℏ 3 8 π ( 1 r − x − i y r 2 ∂ ∂ x r ) = − i ℏ 3 8 π ( 1 r − x − i y r 2 x r ) = − i ℏ 3 8 π 1 r 3 ( r 2 − x 2 + i x y ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0670.tif"/> p ^ y Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ y x − i y r = − i ℏ 3 8 π ( − i r − x − i y r 2 ∂ ∂ y r ) = − i ℏ 3 8 π ( − i r − x − i y r 2 y r ) = − i ℏ 3 8 π 1 r 3 ( − i r 2 − x y + i y 2 ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0671.tif"/> L ^ z Y → 1 , − 1 : = ( x p ^ y − y p ^ x ) Y 1 , − 1 = − i ℏ 3 8 π 1 r 3 ( ( − i x r 2 − x 2 y + i x y 2 ) − ( y r 2 − y x 2 + i x y 2 ) ) = − i ℏ 3 8 π 1 r 3 ( − i x r 2 − y r 2 ) = − ℏ 3 8 π 1 r { x − i y } = − ℏ Y 1 , − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0672.tif"/> 203 Next we have p ^ x Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ x z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ x r ) = i ℏ 3 4 π z x r 3 . p ^ y Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ y z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ y r ) = i ℏ 3 4 π z y r 3 . L ^ z Y → 1 , 0 : = ( x p ^ y − y p ^ x ) Y 1 , 0 = x ( i ℏ 3 4 π z y r 3 ) − y ( i ℏ 3 4 π z x r 3 ) = 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0673.tif"/> Finally we have p ^ x Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ x x + i y r = − i ℏ 3 8 π ( 1 r − x + i y r 2 ∂ ∂ x r ) = i ℏ 3 8 π ( 1 r − x + i y r 2 x r ) = i ℏ 3 8 π 1 r 3 ( r 2 − x 2 − i x y ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0674.tif"/> p ^ y Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ y x + i y r = i ℏ 3 8 π ( i r − x + i y r 2 ∂ ∂ y r ) = i ℏ 3 8 π ( i r − x + i y r 2 y r ) = i ℏ 3 8 π 1 r 3 ( i r 2 − x y − i y 2 ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0675.tif"/> L ^ z Y → 1 , 1 : = ( x p ^ y − y p ^ x ) Y 1 , 1 = i ℏ 3 8 π 1 r 3 { ( i x r 2 − x 2 y − i x y 2 ) = − ( y r 2 − y x 2 − i x y 2 ) } = i ℏ 3 8 π 1 r 3 { i x r 2 − y r 2 } = − ℏ 3 8 π 1 r ( x + i y ) = ℏ Y 1 , 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0676.tif"/> 204 SQ36(1)(b) (1) Consider a transformation from the original coordinate system ( x , y , z ) to a new coordinate system ( x ′, y ′, z ′): x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0677.tif"/> We can see that the new z ′-axis coincides with the original x -axis, and that r = x 2 + y 2 + z 2 = x ′ 2 + y ′ 2 + z ′ 2 = r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0678.tif"/> Physically we expect the angular momentum about the x -axis to be the same as the angular momentum about the z′ -axis. We can confirm this mathematically by showing that L ^ x = y p ^ z − z p ^ y = − i ℏ ( y ∂ ∂ z − z ∂ ∂ y ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0679.tif"/> is equal to L ^ z ′ = x ′ p ^ y ′ − y ′ p ^ x ′ = − i ℏ ( x ′ ∂ ∂ y ′ − y ′ ∂ ∂ x ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0680.tif"/> This is obvious since x′ = y , y′ = z . It follows that the eigenfunctions and eigenvalues of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1700.tif"/> are the same in terms of the dashed coordinates as that of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1701.tif"/> . (2) We know the eigenfunctions Y ℓ , m ℓ ′ ( x ′ , y ′ , z ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1702.tif"/> and eigenvalues of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1703.tif"/> in the transformed coordinates (x′, y′, z′) as they are of the same form as those of the eigenfunctions and eigenvalues of L ^ z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1704.tif"/> in the original coordinates, i.e., we have Y 1 , − 1 ′ ( x ′ , y ′ , z ′ ) = 3 8 π x ′ − i y ′ r ′ , Y 1 , 0 ′ ( x ′ , y ′ , z ′ ) = 3 4 π z ′ r ′ Y 1 , 1 ′ ( x ′ , y ′ , z ′ ) = − 3 8 π x ′ + i y ′ r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0681.tif"/> 205 Since L ^ z ′ = L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1705.tif"/> these are also eigenfunctions of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1706.tif"/> Written in terms of the original coordinates these functions become Y 1 , − 1 ′ = 3 8 π y − i z r , Y 1 , 0 ′ = 3 4 π x r , Y 1 , 1 ′ = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0682.tif"/> We can conclude that the required eigenfunctions L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1707.tif"/> denoted more conveniently by X 1,1 , X 1,0 , X 1, −1 are X 1 , − 1 = 3 8 π y − i z r , X 1 , 0 = 3 4 π x r , X 1 , 1 = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0683.tif"/> Q36(2) Suppose L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1708.tif"/> and L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1709.tif"/> are measured giving the eigenvalues 2 ħ 2 and – ħ , respectively. A measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1710.tif"/> is then made. What are the possible results of the measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1711.tif"/> ? Find the probability of each of these possible results.

•Aug 27, 2020•0 citations
Assume that system 1 evolves according to the dynamics ρ ′ ( t ) ⁢ = i [ H 1 , ρ ( t ) ] + ⁢ θ 1 ( ρ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math407.tif"/> 78 and system 2 evolves according to σ ′ ( t ) = − i [ H 2 , σ ( t ) ] + ⁢ θ 2 ( σ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math408.tif"/> where θ 1 , θ 2 are Lindblad maps: θ 1 ( ρ ) ⁢ = ⁢ ( - 1 / 2 ) Σ k = 1 N ( L k * L k ρ + ρ L k * L k - 2 L k ρ L k * ) θ 2 ( σ ) ⁢ = ⁢ ( - 1 / 2 ) ⁢ Σ k = 1 M ⁢ ( P k * P k σ ⁢ + ⁢ σ P k * P k - 2 P k σ P k * ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math409.tif"/> Then calculate the rate of change of the relative entropy between the two states at time t : H ( ρ ( t ) , σ ( t ) ) ⁢ = T r ( ρ ( t ) ( l o g ( ρ ( t ) ) ⁢ − l o g ( σ ( t ) ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math410.tif"/> hint: Use the formula for the differential of the exponential map in the form: d d t e x p ( Z ( t ) ) ⁢ = e x p ( Z ( t ) ) g ( a d ( Z ( t ) ) − 1 ( Z ′ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math411.tif"/> where g ( z ) = z 1 − exp ⁡ ( − z ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math412.tif"/> Equivalently, Z ′ ( t ) ⁢ = g ( α d ( Z ( t ) ) ( e x p ( − Z ( t ) ) . d d t e x p ( Z ( t ) ) ) f https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math413.tif"/> Take ρ ( t )= exp ( Z ( t )) and deduce that d d t l o g ( ρ ( t ) ) ⁢ = g ( a d ( l o g ( ρ ( t ) ) ) ) ( ρ ( t ) − 1 ρ ′ ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math414.tif"/> Now note that g ( z ) ⁢ = 1 1 − z / 2 ! + z 2 / 3 ! + … = - 1 + Σ k ≥ 1 c [ k ] z k https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math415.tif"/> Note that ρ(t) has eigenvalues in the range [0,1]. So log(ρ(t)) has all eigenvalues in the range (−∞,0]. Hence, ad(log(ρ(t)) has eigenvalues in the range ℝ = (−∞,∞). So in order that we be able to substitute ad(log(ρ(t))) for z in the above equation, we require that the Taylor series for g(z) be convergent for all z ∈ ℝ which may not be the case. Note that Tr(ρ ' (t)) = 0 and that for k ≥ 1, T r ( ρ ( t ) a d ( l o g ( ρ ( t ) ) k ( ρ ( t ) − 1 ρ ′ ( t ) ) ) ⁢ - T r ( a d ( l o g ( ρ ( t ) ) ) k ( ρ ′ ( t ) ) ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math416.tif"/> since the commutator of two operators has trace zero (provided that the operators are bounded). Thus, we get T r ( ρ ( t ) d d t l o g ( ρ ( t ) ) ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math417.tif"/> 79 and so d d t T r ( ρ ( t ) . ⁢ l o g ( ρ ( t ) ) ) ⁢ = T r ( ρ ′ ( t ) . ⁢ l o g ( ρ ( t ) ) ) - T r ( T ( ρ ( t ) ) . ⁢ l o g ( ρ ( t ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math418.tif"/> where T ( X ) ⁢ = - i [ H 1 , X ] + θ 1 ( X ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math419.tif"/> Now, T r ( [ H 1 , ρ ] . ⁢ l o g ( ρ ) ) ⁢ = T r ( [ H 1 ⁢ . ⁢ l o g ( ρ ) , ρ ] ) ⁢ = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math420.tif"/> so d d t ⁢ T r ( ρ ( t ) ⁢ . ⁢ l o g ( ρ ( t ) ) ) ⁢ = T r ( θ 1 ( ρ ( t ) ) ⁢ . ⁢ l o g ( ρ ( t ) ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math421.tif"/>
Let ρ, σ be two states and let spectral be https://www.w3.org/1998/Math/MathML"> ρ = ∑ a | e a > p ( a ) < e a | , σ = ∑ a | f a > q ( a ) < f a | https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003353553/1603a02c-0268-4c9d-8b97-09135b05559e/content/math10_129_1_B.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/> so that < e a | e b >= δ ( a, b ), < f a | f b >= δ ( a, b ). The relative entropy between these two states is https://www.w3.org/1998/Math/MathML"> D ( ρ | σ ) = T r ( ρ . ( log ( ρ ) − log ( σ ) ) ) = ∑ a p ( a ) log ( p ( a ) ) − ∑ a , b p ( a ) log ( q b ) | < e a | f b > | 2 = ∑ a , b ( p ( a ) | < e a | f b > | 2 ( log ( p ( a ) | < e a | f b > | 2 ) − log ( q b | < e a | f b > | 2 ) ) = ∑ a , b ( P ( a , b ) log ( P ( a , b ) ) − log ( Q ( a , b ) ) ) = D ( P | Q ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003353553/1603a02c-0268-4c9d-8b97-09135b05559e/content/math10_129_2_B.tif" xmlns:xlink=" https://www.w3.org/1999/xlink "/>
We wish j t ( X ) to track the noiseless trajectory X d ( t )∈ L ( h ). Assume non-demoltion measurements Y o ( t )= U ( t )* Y i ( t ) U ( t ) are made and the Belavkin filter for π t ( X ) =  j t ( X ) | η t ) , η t = σ ( Y 0 ( s ) : s ≤ t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1006.tif"/> has been constructed as d π t ( X ) = F t ( X ) d t + G t ( X ) d Y o ( t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1007.tif"/> trajectory estimation error X d ( t )− π t ( X ) is given as feedback to the state equations in the form d j t ( X ) = j t ( θ 0 ( X ) ) d t ⁢ + ⁢ j t ( θ 1 ( X ) ) d A ( t ) + j t ( θ 2 ( X ) ) d A ( t ) * + ⁢ j t ( θ 3 ( X ) ) d Λ ( t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1008.tif"/> 192 + ⁢ K ( t ) ( X d ( t ) − π t ( X ) ) d t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1009.tif"/> Alternately, if we assume that X d follows a noiseless trajectory, ie, it evolves according to the equation X d ( t ) , = j t ( 0 ) ( X d ) = U 0 ( t ) * X d U 0 ( t ) , U 0 ( t ) = e x p ( − i t H 0 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1010.tif"/> then we can give as feedback the estimation error E ( t ) = X d ( t ) − T r 2 ( π t ( X ) ( I ⊗ | ϕ ( u ) > < ϕ ( u ) | ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1011.tif"/> which is a system observable. This error feedback is given to the above state equations, ie, to the Evans-Hudson flow resulting in the dynamics d j t ( X ) = j t ( θ 0 ( X ) ) d t + j t ( θ 1 ( X ) ) d A ( t ) + j t ( θ 2 ( X ) ) d A ( t ) * + j t ( θ 3 ( X ) ) d Λ ( t ) + K ( t ) E ( t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1012.tif"/> Alternately, we can incorporate this error feedback in the original HP equation as d U ( t ) ⁢ = ⁢ ( - ( i H + ⁢ P + ⁢ K ( t ) E ( t ) ) d t + L 1 d A ( t ) ⁢ + ⁢ L 2 d A ( t ) * ⁢ + ⁢ S d Λ ( t ) ) U ( t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1013.tif"/> resulting in the Evans-Hudson flow the same as above but with θ 0 (X) replaced by θ 0 ( X )+ iK ( t )[ E ( t ), X ]. Luc-Bouten’s method of control is slightly different. Here, we choose a system observable Z and give an infinitesimal control unitary U c ( t , t ⁢ + ⁢ d t ) ⁢ = ⁢ U ( t ⁢ + ⁢ d t ) * e x p ( - i Z d Y i ( t ) ) U ( t + d t ) ⁢ = ⁢ e x p ( i Z ( t + d t ) d Y o ( t ) ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1014.tif"/> Note that Y o ( t ) ⁢ = ⁢ U ( t + d t ) * Y i ( t ) U ( t + d t ) , Y o ( t + d t ) ⁢ = ⁢ U ( t + d t ) * Y i ( t + d t ) U ( t + d t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1015.tif"/> and hence, forming the difference, we get d Y 0 ( t ) = U ( t + d t ) * d Y i ( t ) U ( t + d t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1016.tif"/> Also, Z commutes with dY i ( t ) and hence Z ( t + dt )= U ( t + dt )* ZU ( t + dt ) commutes with U ( t + dt )* dY o ( t ) U ( t + dt ). Let ρ c ( t ) denote the controlled state at time t, ie after applying the Belavkin filter and control upto time t to the HP evolved state. Then, we apply the Belavkin filter from t to t + dt , via ρ B ( t + d t ) - ρ c ( t ) + δ ρ B ( t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1017.tif"/> where δ ρ B ( t ) = L t * ( ρ c ( t ) ) d t + ( M t ρ c ( t ) + ρ c ( t ) M t * - T r ( M t + M t * ) ρ c ( t ) ) ρ c ( t ) ) ( d Y o ( t ) - T r ( ρ c ( t ) ( M t + M t * ) ) d t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1018.tif"/> Finally, the filtered and controlled state at time t + dt is given by ρ c ( t + d t ) - ρ c ( t ) ⁢ + ⁢ δ ρ c ( t ) = https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1019.tif"/>
H a , a =1,2,..., N are identical copies of a one particle Hamiltonian acting in different Hilbert spaces and V ab , a < b are identical copies of the interaction Hamiltonian between two particles We try a wave function | ψ t > = ⊗ k - 1 N | ψ k t > https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1428.tif"/> Substituting this into the Schrodinger equation gives Σ k ⁢ ψ 1 t ⊗ … ⊗ ( i d ψ k t / d t ) ⊗ … ⊗ ψ n t = = Σ k ⁢ ψ 1 t ⊗ ... ⊗ H k ψ k t ⊗ ... ⊗ ψ n t + + ⁢ Σ a < b ψ 1 t ⊗ ... ⊗ V a b ψ a t ⁢ ... ⊗ ψ b t ⊗ ... ⊗ ψ n t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1429.tif"/> Taking the inner product on both sides with ψ 1 t ⨶.. ψ kt ⨶...⨶ ψ nt where a hat above a symbol means omission of that symbol gives the approximate equations i d | ψ k t > / d t = H k | ψ k t > + Σ m ≠ k < I ⊗ ψ m t | V k m | ψ k t ⊗ ψ m t > https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1430.tif"/> 256 In terms of kernels, we can express this as i d ψ k t ( r ) / d t = H k ψ k t ( r ) + ( Σ m ≠ k ∫ V ( r , r ′ ) | ψ m t ( r ′ ) | 2 d 3 r ′ ) ψ k t ( r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1431.tif"/> This equation however does not take into account the Pauli exclusion principle. For taking that into account, we must try an antisymmetrized wave function | ψ t > = Σ σ ∈ S n ⁢ | ψ σ 1 , t > ⊗ ... ⊗ | ψ σ n , t > https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1432.tif"/> Substituting and taking the inner products after assuming orthonormality of the component wave functions gives us i d | ψ k t > / d t = H k | ψ k t > + Σ m ≠ k < I ⊗ ψ m t | V | ψ k t ⊗ ψ m t > − Σ m ≠ k < ψ m t ⊗ I | V | ψ k t ⊗ ψ m t > https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1433.tif"/> which in coordinate form can be expressed as i d ψ k t ( r ) / d t = H k ψ k t ( r ) + Σ m ≠ k ⁢ ( ∫ V ( r , r ′ ) | ψ m t ( r ′ ) | 2 d 3 r ′ ) ψ k t ( r ) − Σ m ≠ k ⁢ ( ∫ V ( r , r ′ ) ψ ¯ m t ( r ′ ) ψ k t ( r ′ ) d 3 r ′ ) ψ m t ( r ) = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1434.tif"/> These equations are special cases of nonlinear Schrodinger equations of the form i d ψ k t ( r ) / d t = ∇ 2 ψ k t ( r ) / 2 m + V 0 ( r ) ψ k t ( r ) + Σ m ⁢ V k m ⁢ ( r , ψ 1 t , ... , ψ N t ) ψ m t ( r ) , k = 1 , 2 , ... , N https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1435.tif"/> The charge density is ρ ( t , r ) = − e ⁢ Σ k ⁢ | ψ k t ( r ) | 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1436.tif"/> and its rate of increase is given by ρ t ( r ) = - e ⁢ Σ k ( ψ ¯ k t , t ( r ) ⁢ . ⁢ ψ k t ( r ) + ψ ¯ k t ( r ) ⁢ . ⁢ ψ k t , t ( r ) = ( i e / 2 m ) ⁢ Σ k [ ψ k t ( r ) ∇ 2 ψ ¯ k t ( r ) - ψ ¯ k t ( r ) ∇ 2 ψ k t ( r ) ] = d i v ( ( i e / 2 m ) ⁢ Σ k ⁢ [ ψ k t ( r ) ∇ ψ ¯ k t ( r ) - ψ ¯ k t ( r ) ∇ ψ k t ( r ) ] ) = - d i v J ( t , r ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1437.tif"/> 257 where the current density J ( t,r ) is given by J ( t , r ) = ( − i e / 2 m ) Σ k [ ψ k t ( r ) ∇ ψ ¯ k t ( r ) − ψ ¯ k t ( r ) ∇ ψ k t ( r ) ] https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1438.tif"/> It should be noted that this expression for the current density is the same as that derived using the linear Schrodinger equations but the wave functions ψ kt are calculated by solving the nonlinear Schrodinger equation. It should be noted that in deriving the above formula for the current density, we have made used of the relation V ¯ k m ( r , ψ 1 , … , ψ N ) = V m k ( r , ψ 1 , … , ψ N ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9781003163626/7a1b5c8e-5b02-4966-b7fd-03fb766469c4/content/math1439.tif"/> which is readily verified.
Q36(1) The spherical harmonics Y 1,1 ( θ , φ ), Y 1,0 ( θ , φ ), Y 1, −1( θ , φ ) are given by Eqs. (16.65) to (16.67). In Cartesian coordinates, these functions are given by 1 Y 1 , − 1 ( x , y , z ) = 3 8 π x − i y r , Y 1 , 0 ( x , y , z ) = 3 4 π z r , Y 1 , 1 ( x , y , z ) = − 3 8 π x + i y r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0667.tif"/> (a) Using the expression in Cartesian coordinates for the quantised orbital angular momentum operator L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1694.tif"/> in Eq. (27.86), verify by explicit calculations that Y 1,1 ( x , y , z ), Y 1,0 ( x , y , z )and Y 1,-1 ( x , y , z ) are the eigenfunctions of L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1695.tif"/> corresponding to eigenvalues ħ , 0 and – ħ . (b) Consider the following coordinate transformations: x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0668.tif"/> (1) 202 Show that the component of the orbital angular momentum operator along the z ′-direction is the same as that along the x -direction, i.e., show that L ^ c z ′ = L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1696.tif"/> . (2) Show that the simultaneous eigenfunctions of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1697.tif"/> and L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1698.tif"/> corresponding to eigenvalues of L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1699.tif"/> equal to 2 ħ 2 are given in Cartesian coordinates by X 1 , − 1 ( x , y , z ) = 3 8 π y − i z r , X 1 , 0 ( x , y , z ) = 3 4 π x r , X 1 , 1 ( x , y , z ) = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0669.tif"/> SQ36(1)(a) In Cartesian coordinates we have 2 : p ^ x Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ x x − i y r = − i ℏ 3 8 π ( 1 r − x − i y r 2 ∂ ∂ x r ) = − i ℏ 3 8 π ( 1 r − x − i y r 2 x r ) = − i ℏ 3 8 π 1 r 3 ( r 2 − x 2 + i x y ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0670.tif"/> p ^ y Y → 1 , − 1 : = − i ℏ 3 8 π ∂ ∂ y x − i y r = − i ℏ 3 8 π ( − i r − x − i y r 2 ∂ ∂ y r ) = − i ℏ 3 8 π ( − i r − x − i y r 2 y r ) = − i ℏ 3 8 π 1 r 3 ( − i r 2 − x y + i y 2 ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0671.tif"/> L ^ z Y → 1 , − 1 : = ( x p ^ y − y p ^ x ) Y 1 , − 1 = − i ℏ 3 8 π 1 r 3 ( ( − i x r 2 − x 2 y + i x y 2 ) − ( y r 2 − y x 2 + i x y 2 ) ) = − i ℏ 3 8 π 1 r 3 ( − i x r 2 − y r 2 ) = − ℏ 3 8 π 1 r { x − i y } = − ℏ Y 1 , − 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0672.tif"/> 203 Next we have p ^ x Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ x z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ x r ) = i ℏ 3 4 π z x r 3 . p ^ y Y → 1 , 0 : = − i ℏ 3 4 π ∂ ∂ y z r = − i ℏ 3 4 π ( − z r 2 ∂ ∂ y r ) = i ℏ 3 4 π z y r 3 . L ^ z Y → 1 , 0 : = ( x p ^ y − y p ^ x ) Y 1 , 0 = x ( i ℏ 3 4 π z y r 3 ) − y ( i ℏ 3 4 π z x r 3 ) = 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0673.tif"/> Finally we have p ^ x Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ x x + i y r = − i ℏ 3 8 π ( 1 r − x + i y r 2 ∂ ∂ x r ) = i ℏ 3 8 π ( 1 r − x + i y r 2 x r ) = i ℏ 3 8 π 1 r 3 ( r 2 − x 2 − i x y ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0674.tif"/> p ^ y Y → 1 , 1 : = i ℏ 3 8 π ∂ ∂ y x + i y r = i ℏ 3 8 π ( i r − x + i y r 2 ∂ ∂ y r ) = i ℏ 3 8 π ( i r − x + i y r 2 y r ) = i ℏ 3 8 π 1 r 3 ( i r 2 − x y − i y 2 ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0675.tif"/> L ^ z Y → 1 , 1 : = ( x p ^ y − y p ^ x ) Y 1 , 1 = i ℏ 3 8 π 1 r 3 { ( i x r 2 − x 2 y − i x y 2 ) = − ( y r 2 − y x 2 − i x y 2 ) } = i ℏ 3 8 π 1 r 3 { i x r 2 − y r 2 } = − ℏ 3 8 π 1 r ( x + i y ) = ℏ Y 1 , 1 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0676.tif"/> 204 SQ36(1)(b) (1) Consider a transformation from the original coordinate system ( x , y , z ) to a new coordinate system ( x ′, y ′, z ′): x → z ′ , y → x ′ , z → y ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0677.tif"/> We can see that the new z ′-axis coincides with the original x -axis, and that r = x 2 + y 2 + z 2 = x ′ 2 + y ′ 2 + z ′ 2 = r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0678.tif"/> Physically we expect the angular momentum about the x -axis to be the same as the angular momentum about the z′ -axis. We can confirm this mathematically by showing that L ^ x = y p ^ z − z p ^ y = − i ℏ ( y ∂ ∂ z − z ∂ ∂ y ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0679.tif"/> is equal to L ^ z ′ = x ′ p ^ y ′ − y ′ p ^ x ′ = − i ℏ ( x ′ ∂ ∂ y ′ − y ′ ∂ ∂ x ′ ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0680.tif"/> This is obvious since x′ = y , y′ = z . It follows that the eigenfunctions and eigenvalues of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1700.tif"/> are the same in terms of the dashed coordinates as that of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1701.tif"/> . (2) We know the eigenfunctions Y ℓ , m ℓ ′ ( x ′ , y ′ , z ′ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1702.tif"/> and eigenvalues of L ^ z ′ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1703.tif"/> in the transformed coordinates (x′, y′, z′) as they are of the same form as those of the eigenfunctions and eigenvalues of L ^ z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1704.tif"/> in the original coordinates, i.e., we have Y 1 , − 1 ′ ( x ′ , y ′ , z ′ ) = 3 8 π x ′ − i y ′ r ′ , Y 1 , 0 ′ ( x ′ , y ′ , z ′ ) = 3 4 π z ′ r ′ Y 1 , 1 ′ ( x ′ , y ′ , z ′ ) = − 3 8 π x ′ + i y ′ r ′ . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0681.tif"/> 205 Since L ^ z ′ = L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1705.tif"/> these are also eigenfunctions of L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1706.tif"/> Written in terms of the original coordinates these functions become Y 1 , − 1 ′ = 3 8 π y − i z r , Y 1 , 0 ′ = 3 4 π x r , Y 1 , 1 ′ = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0682.tif"/> We can conclude that the required eigenfunctions L ^ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1707.tif"/> denoted more conveniently by X 1,1 , X 1,0 , X 1, −1 are X 1 , − 1 = 3 8 π y − i z r , X 1 , 0 = 3 4 π x r , X 1 , 1 = − 3 8 π y + i z r . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/eqn0683.tif"/> Q36(2) Suppose L ^ c 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1708.tif"/> and L ^ c z https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1709.tif"/> are measured giving the eigenvalues 2 ħ 2 and – ħ , respectively. A measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1710.tif"/> is then made. What are the possible results of the measurement of L ^ c x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429296475/bdc73243-97dc-4cf5-9624-109e4c3275b1/content/ieq1711.tif"/> ? Find the probability of each of these possible results.