# Test2 - PHY4221 Quantum Mechanics I Fall Term of 2005 Test...

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Unformatted text preview: PHY4221 Quantum Mechanics I Fall Term of 2005 Test 2 (Suggested Solution) Question 1: (1): We know that the result should be independent on the position of the potential, that is, independent on the exact value of a, so we could simply choose a to be 0. Then the potential is given by: ⎧∞ , x ≤ 0 V ( x) = ⎨ 2 ⎩kx , x > 0 (1) In the region of a>0, this is a simple harmonic oscillator (SHO) whose eigen energy and eigen function we have been known. Because of the infinite high wall at a=0, the wave functions at this point should vanish, so only those eigen functions of SHO which vanish at a=0 could survive due to the modification to the SHO potential. On the other hand, we know that only the eigen functions of SHO who have odd parities will vanish at a=0, so only these eigen functions survive and the even-parity ones are killed. We then conclude that the eigenfunctions and eigenenergies for the given potential are: ψ n ( x) = ⎧ 1 ⎡ x − a ⎤2 ⎫ ⎛ x−a⎞ ⎪ ⎪ Hn ⎜ ⎟ exp ⎨ − ⎢ ⎥ ⎬ , x0 = n mω , ⎝ x0 ⎠ 2 n ! π x0 ⎪ 2 ⎣ x0 ⎦ ⎪ ⎩ ⎭ 1 (2) 1⎤ ⎡ En = ⎢ n + ⎥ ω , n = 1, 3, 5, ⋅⋅⋅ 2⎦ ⎣ where we have rescaled x to be (x-a). (2): Assume such a state exists, and it is a combination of the eigen states of SZ as: ψ = a ↑ +b ↓ , (3) then since <SX> = <SY> =0, we have <S+> = <S-> =0, i.e.: S+ = ψ S+ ψ = 0 S− = ψ S− ψ = 0 . (4) Obviously the equations (4) hold only when a=0 or b=0. But in any case <SZ> is nonzero (and in the case of a=b=0 the state is nonsense). So the state in which <SX> = <SY> = <SZ> =0 does not exist for spin ½ particles. (3): Let us prove the following proposition: in any eigen state of SZ we have <SX> = <SY> =0. Proof: Let ψ be an eigen state of SZ with eigen value sz, then we have: 1 ψ SY S Z − S Z SY ψ i 1 1 = ψ SY S Z ψ − ψ S Z SY ψ , i i sz sz = SY − SY i i =0 SX = where we used the fact that the eigen value of sz is real. Similarly we have <SY> =0. (5) As a result, in the eigen state of SZ which has an eigen value sz=0, we have <SX> = <SY> = <SZ> =0. This state of course exists for spin-1 particles: l , m = 1, 0 . Question 2: The energy calculated using the given trial wave function could be expressed as a function of the parameter α : E (α ) = ∫ ∞ 0 ∫ ˆ r 2 drϕ0 ( r ) H ϕ0 ( r ) ∞ 0 − = r 2 drϕ0 ( r ) ϕ0 ( r ) 2 2µ ∫ ∞ 0 r 2 dr exp ( −α r 2 ) ⎡∇ 2 exp ( −α r 2 ) ⎤ − e 2 ∫ drr exp ( −2α r 2 ) ⎣ ⎦ ∞ 0 ∫ ∞ 0 − = 2 2µ ∫ ∞ 0 drr exp ( −α r 2 ) = α − ) . (1) ∂⎡ r exp ( −α r 2 ) ⎤ − e 2 ∫ drr exp ( −2α r 2 ) ⎦ 0 ∂r 2 ⎣ ∞ 0 2π 2 2 ∫ 32 32 µ drr exp ( −2α r 2 ∞ drr 2 exp ( −2α r 2 ) e2 4α 1 2π 16 α 3 The minimum of E(alpha) is given by the condition: α= 8e4 µ 2 9π 4 . (2) Substituting equation (2) into (1) we obtain an upper bound of the ground state energy: 4e 4 µ ˆ E=− 3π 2 . (3) The exact solution is: E=− e4 µ 22 . Note that in equation (1) the angular variables are not included in the calculation because the trial wave function is independent on them. Question 3: (a): Perturbation solution. The eigen values and eigen functions are: ( En0) = ( n + 1) ω ( ( ( ψ n0,)n = ψ n0) ( x )ψ n0) ( y ) 1 2 in which: 1 2 , (1) n = n1 + n2 n1 , n2 = 0, 1, 2, ⋅ ⋅ ⋅ . (2) First consider the ground state energy (n1 = n2 = 0) and its perturbation: ( ( ( En ) = ψ 0,0) H ′ ψ 0,0) 1 0 0 ( ( = λ ψ 0,00) xy ψ 000) , , (3) ( ( ( ( = λ ψ 00 ) x ψ 00 ) ψ 00 ) y ψ 00 ) =0 in which the last step is because (the subscript x denote the lowering/raising operators with x and px): 1 ( ( + ψ 0 0 ) a x + a x ψ 00 ) 2 1 1 (0 (0 ( ( = axψ 0 ) ψ 0 ) + ψ 00) axψ 00) . 2 2 =0 ( ( ψ 00 ) x ψ 00 ) ∝ (4) So the first order perturbation to the ground state energy vani shes and we need to seek the second order perturbations: ( En ) = ( ( ψ n0,)n H ′ ψ 000) , ∑ 2 1 ( n1 , n2 ) ≠( 0,0) 2 , 2 ( 0) ( E0=0+0 − En =)n1 + n2 0 (5) in which ( ( ψ n0,)n H ′ ψ 000) , 1 2 ( 0) ( = λ ψ n1 ,n2 xy ψ 0,0) 0 ( ( ( ( = λ ψ n10) x ψ 00) ψ n20) y ψ 00) λ ( ( ( ( + + ψ n0) ax + ax ψ 00) ψ n0) a y + a y ψ 00) 2mω λ ( ( ( ( + + = ψ n0) ax ψ 00) ψ n0) a y ψ 00) 2mω λ = δ n ,1δ n ,1 2mω = 1 (6) 2 1 1 , 2 2 as a result only the contribution from n1 = n2 =1 is nonzero and we have: ( En2) = ( ( ψ 1,0) H ′ ψ 000) 1 , ( ( E00) − E20) ⎡ λ⎤ ⎢ 2mω ⎥ ⎦ =−⎣ 2ω =− 2 2 . (7) λ2 8m2ω 3 Equation (7) is the leading order perturbation to the ground state energy. Next let us turn to the perturbations to the first excited state energy: This state is degenerate: n = n1 + n2 =1 so that n1 = 1, n2 =0 or n1 = 0, n2 =1. So we need to consider ( 0) ( 0) the problem in the subspace extended byψ 1,0 andψ 0,1 . The matrix elements of H’ are given as: ( ( ( ⎡ ψ 1,0) H ′ ψ 1(,0) ψ 1,00) H ′ ψ 001) 0 , ⎢0 H′ = ( ( ⎢ ψ ( 0) H ′ ψ ( 0) ψ 0,01) H ′ ψ 001) 1,0 , ⎢ 0,1 ⎣ ( ( ⎡ ψ 1( 0) x ψ 1( 0) ψ 00) y ψ 00) ⎢ =λ ⎢ ψ ( 0) x ψ ( 0) ψ ( 0) y ψ ( 0) 1 1 0 ⎢0 ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ( ( ψ 1( 0) x ψ 00) ψ 00) y ψ 1( 0) ⎤ ( 0) ( 0) ψ0 x ψ0 ( 0) ( 0) + ( 0) ⎡ ( 0) + λ ⎢ ψ 1 ax + a x ψ 1 ψ 0 a y + a y ψ 0 = 2mω ⎢ ψ ( 0) a + + a ψ ( 0) ψ ( 0) a + + a ψ ( 0) 1 1 x x y y 0 ⎢0 ⎣ = ( 0) ψ1 ( 0) y ψ1 ⎥ ⎥ ⎥ ⎦ , ( ( + + ψ 1( 0) ax + ax ψ 00) ψ 00) a y + a y ψ 1( 0) ⎤ ( 0) ( 0) ψ 0 a + ax ψ 0 + x ( 0) ( 0) ψ 1 a + ay ψ 1 + y ⎥ ⎥ ⎥ ⎦ λ ⎡0 1 ⎤ 2mω ⎢1 0 ⎥ ⎣ ⎦ (8) The diagonalization of equation (8) gives the eigen values to b e: λ λ . ,− 2mω 2mω (9) These are the (leading order) corrections to the first excited state energy. (b): With the given unitary transformation we have: H new = U + H oldU L⎤ ⎡ ⎡ L⎤ . = exp ⎢ −iθ z ⎥ [ H 0 + H ′] exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ L⎤ L⎤ ⎡ ⎡ L⎤ ⎡ ⎡ L⎤ = exp ⎢ −iθ z ⎥ H 0 exp ⎢iθ z ⎥ + exp ⎢ −iθ z ⎥ H ′ exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Note that we have [ H 0 , Lz ] = ⎡ ⎢ 1 ( px2 + p y2 ) + 1 mω 2 ( x 2 + y 2 ) , Lz ⎤ ⎥ 2 ⎣ 2m ⎦ 1 1 1 1 2 2 2 2 2 2 ⎡ px , Lz ⎤ + = ⎣ ⎦ 2m ⎡ p y , Lz ⎤ + 2 mω ⎡ x , Lz ⎤ + 2 mω ⎡ y , Lz ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 2m −i i = px p y + px p y − i mω 2 xy + i mω 2 xy m m =0 as expected, so equation (10) changes into: (10) L⎤ L⎤ ⎡ ⎡ L⎤ ⎡ ⎡ L⎤ H new = exp ⎢ −iθ z ⎥ H 0 exp ⎢iθ z ⎥ + exp ⎢ −iθ z ⎥ H ′ exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ L⎤ ⎡ ⎡ L⎤ = H 0 + λ exp ⎢ −iθ z ⎥ xy exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ . (11) L⎤ L⎤ ⎡ ⎡ L⎤ ⎡ ⎡ L⎤ = H 0 + λ exp ⎢ −iθ z ⎥ x exp ⎢iθ z ⎥ exp ⎢ −iθ z ⎥ y exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ We could calculate the terms in the last line of equation (11) as L⎤ ⎡ ⎡ L⎤ exp ⎢ −iθ z ⎥ x exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ = x+ iθ [ x, Lz ] + 1 ⎡ iθ ⎤ ⎡[ x, Lz ] , Lz ⎤ + ⋅ ⋅ ⋅ ⎦ 2! ⎢ ⎥ ⎣ ⎣⎦ 2 12 θ x + ⋅⋅⋅ 2! = x cos θ + y sin θ = x +θ y − L⎤ ⎡ ⎡ L⎤ exp ⎢ −iθ z ⎥ y exp ⎢iθ z ⎥ ⎣ ⎦ ⎣ ⎦ = y+ iθ [ y, Lz ] + 1 ⎡ iθ ⎤ ⎡[ y, Lz ] , Lz ⎤ + ⋅ ⋅ ⋅ ⎦ 2! ⎢ ⎥ ⎣ ⎣⎦ 2 12 θ y + ⋅⋅⋅ 2! = y cos θ − x sin θ = y −θ x − so that equation (11) becomes: H new = H 0 + λ [ x cos θ + y sin θ ][ y cos θ − x sin θ ] = H 0 + λ ⎡ xy ( cos 2 θ − sin 2 θ ) + ( y 2 − x 2 ) sin θ cos θ ⎤ ⎣ ⎦ We could choose the value of theta such that sin θ . (12) = cos θ = 2 / 2 and the equation (12) turns to: H new = H 0 + λ [ x cos θ + y sin θ ] [ y cos θ − x sin θ ] λ ⎡ y 2 − x2 ⎤ ⎦ 2⎣ 2 2 p + p y mω 2 − λ 2 mω 2 + λ 2 x+ y =x + 2m 2 2 2 2 2 px + p y mω12 2 mω2 2 x+ y ≡ + 2m 2 2 = H0 + . (13) So as long as λ<mω2 (this should be satisfied as the perturbation condition), ther e will be bound states. The ground state energy of (13) is of course: EGround = 1 2 (ω1 + ω2 ) λ λ⎤ 1⎡ + ω 1+ ⎢ω 1 − ⎥ 2 2⎣ mω mω 2 ⎦ , 2 2 ⎡⎛ ⎞ ⎛ ⎞⎤ λ λ 1 1λ 1λ = − − ⎢ω ⎜1 − ⎟ + ω ⎜1 + ⎟⎥ 2 ⎣ ⎝ 2mω 2 8 m 2ω 4 ⎠ 2mω 2 8 m 2ω 4 ⎠ ⎦ ⎝ = = ω− (14) 1 λ2 8 m 2ω 3 so we see that the first order perturbation vanishes and the le ading order is the second order, and the perturbation obtained here is the same as that in equat ion (7). The first excited state energy of (13) is: 3 1 ω1 + ω2 2 2 λ λ 3 1 = ω 1− + ω 1+ 2 mω mω 2 2 2 E1,0 = ⎛ ⎛ λ λ 3 1 λ2 ⎞ 1 1 λ2 ⎞ ω ⎜1 − − + ω ⎜1 + − 2 2 4⎟ 2 2 4⎟ 2 ⎝ 2mω 8 m ω ⎠ 2 ⎝ 2mω 8 m ω ⎠ 1 λ 1 λ2 =2 ω− − 2 mω 4 m 2ω 3 1 3 E0,1 = ω1 + ω2 2 2 λ λ 1 3 ω 1− = + ω 1+ 2 2 mω 2 mω 2 = ⎛ ⎛ λ λ 1 1 λ2 ⎞ 3 1 λ2 ⎞ ω ⎜1 − − − ⎟ + ω ⎜1 + 2 24 2 2 4⎟ 2 ⎝ 2mω 8 m ω ⎠ ⎝ 2mω 8 m ω ⎠ 2 1 λ 1 λ2 =2 ω+ − . 2 mω 4 m 2ω 3 = (15) So we see here that the leading order is the first order and th e corrections we obtain here are the same as in equation (9). ...
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## This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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