苏汝铿量子力学课后习题及答案

AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON Exercises for the Course of Quantum Mechanics, 2007 Prof. Ru-Keng Su Shaoyu Yin Jia Zhou & Yi Li Department of Physics, Fudan University, Shanghai 200433, China 1 enjoy 仅供参考 All comments are welcomed. Please email to 051019008@fudan.edu.cn or consult in 2301 the East Building of Guanghua Tower enjoy 机密 enjoy 已修订 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON I. CHAPTER 1 1.1. Using light or electrons as probe, if we want to observe an erythrocyte (red blood cell), how is the order of amplitude of the minimum energy needed respectively? What if a atom is observed? Solution: The probe must has the wave length less than the scale of the observed object. erythrocyte: 7 ∼ 8µm; atom: ∼ 1A˚= 0.1nm. wavelength of light: λphoton = hc E ; (1) wavelength of electron (nonrelativistic): λelectron = h p = h√ 2meE ; (2) wavelength of electron (relativistic): λelectron = h¯ p = h¯√ E2k/c 2 + 2mEk , (3) For erythrocyte, Ephoton = hc 7µm = 1.24nm · keV 7µm ' 0.177eV; Eelectron = h 2 2me(7µm)2 ' 3.06 ∗ 10−8eV. (4) For atom, Ephoton = hc 0.1nm = 1.24nm · keV 0.1nm ' 12.4keV; Eelectron = h 2 2me(0.1nm)2 ' 150eV. (5) (nonrelativistic relation is good enough.) II. CHAPTER 2 2.1. (QM book of Su, Ex.2.7.) A 1-d harmonic oscillator is in the state of ψ(x, t) =√ α pi1/2 e− α2x2 2 − iωt 2 . (i). The expectation value of potential energy? (ii) The probability distribution of the momentum. (iii). The expectation value of kinetic energy? Solution: Compared with the standard harmonic oscillator eigenfunction, it is easy to find that this is the ground state and α = √ mω/h¯, 〈U〉 = ∫ ψ∗(x, t) 1 2 mω2x2ψ(x, t)dx = h¯ω 4 , (6) 2 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON C(p, t) = 1√ 2pih¯ ∫ ψ(x, t)e−ipx/h¯dx = 1 (mωpih¯)1/4 e− p2 2mωh¯ − iωt 2 , (7) 〈T 〉 = ∫ C∗(p, t) p2 2m C(p, t)dp = h¯ω 4 = − ∫ ψ∗(x, t) h¯2 2m d2 dx2 ψ(x, t)dx. (8) Or using the Virial theorem (QM book of Su, Chapter 3.8, P117), 〈T 〉 = 1 2 〈xdU dx 〉 = 〈U〉 = 1 2 〈E〉 = h¯ω 4 . (9) 2.2. Discuss the reflection and transmission ratio of a particle with mass m and positive energy E on a Dirac δ barrier, U(x) = V δ(x) (V > 0). How about if it is a δ well (V < 0)? Solution: − h¯ 2 2m d2 dx2 ψ + V δ(x)ψ = Eψ, (10) take k = √ 2mE/h¯, V˜ = 2mV/h¯2, equation becomes d2 dx2 ψ + k2ψ − V˜ δ(x)ψ = 0. (11) Integrate in the range (−², ²), where ² is an infinitesimal value, ψ′(²)− ψ′(−²) + k2 ∫ ² −² ψdx− V˜ ψ(0) = 0. (12) Taking ²→ 0+, notice that ψ is finite at x = 0, we get ψ′(0+)− ψ′(0−) = V˜ ψ(0). (13) Choose the following ansatz ψ(x) =  Aeikx +Be−ikx, x<0;A′eikx, x>0, (14) the connection condition of ψ at x = 0 yields A+B = A′, ikA′ − ikA+ ikB = V˜ A′, (15) the results are A′ = 2ikA 2ik−V˜ = ik ik−mV/h¯2A, B = V˜ A 2ik−V˜ = mV/h¯2 ik−mV/h¯2A. (16) 3 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON So the transmission ratio is D = ∣∣∣∣A′A ∣∣∣∣2 = k2k2 +m2V 2/h¯4 = 11 + mV 2 2h¯2E , (17) the reflection ratio is R = ∣∣∣∣BA ∣∣∣∣2 = m2V 2/h¯4k2 +m2V 2/h¯4 = 11 + 2h¯2E mV 2 = 1−D. (18) Notice that the above derivation is independent on the sign of V , so if the particle is reflected by a δ well instead of a δ barrier, the above derivation remains unchanged except for the substitution of V by −V , but the final result for the reflection and transmission ratio are the same since they only depend on V 2. 2.3. What’s the energy of ground state of a neutron in the gravitation field? Solution: eigenfunction of energy in stationary state: − h¯ 2 2mn d2 dz2 ψ +mngzψ = Eψ, (19) with boundary condition ψ(z = 0) = 0. Introduce dimensionless parameters ξ = (z − λ)/l, (20) where l3 = h¯2 2m2ng , λ = E mng . (21) The equation becomes d2 dξ2 ψ − ξψ = 0, (22) with a boundary condition ψ(ξ = −λ l ) = 0. This is a 1/3 order Bessel function, whose solution is the Airy function. Corresponding to the ground state, the first zero appears at ξ1 = −2.338, so Eground = mngλ1 = 2.338mng ( h¯2 2m2ng )1/3 ' 1.41 ∗ 10−12eV. (23) 2.4. (QM book of Su, Ex.2.14.) The state of electron in Hydrogen atom is ψ = 1√ pia30 e−r/a0 , where a0 is the Bohr radius. Try to find: (i) The expectation value of r. 4 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON (ii) The expectation value of potential −e2/r. (iii) The probability distribution function of momentum. Solution: (i). Check that the wave function ψ = 1√ pia30 e−r/a0 is normalized: 4pi ∫ ∞ 0 ψ†ψr2dr = ∫ ∞ 0 4 a30 e−2r/a0r2dr = 4 a30 ( a0 2 )3 ∫ ∞ 0 x2e−xdx = 1 2 Γ(3) = 1. (24) So 〈r〉 = 4pi ∫ ∞ 0 ψ†rψr2dr = ∫ ∞ 0 4 a30 e−2r/a0r3dr = 4 a30 ( a0 2 )4 ∫ ∞ 0 x3e−xdx = a0 4 Γ(4) = 3a0 2 . (25) (ii). 〈−e2/r〉 = −e24pi ∫ ∞ 0 ψ† 1 r ψr2dr = −e2 ∫ ∞ 0 4 a30 e−2r/a0rdr = −e2 4 a30 ( a0 2 )2 ∫ ∞ 0 xe−xdx = −e2 a0 Γ(2) = −e 2 a0 . (26) (iii). The probability distribution function of momentum is C(p) = 1 ( √ 2pih¯)3 ∫ ∞ 0 1√ pia30 e−r/a0e− i h¯ ~p·~rr2dr sin θdθdφ = 1√ pia30 1 ( √ 2pih¯)3 ∫ ∞ 0 e−r/a0e− i h¯ pr cos θr2dr sin θdθdφ = 2pi√ pia30 1 ( √ 2pih¯)3 ∫ ∞ 0 e−r/a0 h¯ ip (e− i h¯ pr cos θ|pi0 )rdr = 2pi√ pia30 1 ( √ 2pih¯)3 2h¯ p ∫ ∞ 0 e−r/a0 sin pr h¯ rdr = 2pi√ pia30 1 ( √ 2pih¯)3 2h¯ p Im[ ∫ ∞ 0 e (−1 a0 + ip h¯ )r rdr] = 2pi√ pia30 1 ( √ 2pih¯)3 2h¯ p Im[( −1 a0 + ip h¯ )−2 ∫ ∞ 0 ezzdz] = 2pi√ pia30 1 ( √ 2pih¯)3 2h¯ p 2pa30h¯ 3 (h¯2 + a20p 2)2 = 1√ pia30 1 ( √ 2pih¯)3 8pih¯4a30 (h¯2 + p2a20) 2 . (27) 5 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON 2.5. (QM book of Su, Ex.2.24.) A nonrelativistic particle with mass m moves in a potential U(x, y, z) = A(x2 + y2 + 2λxy) +B(z2 + 2µz), where A > 0, B > 0, |λ| < 1, µ is arbitrary. Now try to find: (i). The eigenvalue of energy. (ii). If now the potential changes to Unew = { U, z > −µ +∞, z < −µ , how about the energy of ground state? Solution: (i). First perform the coordinate transformation: x = 1√ 2 (u+ w), y = 1√ 2 (u− w). (28) Then the potential becomes U(u,w, z) = A( 1 2 (u+ w)2 + 1 2 (u− w)2 + λ(u+ w)(u− w)) +B(z2 + 2µz + µ2 − µ2) = A(1 + λ)u2 + A(1− λ)w2 +B(z + µ)2 −Bµ2, (29) when |λ| < 1, U is a confinement potential in both u and w direction. And ∂2 ∂2x + ∂2 ∂2y = ∂2 ∂2u + ∂2 ∂2w , (30) so by separation of variable, the Schro¨dinger equation (−(h¯2/2m)∇2 + U)ψ = Eψ can be decomposed into three 1-d harmonic oscillators explicitly: (−(h¯2/2m)∂2u + A(1 + λ)u2)ψ(u) = Euψ(u), (−(h¯2/2m)∂2w + A(1− λ)w2)ψ(w) = Ewψ(w), (−(h¯2/2m)∂2z +B(z + µ)2 −Bµ2)ψ(z) = Ezψ(z). (31) ⇒ ωu = √ 2A(1 + λ) m , ωw = √ 2A(1− λ) m , ωz = √ 2B m . 6 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON So Eu = (nu + 1 2 )h¯ωu = (nu + 1 2 )h¯ √ 2A(1 + λ) m , Ew = (nw + 1 2 )h¯ωw = (nw + 1 2 )h¯ √ 2A(1− λ) m , Ez = (nz + 1 2 )h¯ωz −Bµ2 = (nz + 1 2 )h¯ √ 2B m −Bµ2. (nu, nv, nw = 0, 1, 2, · · · ) (32) The eigenvalues of energy are E = Eu + Ew + Ez = (nu + 1 2 )h¯ √ 2A(1 + λ) m + (nw + 1 2 )h¯ √ 2A(1− λ) m + (nz + 1 2 )h¯ √ 2B m −Bµ2. (nu, nv, nw = 0, 1, 2, · · · ) (33) (ii). If now the potential changes to Unew = { U, z > −µ +∞, z < −µ , z = −µ is the position of the center of this potential, so only the states with odd parity in z-direction can survive in Unew. Then in the ground state, nu = nw = 0, nz = 1, its energy is Enew0 = 1 2 h¯ √ 2A(1 + λ) m + 1 2 h¯ √ 2A(1− λ) m + 3 2 h¯ √ 2B m −Bµ2. (34) III. CHAPTER 3 3.1. (QM book of Su, Ex.3.2.) Prove: eLˆAˆe−Lˆ = Aˆ+ 1 n! ∑ n=1[ n times︷ ︸︸ ︷ Lˆ · · · · · · [Lˆ, [Lˆ, Aˆ]] · · · ]. Prove: From f(t) = etLˆAˆe−tLˆ, we have df(t) dt = etLˆLˆAˆe−tLˆ − etLˆAˆLˆe−tLˆ = etLˆ[Lˆ, Aˆ]e−tLˆ, d2f(t) dt2 = etLˆLˆ[Lˆ, Aˆ]e−tLˆ − etLˆ[Lˆ, Aˆ]Lˆe−tLˆ = etLˆ[Lˆ, [Lˆ, Aˆ]]e−tLˆ, · · · dnf(t) dtn = etLˆ[ n times︷ ︸︸ ︷ Lˆ · · · · · · [Lˆ, [Lˆ, Aˆ]] · · · ]e−tLˆ, (35) 7 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON so expand f(1) at f(0), eLˆAˆe−Lˆ = f(1) = f(0) + ∑ n=1 1 n! ( dnf(t) dtn ) t=0 = Aˆ+ 1 n! ∑ n=1 [ n times︷ ︸︸ ︷ Lˆ · · · · · · [Lˆ, [Lˆ, Aˆ]] · · · ]. (36) Proved. 3.2 (QM book of Su, Ex.3.4.) Prove: Skipped. 3.3. (QM book of Su, Ex.3.10.) Explore the uncertainty principle to estimate the energy in ground state of hydrogen atom. Solution: The potential of hydrogen atom is V (r) = −e2/r, the Hamiltonian can be written in the radial coordinate as Hˆ = pˆ2 2m − e 2 r = pˆ2r 2m + lˆ2 2mr2 − e 2 r = pˆ2r 2m − e 2 r , (37) where pˆr = −ih¯( ∂∂r + 1r ) and we have used that 〈lˆ2〉 = 0 for ground state. The total energy is 〈H〉 = 〈 pˆ 2 r 2m 〉 − 〈e 2 r 〉 ≈ 〈pˆ 2 r〉 2m − e 2 〈r〉 . (38) Remember the commutator [r, pˆr] = ih¯, we have the uncertainty principle 〈pˆ2r〉〈r2〉 ≥ h¯ 2 4 (rather than 〈∆pˆ2r〉〈∆r2〉 ≥ h¯ 2 4 since now 〈r〉 6= 0 although 〈~r〉 = 0), one gets 〈pˆ2r〉 ∼ h¯ 2 4〈r2〉 ∼ h¯2 4〈r〉2 , where we have already changed the sign ”≥” into ”∼”. Accordingly, 〈H〉 ∼ h¯ 2 8m〈r〉2 − e2 〈r〉 . (39) For the energy in ground state is the lowest one, the derivative ∂〈H〉/∂〈r〉2 should be zero to meet the extremum condition. Therefore one obtains 〈r〉 = h¯2 4me2 (real Bohr radius a0 = h¯2 me2 ). Then substitute the value of 〈r〉 into the expression of 〈H〉, we can get the energy in ground state that 〈H〉ground ≈ −2me 4 h¯2 . (40) Compare with the real ground energy level E0 = −me42h¯2 , we see that the result we obtained by uncertainty principle is lower than the real case. The inaccuracy comes from the 8 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON approximations 〈1 r 〉 ≈ 1〈r〉 and 〈r2〉 ≈ 〈r〉2 we take during calculation. The ground state is classically corresponding to a circular orbit, so these approximations will introduce not too much inaccuracy and is acceptable. 3.4. (QM book of Su, Ex.3.17.) When t = 0, the wave function for a single particle involving in the harmonic oscillator potential U = 1 2 kx2 is written as ψ(x, 0) = Ae−(αx) 2/2[cos βH0(αx) + sin β 2 √ 2 H2(αx)], (41) where β and A are real constants respectively, α2 = √ mk/h¯2, and the normalization condi- tion of Hermitian Polynomial is∫ +∞ −∞ e−α 2x2 [Hn(αx)] 2dx = √ pi α 2n · n!. (42) (i). Write down the expression of ψ(x, t); (ii). Determine the measured particle’s probable energy values and their respective prob- abilities in state ψ(x, t); (iii). Calculate 〈x〉 at t = 0, and check whether 〈x〉 changes with time t? Solution: (i). First the normalization condition is (the orthonormal relation of Hermite polynomial is applied) ∫ ∞ −∞ |ψ|2dx = |A|2 ∫ ∞ −∞ e−(αx) 2 [cos2 βH20 (αx) + sin2 β 8 H22 (αx)]dx = |A|2 cos2 √ pi α + |A|2 sin 2 β 8 √ pi α 22 ∗ 2! = |A|2 √ pi α = 1, (43) so taking the real value we have A = √ α√ pi . Using the time-dependent unitary operator exp(−iHˆt/h¯) on the state ψ(~x, 0)), one can obtain the time evolution of the state: exp(−iHˆt/h¯)|ψ(~x, 0)〉 = |ψ(~x, t)〉, (44) Notice the initial wavefunction are the superposition of two energy eigenstates of the har- 9 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON monic oscillator corresponding to n = 0 and n = 2 respectively, so |ψ(x, t)〉 = exp(−iHˆt/h¯)ψ(x, 0) = √ α√ pi e− α2x2 2 [cos βH0(αx) exp(−iE0t/h¯) + sin β 2 √ 2 H2(αx) exp(−iE2t/h¯)] = cos β [√ α√ pi e− α2x2 2 H0(αx) exp(−iE0t/h¯) ] + sin β [√ α 8 √ pi e− α2x2 2 H2(αx) exp(−iE2t/h¯) ] , (45) where E0 = h¯ 2 √ k m , E2 = 5 2 h¯ √ k m . (ii). From the expression of ψ(x, t) above, one can immediately recognize that the energy value E0 and E2 are just the probable values of the system’s energy, and P (E0) = cos 2 β, P (E2) = sin 2 β, (46) (iii). 〈x〉 = ∫ ∞ −∞ ψ†(x, 0)xψ(x, 0)dx = 0, (47) since the functions exp(−α2x2), H0(αx) = 1 and H2(αx) = 4α2x2 − 2 are all even functions with respect to x, the total wavefunction is an even function of x. Furthermore, the result is independent on t, because each component evolves as stationary state with a certain phase velocity and the total wavefunction is always an even function of x, therefore 〈x〉 ≡ 0 does not changes with time t. 3.5. (QM book of Su, Ex.3.21.) The wavefunction of a particle is ψ = k(x+ y+2z)e−αr, where r = √ x2 + y2 + z2 and k and α are real constants. (i). What’s the angular momen- tum? (ii). What’s the expectation value of the z-component of the angular momentum? (iii). What’s the probability to obtain h¯ when Lz is measured? (iv). What’s the probability to find the particle in the spherical angle dΩ along the direction θ and φ, where θ and φ are normal directional angles in spherical coordinate? Solutions: (i). x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ, → ψ(r, θ, φ) = kre−αr(sin θ(cosφ + sinφ) + 2 cos θ). Remembering that Y10 = √ 3 4pi cos θ, Y11 = √ 3 8pi sin θeiφ, Y1−1 =√ 3 8pi sin θe−iφ, we can rewrite ψ as ψ = kre−αr( √ 2pi 3 (Y11(1 − i) + Y1−1(1 + i)) + 4 √ pi 3 Y10). 10 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON From this expression we see that the only possible value of the angular momentum quantum number is 1 and L2 = 2h¯2, L = √ 2h¯. (ii). And we see that the possible value of Lz is 0, ±h¯, where h¯ and −h¯ corresponding to Y11 and Y1−1 have the same possibility (square of module of the possibility amplitude), so 〈Lz〉 = 0. (iii). The possibility of Lz = h¯ is |2pi3 (1− i)|2/(|2pi3 (1− i)|2 + |2pi3 (1 + i)|2 + (4 √ pi 3 )2) = 1 6 . (iv). Normalization of the angular part gives Y (θ, φ) = ( √ 2pi 3 (Y11(1− i) + Y1−1(1 + i)) + 4 √ pi 3 Y10)/ √ (|2pi 3 (1− i)|2 + |2pi 3 (1 + i)|2 + (4√pi 3 )2) = √ 1 8pi (sin θ(cosφ + sinφ) + 2 cos θ). So the possibility in the direction θ, φ is independent of the radial part, dP = |Y |2dΩ = 1 8pi (sin θ(cosφ+ sinφ) + 2 cos θ)2dΩ. IV. CHAPTER 4 4.1. (QM book of Su, Ex.4.2.) Suppose the Hermitian operators Aˆ and Bˆ satisfying A2 = B2 = 1, AˆBˆ + BˆAˆ = 0, request: (i). Matrix expression of the operators Aˆ and Bˆ in the representation of Aˆ; (ii). Matrix expression of the operators Aˆ and Bˆ in the representation of Bˆ; (iii). The eigenvalue and eigenfunction of operators Bˆ in the representation of Aˆ; (iv). The eigenvalue and eigenfunction of operators Aˆ in the representation of Bˆ; (v). The unitary translation matrix Sˆ from Aˆ representation to Bˆ representation. Solution: (i). The operators Aˆ and Bˆ are hermite, thus Aˆ† = Aˆ and Bˆ† = Bˆ. According to the condition A2 = B2 = 1, for their respective eigen-matrices X and Y , we have A2X = AˆλAˆX = λ 2 Aˆ X and B2Y = BˆλBˆY = λ 2 Bˆ Y , where λAˆ and λBˆ are their corresponding eigenvalues. Therefore, one obtains λAˆ = ±1 as well as λBˆ = ±1 respectively. Because the matrices expression as for operator Aˆ in its own representation is diagonal, with out considering the degeneracy, the matrix of Aˆ is  1 0 0 −1 . Assume the matrix 11 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON expression of Bˆ is  a b c d . Thus, AˆBˆ + BˆAˆ =  2a 0 0 −2d  = 0 (48) ⇒ a = d = 0. B2 = Bˆ†Bˆ =  c∗c 0 0 b∗b  = I2. (49) Since Bˆ† = Bˆ, we get c = b∗. If let b = eiα, finally we get  0 eiα e−iα 0 , where α is the undefined free phase parameter. If there is degeneracy, notice that from AˆBˆ = −BˆAˆ, we have Tr(Aˆ) = −Tr(BˆAˆBˆ−1) = −Tr(Aˆ) = 0, so the numbers of eigenvalue 1 and −1 are the same, the matrix for Aˆ can then be properly expressed by rearranging the order of the eigenvectors as 1 0 0 −1 1 0 0 −1 1 0 0 −1 . . .  ; (50) while the matrix for Bˆ becomes 0 eiα e−iα 0 0 eiα e−iα 0 0 eiα e−iα 0 . . .  . (51) So there are many duplications of the 2 × 2 matrices and their properties are exactly the same, so the discussion for one 2× 2 matrices is enough. 12 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON (ii). According to the symmetry between Aˆ and Aˆ, we can simply exchange the expres- sions of Aˆ and Bˆ. (iii). Suppose Y =  y1 y2 , when λBˆ = 1, Y = 1√2  1 e−iα ; when λBˆ = −1, Y = 1√ 2  1 −e−iα . (May have a difference on the total phase angle.) (iv). In the representation of Bˆ, X = 1√ 2  1 e−iα  forλAˆ = 1; X = 1√2  1 −e−iα  for λAˆ = −1. (The same as in (iii) due to the symmetry of Aˆ and Bˆ.) (v). Assume Sˆ =  a b c d , then Sˆ  1 0 0 −1  Sˆ−1 =  0 eiα e−iα 0  (52) ⇒  a −b c −d  =  ceiα deiα ae−iα be−iα . This supplies only two constraints for four variables, so we can set a = b = 1, then c = eiα and d = −e−iα. Because Sˆ is unitary, Sˆ† = Sˆ−1, finally one gets Sˆ = 1√ 2  1 1 e−iα −e−iα . (May have a difference on the total phase angle.) This is just the combination of the two eigenvectors of Bˆ in the representation of Aˆ. 4.2. Derive the eigenstates and eigenvalues of the annihilation operator aˆ in particle number representation (energy representation). 13 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON Solution: Assume aˆ |α〉 = α |α〉 , |α〉 =∑∞n=0 cn |n〉 = c0 |0〉+∑∞n=1 cn |n〉 . aˆ |α〉 = α |α〉 ⇒ ∞∑ n=0 cnaˆ |n〉 = ∞∑ n=0 αcn |n〉 ⇒ ∞∑ n=1 cn √ n |n− 1〉 = ∞∑ n=0 αcn |n〉 ⇒ ∞∑ n=0 cn+1 √ n+ 1 |n〉 = ∞∑ n=0 αcn |n〉 ⇒ cn+1 √ n+ 1 = αcn ⇒ cn+1 = α√ n+ 1 cn ⇒ c1 = α√ 1 c0, c2 = α√ 2 c1 = α2√ 2 √ 1 c0, ... cn = αn√ n! c0 ⇒ |α〉 = ∞∑ n=0 cn |n〉 = ∞∑ n=0 αn√ n! c0 |n〉 . The normalization condition: 1 = 〈α|α〉 = ∞∑ m,n=0 α†mαn√ m! √ n! |c0|2 〈m|n〉 = ∞∑ n=0 |α|2n n! |c0|2 = |c0|2 e|α|2 . (53) So c0 = e −|α|2/2+iθ. |α〉 =∑∞n=0 αn√n!e−|α|2/2+iθ |n〉. 4.3. (QM book of Su, Ex.4.5.) The infinite depth well of width a, please give the matrix form of coordinate x and momentum p in the energy representation. 14 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON Solution: The eigenstates are ψn (x) = 1√ a 2 sin ( npi a x ) . (0 ≤ x ≤ a, n = 1, 2, ...). 〈n′|x|n〉 = ∫ a 0 dxψ†n′(x)xψn (x) = 2 a ∫ a 0 dx sin ( n′pi a x ) x sin (npi a x ) = 1 a ∫ a 0 [ x cos ( (n− n′)pix a ) − x cos ( (n+ n′)pi x a )] dx = 1 a [∫ a 0 x cos ( (n− n′)pix a ) dx− ∫ a 0 x cos ( (n+ n′)pi x a ) dx ] = (−1)n−n′ a [∫ 1 0 x cos ((n− n′)pix) dx− ∫ 1 0 x cos ((n+ n′)pix) dx ] . If n = n′, 〈n|x|n〉 = a [∫ 1 0 x dx− ∫ 1 0 x cos (2npix) dx ] = a [ 1 2 − 1 2npi ∫ 1 0 xd sin (2npix) ] = a ( 1 2 − 1 2npi [ x sin (2npix) |10 − ∫ 1 0 sin (2npix) dx ]) = a ( 1 2 + 1 2npi ∫ 1 0 sin (2npix) dx ) = a 2 If n 6= n′, 〈n′|x|n〉 = a{ 1 (n− n′)pi ∫ 1 0 xd sin ((n− n′)pix) − 1 (n+ n′)pi ∫ 1 0 xd sin ((n+ n′)pix) } = a{ 1 (n− n′)pi [ x sin ((n− n′)pix) |10 − ∫ 1 0 sin ((n− n′)pix) dx ] − 1 (n+ n′)pi [ x sin ((n+ n′)pix) |10 − ∫ 1 0 sin ((n+ n′)pix) dx ] } = a [ 1 (n− n′)2 pi2 ∫ 1 0 d cos ((n− n′)pix)− 1 (n+ n′)2 pi2 ∫ 1 0 d cos ((n+ n′)pix) ] = a [ 1 (n− n′)2 pi2 ( (−1)(n−n′) − 1 ) − 1 (n+ n′)2 pi2 ( (−1)(n+n′) − 1 )] = ( a (n− n′)2pi2 − a (n+ n′)2pi2 ) ((−1)n−n′ − 1) = 4a pi2 nn′ (n2 − n′2)2 ( (−1)n−n′ − 1 ) . 15 AL L RI GH TS R ES ER VE D, B Y SH AO -Y U YI N, Y I LI , JI A ZH OU NO T FO R DI ST RI BU TI ON So xn′n vanishes if n and n ′ has the same parity; while equals − 8an