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Exercise ( 第二章 波函数 )

Exercise ( 第二章 波函数 ). 1. Let the wave function. ( is constant). Problem: solve the normalization constant A. 2. Let the wave function. Problem: (a) the probability distribution of particle’s position. (b) Is the wave function normalized?.

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Exercise ( 第二章 波函数 )

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  1. Exercise (第二章 波函数) 1. Let the wave function ( is constant) Problem: solve the normalization constant A 2. Let the wave function Problem: (a) the probability distribution of particle’s position (b) Is the wave function normalized? 3. In the spherical coordinate, the wave function can be expressed by Problem: (a) the probability of finding a particle in (r, r+dr). (b) the probability of finding a particle in solid angle d (立体角) of the direction (, ).

  2. 1. Let the wave function ( is constant) Problem: solve the normalization constant A Solution: According to the property of wave function So, we can obtain Due to So

  3. 2. Let the wave function Problem: (a) the probability distribution of particle’s position (b) Is the wave function normalized? Solution: (a) The probability is (b) The wave function is not normalized.

  4. 3. In the spherical coordinate, the wave function can be expressed by Problem: (a) the probability of finding a particle in (r, r+dr). (b) the probability of finding a particle in solid angle d (立体角) of the direction (, ). Solution: (a) the probability of finding a particle in the whole space is

  5. The probability of finding a particle in (r, r+dr) is (b) the probability of finding a particle in solid angle d of the direction (, ) is

  6. Exercise one(第四章 量子力学中的力学量的算符表示) 1. Prove the following formulas: (1) (2) [L, p2]=0 • 2. Translate the text contents on the page P68~69.

  7. 1. Prove the following formulas: (1) (2) [L, p2]=0 Solution:

  8. So Hence we can obtain

  9. (2) [L, p2]=0 Consider its component x

  10. Using the same deducing, we can obtain the value of other two component y and z is also zero. Therefore, the following relation is true. [L, p2]=0 consider

  11. Exercise two(第四章 量子力学中的力学量的算符表示) 3. Set the mechanics quantity satisfy the most simple algebra (代数) equation: Where C1, C2, ……, Cnare constant coefficients. Please prove that the number of engenvalue of is n, and they are the roots of equation f(x) = 0 • Using uncertainty principle, please estimate the ground state energy of helium atom (氦原子). • Prove the following equations

  12. Imply: set ,prove , then integrate.

  13. -e -e r12 r1 r2 2e • Using uncertainty principle, please estimate the ground state energy of helium atom (氦原子). Solution: The whole Hamiltonian of helium atom is Set the radius which the probability finding atom is maximum is R, (最可几半径)

  14. According to uncertainty principle, So the energy of ground state is The value R must be satisfy the condition that E is minimum extremum, i.e.

  15. So we get a0 is Bohr radius The energy of ground state E is approximate

  16. 2. Prove the following equations Solution: (1) Using the basic commuting relation

  17. So we can obtain (2) So

  18. 3. Set the mechanics quantity satisfy the most simple algebra (代数) equation: Where C1, C2, ……, Cnare constant coefficients. Please prove that the number of engenvalue of is n, and they are the roots of equation f(x) = 0 Solution: Set  is an arbitrary wavefunction, we can get If  is the eigenfunction of operator A, so Where a is eigenvalue of operator A.

  19. So (1) The number of roots of equation (1) should be n, i.e, therefore the number of eigenvalue of operator A is n. If the eigenvalue of A is a1, a2, …, an, the arbitrary eigenfunction of A will satisfy the following equation (2)

  20. Since an arbitrary wavefunction is always described by the linear superposition of eigenfunction of A, equation (2) is true for an arbitrary wavefunction. Therefore

  21. 4. Set Prove Glauber formula Solution: According to the results of above example, we get

  22. Set So So we obtain (1) We introduce parameter , and set (2) So (3) By means of equation (1)

  23. We get Integrate

  24. Hence Set We finally get

  25. Exercise one(第六章 薛定谔方程) 1. Set a particle is defined in infinite potential well Problem: solve its energy eigenvalue, energy eigenfunction and momentum probability distribution. 2. A particle motions in half-infinite potential well V(r) Consider bound state (V2 > E > 0) only Problem: (1) solve the energy level (2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution.

  26. V x a 0 Potential along the x-axis 1. Set a particle is defined in infinite potential well Problem: (a) solve its energy eigenvalue, energy eigenfunction. (b) set the particle stays in ground state, solve the momentum probability distribution. Solution: (1) Inside well Set

  27. The solutions of above equations are (1) C is the normalization constant, δis the phase which must be determined. (2) Outside well According to boundary condition

  28. Insert boundary condition into equation (1), we can get According the normalization condition Therefore we can obtain energy eigenfunction and eigenvalue,

  29. (b) The distribution of momentum probability of a particle staying the ground state The probability finding a particle in momentum range (p, p+dp) is given by Due to the particle staying ground state, the wavefunction is Using Fourier transformation, we can obtain

  30. Using the method of integrate step by step, we can get So

  31. 2. A particle motions in half-infinite potential well V(r) V2 V1 E a x 0 Consider bound state (V2 > E > 0) only Problem: (1) solve the energy level (2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution. Solution:

  32. In the case of only considering the solution of bound state (V1>V2 > E > 0), and by means of Schordinger equation, we can obtain the wave function in different zone, where

  33. In the case of x=0, x=a, the wavefunction and its differential are continuous, i.e. is continuous. So we get Above two equations are equivalent to the following,

  34. Throw off , and obtain the equation of energy level If we can obtain the root kn from above equation, the corresponding energy eigenvalue Enis In general, for every n, we can get a root kn (En).

  35. In symmetrical potential well, there is at least a solution of bound state, but in non-symmetrical potential well, is there a solution of bound state? Now we consider the ground state (n=1). For a anti-trigonometric function sin-1x, the independent variable x satisfies So When the first bound state level presents (n=1),

  36. So the equation of energy level becomes This is the condition that there exists at least one ground state.

  37. (2) If V2=1 eV, V1=1.5 eV, a = 2 nm, the mass of electron is 9.110-31 kg, calculate energy level of electron using the program of Mathematic and draw the sketch of energy level distribution. The equation of energy level is The approximate solution knof above equation can be obtained by the graphic method. The solutions is given by the points intersection of linear line y1 and curve y2,

  38. Insert the values given into above equation, get Set The above equation becomes

  39. Using the program of Mathematic, we can obtain the points of intersection of y1 and y2. In the case of n =1,

  40. In the case of n =2, When n =2, 3, …, the points of intersection of y1 and y2 do not exist. So there is only one bound energy level, i.e., ground state,

  41. Exercise two(第六章 薛定谔方程) 1. A particle with kinetic energy E motions in the delta potential V(x) Solve the transmission coefficient T 2. A particle motions in the double delta potential V(x) (1) Solve the formula of energy level of bound state. (2) This model can be used to approximately simulate the energy level of hydrogen molecule ion (H2+). Set a = 10-5 nm, V0 = 0.1 meV, calculate the energy level of electron using Mathematic program.

  42. V(x) E 0 x 1. A particle with kinetic energy E motions in the delta potential V(x) Solve the transmission coefficient T Solution: The wavefunction satisfies Schordinger equation, At x =0, (x) is continuous, but its differential is not continuous, and satisfies the following equation,

  43. Set the particle moves toward x direction from left side of x = 0, so the incident wavefunction can be expressed by , and the whole wavefunction is given by (1) x 0 Above equation have two linear independent solutions, i.e.

  44. Where is reflection term, is transmission term. According to the continuous condition of (x) at x =0, By means of equation (1), So we obtain

  45. Therefore transmission coefficient and reflection coefficient are

  46. 2. A particle motions in the double delta potential V(x) (1) Solve the formula of energy level of bound state. (2) This model can be used to approximately simulate the energy level of hydrogen molecule ion (H2+). Set a = 2 nm, V0 = 0.1 eV.nm, calculate the energy level of electron using Mathematic program. Solution: For the bound state, E < 0. The wavefunction satisfies Schordinger equation,

  47. At x =a and -a, (x) is continuous, but its differential is not continuous, and satisfies the following equation, (1) (2) Where Set When , Schordinder equation can be expressed by

  48. Above equation have two linear independent solutions, i.e. Due to V(x)=V(-x), wavefunction has certain parity. In the following, we will discuss it in case of the even parity and the odd parity respectively. (a) In the case of even parity The solution of wavefunction is given by

  49. By means of the continuous condition of wavefunction at x = a, and equation (1), we can get (3) Since wavefunction possesses certain parity, the same result will be given using the continuous condition of wavefunction at x = -a, and equation (2). Equation (3) is equivalent to the following equation This is the equation of energy level.

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