Download presentation

Presentation is loading. Please wait.

Published byJamir Lightell Modified over 2 years ago

1
Quantum One: Lecture 4

3
Schrödinger's Wave Mechanics for a Free Quantum Particle

4
In the last lecture we explored implications of Schrödinger's Mechanics for the case in which the Hamiltonian is independent of time. Using the method of separation of variables we obtained separable solutions to the Schrödinger equation that arise when the initial wave function is itself an energy eigenfunction:

5
We discussed appropriate boundary conditions for bound states and continuum states, and excluded those solutions that diverge at infinity. Then, using the fact that the Schrödinger equation is first order in time, and linear, we deduced a 3-step prescription for solving the initial value problem: Given: the Hamiltonian H = T + V (i.e., given V, independent of time) and an arbitrary initial state

6
To find the wave function for times t > 0 1)Solve: 2)Find the initial amplitudes λ n 3)Evolve:

7
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

8
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

9
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m. Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

10
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m. Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

11
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m. Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

12
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m. Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation

13
In this lecture we try to make these ideas more concrete by considering the simplest possible scalar potential energy field the particle could move in, i.e., which corresponds to a free quantum mechanical particle of mass m. Here free means "force free", since the classical force on the particle obviously vanishes if V = 0. In this limit the total energy operator H consists only of the kinetic energy operator so the energy eigenvalue equation reduces for a free particle to the explicit form

14
To simplify this energy eigenvalue equation for the free particle multiply through by. Then, introducing the constant, so that the energy eigenvalue equation above takes the form Thus, the free particle energy eigenvalue equation reduces to what is called Helmholtz equation.

15
To simplify this energy eigenvalue equation for the free particle multiply through by. Then, introducing the constant, so that the energy eigenvalue equation above takes the form Thus, the free particle energy eigenvalue equation reduces to what is called Helmholtz equation.

16
To simplify this energy eigenvalue equation for the free particle multiply through by. Then, introducing the constant, in terms of which the energy eigenvalue equation above takes the form Thus, the free particle energy eigenvalue equation reduces to what is called Helmholtz equation.

17
To simplify this energy eigenvalue equation for the free particle multiply through by. Then, introducing the constant, in terms of which the energy eigenvalue equation above takes the form Thus, the free particle energy eigenvalue equation reduces to what is called Helmholtz equation.

18
To simplify this energy eigenvalue equation for the free particle multiply through by. Then, introducing the constant, in terms of which the energy eigenvalue equation above takes the form Thus, the free particle energy eigenvalue equation reduces to what is called the Helmholtz equation.

19
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables using Cartesian coordinates where it takes the form: and assume separable solutions of the form As before, we substitute into the Helmholtz and divide by φ=XYZ to obtain

20
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables using Cartesian coordinates where it takes the form: and assume separable solutions of the form As before, we substitute into the Helmholtz and divide by φ=XYZ to obtain

21
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables using Cartesian coordinates where it takes the form: We then assume separable solutions of the form As before, we substitute into the Helmholtz and divide by φ=XYZ to obtain

22
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables in Cartesian coordinates where it takes the form: We then assume separable solutions of the form Substituting in, and dividing by φ=XYZ we find that

23
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables using Cartesian coordinates where it takes the form: We then assume separable solutions of the form Substituting in, and dividing by φ=XYZ we find that

24
The Helmholtz equation can be separated in many different coordinates systems. We will focus on applying the method of separation of variables using Cartesian coordinates where it takes the form: and assume separable solutions of the form Substituting in, and dividing by φ=XYZ we find that

25
The individual terms on the left must each equal a constant that add up to -k² Thus we end up with three ordinary differential equations which have the solutions The product of these gives possible free particle energy eigenfunctions: where

26
The individual terms on the left must each equal a constant that add up to -k² Thus we end up with three ordinary differential equations which have the solutions The product of these gives possible free particle energy eigenfunctions: where

27
The individual terms on the left must each equal a constant that add up to -k² Thus we end up with three ordinary differential equations which have the solutions The product of these gives possible free particle energy eigenfunctions: where

28
The individual terms on the left must each equal a constant that add up to -k² Thus we end up with three ordinary differential equations which have the solutions The product of these gives possible free particle energy eigenfunctions: where

29
The individual terms on the left must each equal a constant that add up to -k² Thus we end up with three ordinary differential equations which have the solutions The product of these gives possible free particle energy eigenfunctions: where

30
So we have found functional forms that satisfy the eigenvalue equation. Next step? We have to ask: for what values of are these solutions acceptable? Recall: We just need to exclude solutions that diverge in any direction. But if the constant has a nonzero imaginary part, then the solution will diverge in some direction. You should verify that

31
So we have found functional forms that satisfy the eigenvalue equation. Next step? We have to ask: for what values of are these solutions acceptable? Recall: We just need to exclude solutions that diverge in any direction. But if the constant has a nonzero imaginary part, then the solution will diverge in some direction. You should verify that

32
So we have found functional forms that satisfy the eigenvalue equation. Next step? We have to ask: for what values of are these solutions acceptable? Recall: We just need to exclude solutions that diverge in any direction. But if the constant has a nonzero imaginary part, then the solution will diverge in some direction. You should verify that

33
So we have found functional forms that satisfy the eigenvalue equation. Next step? We have to ask: for what values of are these solutions acceptable? Recall: We just need to exclude solutions that diverge in any direction. But if the constant has a nonzero imaginary part, then the solution will diverge in some direction. You should verify that

34
So we have found functional forms that satisfy the eigenvalue equation. Next step? We have to ask: for what values of are these solutions acceptable? Recall: We just need to exclude solutions that diverge in any direction. But if the constant has a nonzero imaginary part, then the solution will diverge in some direction. You should verify that

35
If all three components are real (positive or negative), then remains bounded. Thus the complete set of solutions is obtained by considering all possible wavevectors The corresponding energy eigenvalues for the free particle (or for the kinetic energy operator, which is the same thing here) take the form The continuous spectrum includes all positive energies, as in the classical theory.

36
If all three components are real (positive or negative), then remains bounded. Thus the complete set of solutions is obtained by considering all possible wavevectors The corresponding energy eigenvalues for the free particle (or for the kinetic energy operator, which is the same thing here) take the form which includes all positive energies, as in the classical theory.

37
Combining each of our energy eigenfunctions with its corresponding time dependence, we obtain the stationary solutions for the free particle, namely where Thus, the free particle energy eigenstates are plane waves traveling in the direction associated with the wavevector So having solved the energy eigenvalue problem for the free particle we are now almost ready to solve the initial value problem for the free particle. But we still have some work to do, which we will motivate with a few preliminary comments.

38
Combining each of our energy eigenfunctions with its corresponding time dependence, we obtain the stationary solutions for the free particle, namely where Thus, the free particle energy eigenstates are plane waves traveling in the direction associated with the wavevector So having solved the energy eigenvalue problem for the free particle we are now almost ready to solve the initial value problem for the free particle. But we still have some work to do, which we will motivate with a few preliminary comments.

39
Combining each of our energy eigenfunctions with its corresponding time dependence, we obtain the stationary solutions for the free particle, namely where Thus, the free particle energy eigenstates are plane waves traveling in the direction associated with the wavevector So, as usual, before proceeding we make a few comments on these results.

40
Combining each of our energy eigenfunctions with its corresponding time dependence, we obtain the stationary solutions for the free particle, namely where Thus, the free particle energy eigenstates are plane waves traveling in the direction associated with the wavevector So, as usual, before proceeding we make a few comments on these results.

41
Combining each of our energy eigenfunctions with its corresponding time dependence, we obtain the stationary solutions for the free particle, namely where Thus, the free particle energy eigenstates are plane waves traveling in the direction associated with the wavevector So, as usual, before proceeding we make a few comments on these results.

42
First, we observe that these free-particle energy eigenstates (or eigenstates of the kinetic energy operator) are also eigenstates of the momentum operator, which is a vector operator with components The eigenvalue equation for the momentum operator takes the form where for this vector operator the eigenvalue itself is also a vector.

43
First, we observe that these free-particle energy eigenstates (or eigenstates of the kinetic energy operator) are also eigenstates of the momentum operator, which is a vector operator with components The eigenvalue equation for the momentum operator takes the form where for this vector operator the eigenvalue itself is also a vector.

44
First, we observe that these free-particle energy eigenstates (or eigenstates of the kinetic energy operator) are also eigenstates of the momentum operator, which is a vector operator with components The eigenvalue equation for the momentum operator takes the form where for this vector operator the eigenvalue itself is also a vector.

45
First, we observe that these free-particle energy eigenstates (or eigenstates of the kinetic energy operator) are also eigenstates of the momentum operator, which is a vector operator with components The eigenvalue equation for the momentum operator takes the form where the eigenvalue itself is a vector.

46
Our plane wave solutions satisfy the momentum eigenvalue equation, since, e.g., For all three components, this implies that or where

47
Our plane wave solutions satisfy the momentum eigenvalue equation, since, e.g., For all three components, this implies that or where

48
Our plane wave solutions satisfy the momentum eigenvalue equation, since, e.g., For all three components, this implies that or where

49
Our plane wave solutions satisfy the momentum eigenvalue equation, since, e.g., For all three components, this implies that or where

50
Thus, Schrödinger's wave mechanics recovers, as a special case, the hypothesis of deBroglie: With every free material particle of momentum and energy we can associate a plane wave of wavevector wavelength and frequency

51
Thus, Schrödinger's wave mechanics recovers, as a special case, the hypothesis of deBroglie: With every free material particle of momentum and energy we can associate a plane wave of wavevector wavelength and frequency

52
Thus, Schrödinger's wave mechanics recovers, as a special case, the hypothesis of deBroglie: With every free material particle of momentum and energy we can associate a plane wave of wavevector wavelength and frequency

53
Thus, Schrödinger's wave mechanics recovers, as a special case, the hypothesis of deBroglie: With every free material particle of momentum and energy we can associate a plane wave of wavevector wavelength and frequency

54
The second comment, is that the probabilistic predictions contained in the Third Postulate, clearly depend on appropriate normalization conditions imposed upon the wave function (e.g, that it be square normalized) and upon the eigenfunctions of the observable of interest. We have also asserted that eigenfunctions associated with continuous eigenvalues are not square normalizable, so we will need a mathematically appropriate normalization convention to deal with that situation. The energy eigenfunctions of the free particle, which has a positive, continuous energy spectrum, clearly fall into this second class. To proceed further, which we will do in the next lecture, we need to address these generalized normalization conditions for observables with a continuous spectrum.

55
The second comment, is that the probabilistic predictions contained in the Third Postulate, clearly depend on appropriate normalization conditions imposed upon the wave function (e.g, that it be square normalized) and upon the eigenfunctions of the observable of interest. We have also asserted that eigenfunctions associated with continuous eigenvalues are not square normalizable, so we will need a mathematically appropriate normalization convention to deal with that situation. The energy eigenfunctions of the free particle, which has a positive, continuous energy spectrum, clearly fall into this second class. To proceed further, which we will do in the next lecture, we need to address these generalized normalization conditions for observables with a continuous spectrum.

56
The second comment, is that the probabilistic predictions contained in the Third Postulate, clearly depend on appropriate normalization conditions imposed upon the wave function (e.g, that it be square normalized) and upon the eigenfunctions of the observable of interest. We have also asserted that eigenfunctions associated with continuous eigenvalues are not square normalizable, so we will need a mathematically appropriate normalization convention to deal with that situation. The energy eigenfunctions of the free particle, which has a positive, continuous energy spectrum, clearly fall into this class.

57
To proceed, therefore, we need to consider normalization conventions for free particle eigenfunctions.

58
Once we do so, we will have all the mathematical tools we will need to treat the initial value problem for the free particle.

59
To proceed, therefore, we need to consider normalization conventions for free particle eigenfunctions. Once we do so, we will have all the mathematical tools we will need to treat the initial value problem for the free particle. This critical extension is covered in the next lecture.

Similar presentations

OK

Quantum One: Lecture 5a. Normalization Conditions for Free Particle Eigenstates.

Quantum One: Lecture 5a. Normalization Conditions for Free Particle Eigenstates.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on cloud gaming Ppt on mars one finalists Certificate templates free download ppt on pollution Ppt on standing order action Ppt on source code theft Paper presentation ppt on nanotechnology Ppt on quality education for all Ppt on e commerce business model Ppt on different forms of agriculture Ppt on l&t finance holdings share price