1. Quantum Two

3. Time Dependent Perturbations

4. Time Dependent Perturbations Perturbations That Turn On

5. We now consider another class of problems, one in which the measurement question that is asked is slightly different.
Consider a system subject to a time dependent perturbation
in which a perturbation begins to be applied to the system at some fixed instant of time (say t = 0), but that it takes a certain time T to develop to full strength.
The time for the current to build up in external circuits, for example.
Thus,
where T is a the time that it takes for the perturbation to build up to full strength.

6. We now consider another class of problems, one in which the measurement question that is asked is slightly different.
Consider a system subject to a time dependent perturbation
in which a perturbation begins to be applied to the system at some fixed instant of time (say t = 0), but that it takes a certain time T to develop to full strength.
The time for the current to build up in external circuits, for example.
Thus,
where T is a the time that it takes for the perturbation to build up to full strength.

7. We now consider another class of problems, one in which the measurement question that is asked is slightly different.
Consider a system subject to a time dependent perturbation
in which a perturbation begins to be applied to the system at some fixed instant of time (say t = 0), but that it takes a certain time T to develop to full strength.
For example, the time it takes for the current to build up in external circuits.
Thus,
where T is a the time that it takes for the perturbation to build up to full strength.

8. We now consider another class of problems, one in which the measurement question that is asked is slightly different.
Consider a system subject to a time dependent perturbation
in which a perturbation begins to be applied to the system at some fixed instant of time (say t = 0), but that it takes a certain time T to develop to full strength.
For example, the time it takes for the current to build up in external circuits.
Thus,

9. So the situation looks as schematically illustrated below:

10. So the situation looks as schematically illustrated below:
Borrowing from our previous notation we denote by and the eigenvectors and eigenvalues of and we denote by and the corresponding quantities for the final Hamiltonian
Then, by assumption,
and
We then ask the following question . . .

11. So the situation looks as schematically illustrated below:
Borrowing from our previous notation we denote by and the eigenvectors and eigenvalues of and we denote by and the corresponding quantities for the final Hamiltonian
Then, by assumption,
and
We then ask the following question . . .

12. So the situation looks as schematically illustrated below:
Borrowing from our previous notation we denote by and the eigenvectors and eigenvalues of and we denote by and the corresponding quantities for the final Hamiltonian
Then, by assumption,
and
We then ask the following question . . .

13. If the system is known to be in an eigenstate of at t = 0,

14. If the system is known to be in an eigenstate of at t = 0,
what is the amplitude for it to be in the eigenstate |n′〉 of the final Hamiltonian after the perturbation has fully turned on?
This is clearly a relevant question, since information about the admixture of final eigenstates allows us to predict the subsequent evolution for
So the basic question is, what happens to the system between t = 0 and T as the perturbation increases in strength to its final form?

15. If the system is known to be in an eigenstate of at t = 0,
what is the amplitude for it to be in the eigenstate |n′〉 of the final Hamiltonian after the perturbation has fully turned on?
This is clearly a relevant question, since information about the admixture of final eigenstates allows us to predict the subsequent evolution for
So the basic question is, what happens to the system between t = 0 and T as the perturbation increases in strength to its final form?

16. If the system is known to be in an eigenstate of at t = 0,
what is the amplitude for it to be in the eigenstate |n′〉 of the final Hamiltonian after the perturbation has fully turned on?
This is clearly a relevant question, since information about the admixture of final eigenstates allows us to predict the subsequent evolution for
So the basic question is, what happens to the system between t = 0 and T as the perturbation increases in strength to its final form?

17. In general, of course the answer depends upon the exact way in which the perturbation "turns on“.
But it becomes very simple in two limiting cases:
The first is when the perturbation turns on “instantly”, so that T = 0
Thus, when it turns on very quickly compared to unperturbed evolution times we can use what is referred to as the sudden approximation.

18. In general, of course the answer depends upon the exact way in which the perturbation "turns on“.
But it becomes very simple in two limiting cases:
The first is when the perturbation turns on “instantly”, so that T = 0
Thus, when it turns on very quickly compared to unperturbed evolution times we can use what is referred to as the sudden approximation.

19. In general, of course the answer depends upon the exact way in which the perturbation "turns on“.
But it becomes very simple in two limiting cases:
The first is when the perturbation turns on very quickly compared to unperturbed evolution times
In this limit, we can use what is referred to as the sudden approximation.

20. The second is when the perturbation turns on very, very slowly, compared to evolution times of the unperturbed system
In this limit the transition amplitude can be obtained using something referred to as the adiabatic theorem.

21. As a useful thought-experiment that provides a mental mnemonic for remembering what happens in these two cases, consider what happens when a marble is placed at rest in the bottom (i.e., ground state) of a bowl, that has been raised to some predetermined height.
If the raised bowl is then suddenly lowered, the marble will be left hanging in air, in the "ground state" of the raised bowl, not the lowered one.
Indeed, it finds itself in a highly excited state of the lowered bowl.
It does not have time, under these circumstances to respond to the changing conditions (Hamiltonian) until long after the bowl is in the lowered position.

22. When, on the other hand, the bowl is lowered very slowly, the marble stays in the "instantaneous ground state" of the bowl for each elevation, ultimately sitting in the bottom of the bowl in the final lowered position.
These features also characterize the behavior of quantum mechanical systems.

23. Sudden Perturbations
In keeping with the thought experiment just described, it is possible to show quite generally that the state vector |ψ(t)〉 of a system subject to an instantaneous (but finite) change in its Hamiltonian undergoes no change itself as a result of the instantaneous change in H.

24. Sudden Perturbations
In keeping with the thought experiment just described, it is possible to show quite generally that the state vector |ψ(t)〉 of a system subject to an instantaneous (but finite) change in its Hamiltonian undergoes no change itself as a result of the instantaneous change in H.
That is, the state of the system remains continuous during a discontinuous change in the Hamiltonian H.

25. Sudden Perturbations
Indeed, in the sudden approximation, the Schrödinger equation
can be written in the form

26. Sudden Perturbations
Indeed, in the sudden approximation, the Schrödinger equation
can be written in the form

27. Sudden Perturbations
Indeed, in the sudden approximation, the Schrödinger equation
can be written in the form

28. Sudden Perturbations
To understand what happens to the state vector during this change, we formally integrate across the discontinuity in at t = 0, by multiplying by dt
and integrating both sides over an infinitesimal region from -τ to τ
the left hand side is just the change that occurs in the state.

29. Sudden Perturbations
To understand what happens to the state vector during this change, we formally integrate across the discontinuity in at t = 0, by multiplying by dt
and integrating both sides over an infinitesimal region from -τ to τ
the left hand side is just the change that occurs in the state.

30. Sudden Perturbations
To understand what happens to the state vector during this change, we formally integrate across the discontinuity in at t = 0, by multiplying by dt
and integrating both sides over an infinitesimal region from -τ to τ
The left hand side is the change that occurs in the state.

31. Sudden Perturbations
Breaking the integral on the right hand side into two parts, we write
where within each of these two intervals is continuous and of unit norm.
Now, according to the mean value theorem, if f(a) is continuous on (a, b) there exists a value c ∈ (a, b) such that

32. Sudden Perturbations
Breaking the integral on the right hand side into two parts, we write
where within each of these two intervals is continuous and of unit norm.
Now, according to the mean value theorem, if f(a) is continuous on (a, b) there exists a value c ∈ (a, b) such that

33. Sudden Perturbations
Breaking the integral on the right hand side into two parts, we write
where within each of these two intervals is continuous and of unit norm.
Now, according to the mean value theorem, if f(a) is continuous on (a, b) there exists a value c ∈ (a, b) such that

34. Sudden Perturbations
Breaking the integral on the right hand side into two parts, we write
where within each of these two intervals is continuous and of unit norm.
Now, according to the mean value theorem, if f(a) is continuous on (a, b) there exists a value c ∈ (a, b) such that

35. Sudden Perturbations
Breaking the integral on the right hand side into two parts, we write
where within each of these two intervals is continuous and of unit norm.
Now, according to the mean value theorem, if f(t) is continuous on (a, b) there exists a value c ∈ (a, b) such that

36. Sudden Perturbations
Applying this to our two integrals, we deduce that there exists a time and a time such that
The entire right hand side is proportional to τ, so provided that the strength of V₀ is finite, we find that
Hence |ψ(t)〉 remains continuous across any finite discontinuity in H.

37. Sudden Perturbations
Applying this to our two integrals, we deduce that there exists a time and a time such that
The entire right hand side is proportional to τ, so provided that the strength of V₀ is finite, we find that
Hence |ψ(t)〉 remains continuous across any finite discontinuity in H.

38. Sudden Perturbations
Applying this to our two integrals, we deduce that there exists a time and a time such that
The entire right hand side is proportional to τ, so provided that the strength of V₀ is finite, we find that
Hence |ψ(t)〉 remains continuous across any finite discontinuity in H.

39. Sudden Perturbations
Thus in this limit, if the system is initially in an eigenstate of , it will still be in that state immediately after the change in the Hamiltonian has occurred.
The transition amplitude to find it, after the change, in an eigenstate of is then just the inner product between the eigenstates of these two different Hamiltonians, i.e.,

40. Sudden Perturbations
Thus in this limit, if the system is initially in an eigenstate of , it will still be in that state immediately after the change in the Hamiltonian has occurred.
The transition amplitude to find it, after the change, in an eigenstate of is then just the inner product between the eigenstates of these two different Hamiltonians, i.e.,

41. Sudden Perturbations
Thus in this limit, if the system is initially in an eigenstate of , it will still be in that state immediately after the change in the Hamiltonian has occurred.
The transition amplitude to find it, after the change, in an eigenstate of is then just the inner product between the eigenstates of these two different Hamiltonians, i.e.,

42. Sudden Perturbations
Thus in this limit, if the system is initially in an eigenstate of , it will still be in that state immediately after the change in the Hamiltonian has occurred.
The transition amplitude to find it, after the change, in an eigenstate of is then just the inner product between the eigenstates of these two different Hamiltonians, i.e.,

43. Sudden Perturbations: Beta Decay of Tritium
As an interesting example of this class of problem, consider the beta decay of the tritium atom, which is an isotope of hydrogen with a nucleus consisting of 2 neutrons and 1 proton, so Z = 1.
The single bound electron of this atom, which sees an electric potential identical to that of hydrogen, is initially in its ground state, when the tritium nucleus to which it is bound undergoes beta decay, a process in which the nucleus ejects a neutrino and an electron with high kinetic energy (∼17 KeV), leaving behind a Helium nucleus with 2 protons and 1 neutron.

44. Sudden Perturbations: Beta Decay of Tritium
As a result of the quick ejection of the "nuclear" electron, the bound atomic electron sees the potential in which its moving change very quickly from to
Thus, immediately after the beta decay the electron is in the ground state of Hydrogen,
but is moving in a potential corresponding to singly ionized Helium (He⁺).

45. Sudden Perturbations: Beta Decay of Tritium
As a result of the quick ejection of the "nuclear" electron, the bound atomic electron sees the potential in which its moving change very quickly from to
Thus, immediately after the beta decay the electron is in the ground state of Hydrogen,
but is moving in a potential corresponding to singly ionized Helium (He⁺).

46. Sudden Perturbations: Beta Decay of Tritium
As a result of the quick ejection of the "nuclear" electron, the bound atomic electron sees the potential in which its moving change very quickly from to
Thus, immediately after the beta decay the electron is in the ground state of hydrogen,
but is moving in a potential corresponding to singly ionized Helium (He⁺).

47. Sudden Perturbations: Beta Decay of Tritium
As a result of the quick ejection of the "nuclear" electron, the bound atomic electron sees the potential in which its moving change very quickly from to
Thus, immediately after the beta decay the electron is in the ground state of hydrogen,
but is moving in a potential corresponding to singly ionized Helium (He⁺).

48. Sudden Perturbations: Beta Decay of Tritium
It is, therefore, in a linear combination of Helium ion ground and excited eigenstates What is the probability amplitude that an energy measurement will find the electron in, say, the state of the Helium ion?
It is just the inner product between the ground state of Hydrogen (with Z = 1) and the corresponding state (with Z = 2) for the He ion.
For a hydrogenic atom with Z = 2,
so the relevant transition amplitude is . . .

49. Sudden Perturbations: Beta Decay of Tritium
It is, therefore, in a linear combination of Helium ion ground and excited eigenstates What is the probability amplitude that an energy measurement will find the electron in, say, the state of the Helium ion?
It is just the inner product between the ground state of Hydrogen (with Z = 1) and the corresponding state (with Z = 2) for the He ion.
For a hydrogenic atom with Z = 2,
so the relevant transition amplitude is . . .

50. Sudden Perturbations: Beta Decay of Tritium
It is, therefore, in a linear combination of Helium ion ground and excited eigenstates What is the probability amplitude that an energy measurement will find the electron in, say, the state of the Helium ion?
It is just the inner product between the ground state of Hydrogen (with Z = 1) and the corresponding state (with Z = 2) for the He ion.
For a hydrogenic atom with Z = 2,
so the relevant transition amplitude is . . .

51. Sudden Perturbations: Beta Decay of Tritium
It is, therefore, in a linear combination of Helium ion ground and excited eigenstates What is the probability amplitude that an energy measurement will find the electron in, say, the state of the Helium ion?
It is just the inner product between the ground state of Hydrogen (with Z = 1) and the corresponding state (with Z = 2) for the He ion.
For a hydrogenic atom with Z = 2,
so the relevant transition amplitude is . . .

52. Sudden Perturbations: Beta Decay of Tritium
It is, therefore, in a linear combination of Helium ion ground and excited eigenstates What is the probability amplitude that an energy measurement will find the electron in, say, the state of the Helium ion?
It is just the inner product between the ground state of Hydrogen (with Z = 1) and the corresponding state (with Z = 2) for the He ion.
For a hydrogenic atom with Z = 2,
so the relevant transition amplitude is . . .

53. Sudden Perturbations: Beta Decay of Tritiumso
There is, therefore, a 25% chance of it ending up in this state. Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.

54. Sudden Perturbations: Beta Decay of Tritiumso
There is, therefore, a 25% chance of it ending up in this state. Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.

55. Sudden Perturbations: Beta Decay of Tritiumso
There is, therefore, a 25% chance of it ending up in this state. Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.

56. Sudden Perturbations: Beta Decay of Tritiumso
There is, therefore, a 25% chance of it ending up in this state. Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.

57. Sudden Perturbations: Beta Decay of TritiumSo
So there is a 25% chance of it ending up in this state.
Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.

58. Sudden Perturbations: Beta Decay of TritiumSo
So there is a 25% chance of it ending up in this state.
Such transitions can be detected when the electron emits a photon and decays back to the ground state of the He ion.