Quantum Two
Time Dependent Perturbations
Time Dependent Perturbations Perturbations That Turn On
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.
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.
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.
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,
So the situation looks as schematically illustrated below:
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 . . .
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 . . .
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 . . .
If the system is known to be in an eigenstate of at t = 0,
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?
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?
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?
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.
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.
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.
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.
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.
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.
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.
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.
Sudden Perturbations Indeed, in the sudden approximation, the Schrödinger equation can be written in the form
Sudden Perturbations Indeed, in the sudden approximation, the Schrödinger equation can be written in the form
Sudden Perturbations Indeed, in the sudden approximation, the Schrödinger equation can be written in the form
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.
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.
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.
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
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
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
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
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
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.
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.
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.
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.,
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.,
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.,
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.,
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.
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⁺).
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⁺).
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⁺).
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⁺).
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 . . .
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 . . .
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 . . .
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 . . .
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 . . .
Sudden Perturbations: Beta Decay of Tritium so 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.
Sudden Perturbations: Beta Decay of Tritium so 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.
Sudden Perturbations: Beta Decay of Tritium so 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.
Sudden Perturbations: Beta Decay of Tritium so 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.
Sudden Perturbations: Beta Decay of Tritium So 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.
Sudden Perturbations: Beta Decay of Tritium So 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.