Time-dependent perturbation theory

In summary, the conversation discusses a problem involving two parts, where the first part involves calculating a desired result and understanding Born's rule for transition probability. The second part involves using first-order time-dependent perturbation theory to compute the transition probability. The confusion arises when calculating the matrix elements of the Hamiltonian and how to properly use the eigenstate basis.
  • #1
davon806
148
1

Homework Statement


The problem consists of 2 parts,the first one(I have done it) is on the following website:
https://www.physicsforums.com/threads/transition-probability-from-two-states.804343/

Q1: I calculated the desired result p(t) = sin^2(Ut/h). However,I don't understand why <1,t | 2 > will give the coefficient corresponds to the transition probability from state 1 to state 2 in Time t. My initial guess is similar to that of OP in the above post, but take the scalar product of the general state with |2> .

In the second part,I was asked to compute the transition probability using 1st-order time-dependent perturbation theory.Using the result from my notes:
Q.jpg

where P_mk means the transition probability from energy eigenstate k to m at time t.

H'_mk = < m(0) | H' | k(0) > , where H' denotes the perturbation.

w_mk = w_m - w_k = ( E_m - E_k )/ h

Q2: I was a bit confused when I calculate H'_mk. Using the definition of scalar product in matrix
representation, I have H'_21 = < u_2 | H' | u_1> , where |u_i> is defined as in the above post.
However,(1 -1)(E U U E) (1 1)(forgive me for the matrix notation...) is 0 since |u_1> and |u_2>
are orthogonal. So the P_21 (t) vanishes which is not the thing I want.

Interestingly, if instead, I take the state m and k to be |2> and |1> respectively, I would have
(0 1)(E U U E) (1 0) = U ,and following the formula I found that P_21 (t) = sin^2(Ut/h)

However,(0 1) and (1 0) are NOT energy eigenstates,but H'_mk works with energy eigenstate.Why would they produce the desired result?

2. Homework Equations

Note that I have calculated that E_1 = E+U and E_2 = E-U where E_1 is the (energy) eigenvalue of |u_1> and etc.

For a clearer version of the original question:
F.jpg


The Attempt at a Solution


Incorporated in question
 
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  • #2
davon806 said:
Q1: I calculated the desired result p(t) = sin^2(Ut/h). However,I don't understand why <1,t | 2 > will give the coefficient corresponds to the transition probability from state 1 to state 2 in Time t.
This is basically Born's rule. The probability of finding the system described by ##| \psi \rangle## in state ##| \phi \rangle## is given by ##\left| \langle \phi | \psi \rangle \right|^2##. Since the system is initially in state ##| 1 \rangle##, i.e., ##| \psi (t=0) \rangle = | 1 \rangle##, calculating ##\left| \langle 2 | \psi (t) \rangle \right|^2## will give the transition probability from ##| 1 \rangle## to ##| 2 \rangle## as a function of time.
davon806 said:
Q2: I was a bit confused when I calculate H'_mk. Using the definition of scalar product in matrix
representation, I have H'_21 = < u_2 | H' | u_1> , where |u_i> is defined as in the above post.
However,(1 -1)(E U U E) (1 1)(forgive me for the matrix notation...) is 0 since |u_1> and |u_2>
are orthogonal. So the P_21 (t) vanishes which is not the thing I want.

Interestingly, if instead, I take the state m and k to be |2> and |1> respectively, I would have
(0 1)(E U U E) (1 0) = U ,and following the formula I found that P_21 (t) = sin^2(Ut/h)

However,(0 1) and (1 0) are NOT energy eigenstates,but H'_mk works with energy eigenstate.Why would they produce the desired result?

In the former case, you simply showed that eigenstates of a time-independent Hamiltonian are constants of motion. It is the second approach that you want here.
Once you decide on a basis set representation, you stick to it! If the system is expressed in the ##\left\{ | 1 \rangle, | 2 \rangle \right\}## basis, then the matrix elements of the Hamiltonian are calculated in that basis, not the eigenstate basis. You can of course decide to work in the eigenstate basis, but then you have to change the vector representation of ##| 1 \rangle## and ##| 2 \rangle##.
 
  • #3
DrClaude said:
This is basically Born's rule. The probability of finding the system described by ##| \psi \rangle## in state ##| \phi \rangle## is given by ##\left| \langle \phi | \psi \rangle \right|^2##. Since the system is initially in state ##| 1 \rangle##, i.e., ##| \psi (t=0) \rangle = | 1 \rangle##, calculating ##\left| \langle 2 | \psi (t) \rangle \right|^2## will give the transition probability from ##| 1 \rangle## to ##| 2 \rangle## as a function of time.
In the former case, you simply showed that eigenstates of a time-independent Hamiltonian are constants of motion. It is the second approach that you want here.
Once you decide on a basis set representation, you stick to it! If the system is expressed in the ##\left\{ | 1 \rangle, | 2 \rangle \right\}## basis, then the matrix elements of the Hamiltonian are calculated in that basis, not the eigenstate basis. You can of course decide to work in the eigenstate basis, but then you have to change the vector representation of ##| 1 \rangle## and ##| 2 \rangle##.

Thanks! I was really happy that someone replies my lengthy and messy post! I understand Q1 but not quite for Q2:

1."In the former case, you simply showed that eigenstates of a time-independent Hamiltonian are constants of motion." ... Why did this imply
the eigenstates of H' is a constant of motion?

2.Below is the first page of my notes(the derivation of time-dependent perturbation)
 
Q.jpg


 Notice the red boxes,everything used in the derivation is the energy eigenstates.I don't understand why |1> and |2> are related to the context.
|1> and |2> actually represents the spin-up and spin-down ,so I think there is nothing to do with energy..?

 
  • #4
davon806 said:
1."In the former case, you simply showed that eigenstates of a time-independent Hamiltonian are constants of motion." ... Why did this imply
the eigenstates of H' is a constant of motion?
What you showed is that if the system is in state ##|u_1 \rangle## or ##|u_2 \rangle##, which are eigenstates of the Hamiltonian, then the system will stay forever in those states, which is what you would expect for a time-independent Hamiltonian.

davon806 said:
2.Below is the first page of my notes(the derivation of time-dependent perturbation)
 Notice the red boxes,everything used in the derivation is the energy eigenstates.I don't understand why |1> and |2> are related to the context.
|1> and |2> actually represents the spin-up and spin-down ,so I think there is nothing to do with energy..?
This is all fine in the abstract Dirac notation. But when you go to matrix-vector notation, you have to make a choice of basis. In the problem, it is stated the basis used is ##|1\rangle \rightarrow (1, 0)^T## and ##|2\rangle \rightarrow (0, 1)^T ##. It is in that basis that
$$
\hat{H} \rightarrow \begin{pmatrix} E & U \\ U & E \end{pmatrix}
$$
such that the matrix elements ##H_{1,2} = H_{2,1} = U##.

Note that in terms of the notes you have posted, ##|1 \rangle## and ##|2 \rangle## are eigenstates of ##\hat{H}_0##,
$$
\hat{H}_0 \rightarrow \begin{pmatrix} E & 0 \\ 0 & E \end{pmatrix}
$$
and the coupling is given by
$$
\hat{H}' \rightarrow \begin{pmatrix} 0 & U \\ U & 0 \end{pmatrix}
$$
So you are working with eigenstates of the base Hamiltonian ##\hat{H}_0##.
 
  • #5
DrClaude said:
This is all fine in the abstract Dirac notation. But when you go to matrix-vector notation, you have to make a choice of basis. In the problem, it is stated the basis used is ##|1\rangle \rightarrow (1, 0)^T## and ##|2\rangle \rightarrow (0, 1)^T ##. It is in that basis that
$$
\hat{H} \rightarrow \begin{pmatrix} E & U \\ U & E \end{pmatrix}
$$
such that the matrix elements ##H_{1,2} = H_{2,1} = U##.

Note that in terms of the notes you have posted, ##|1 \rangle## and ##|2 \rangle## are eigenstates of ##\hat{H}_0##,
$$
\hat{H}_0 \rightarrow \begin{pmatrix} E & 0 \\ 0 & E \end{pmatrix}
$$
and the coupling is given by
$$
\hat{H}' \rightarrow \begin{pmatrix} 0 & U \\ U & 0 \end{pmatrix}
$$
So you are working with eigenstates of the base Hamiltonian ##\hat{H}_0##.

Thanks for clarifying the notations and it makes much more sense to me

My last question: Can you explain why can we decompose the matrix as H_0 and H' as you have shown?
And if H' = ( 0 U U 0) instead of ( E U U E) , then H'_21 = (0 1) (0 U U 0) (1 0) = (U) ?
 
  • #6
davon806 said:
My last question: Can you explain why can we decompose the matrix as H_0 and H' as you have shown?
You have that ##\hat{H} = \hat{H}_0 + \hat{H}'##, so you can decompose it any way you want, so long as both ##\hat{H}_0## and ##\hat{H}'## are hermitian. The most natural approach is to take ##\hat{H}_0## to be the diagonal, and ##\hat{H}'## to something that only induces couplings between the eigenstates of ##\hat{H}_0##. (Most often, ##\hat{H}_0## describes the base system and ##\hat{H}'## an external factor).

davon806 said:
And if H' = ( 0 U U 0) instead of ( E U U E) , then H'_21 = (0 1) (0 U U 0) (1 0) = (U) ?
Yes. But note that you are mixing two approaches. Given the basis set ##\left\{ | \phi_i \rangle \right\}##, the state of the system in vector notation is given by
$$
| \psi \rangle \rightarrow \begin{pmatrix} \langle \phi_1 | \psi \rangle \\ \langle \phi_2 | \psi \rangle \\ \vdots \end{pmatrix}
$$
and any operator ##\hat{A}## is described by a matrix ##\mathbf{A}##, with elements obtained as ##A_{i,j} = \langle \phi_i | \hat{A} | \phi_j \rangle##.

If you are given matrix ##\mathbf{A}##, then you can simply find the elements ##A_{i,j}## by reading them off the matrix.
 
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Related to Time-dependent perturbation theory

1. What is time-dependent perturbation theory?

Time-dependent perturbation theory is a mathematical method used to study the behavior of a quantum system that is subjected to an external time-dependent perturbation. It allows us to calculate the changes in the system’s properties over time due to the perturbation.

2. What is the difference between time-dependent and time-independent perturbation theory?

Time-independent perturbation theory is used when the perturbation is time-independent, meaning it does not change over time. In contrast, time-dependent perturbation theory is used when the perturbation changes over time.

3. What are the applications of time-dependent perturbation theory?

Time-dependent perturbation theory is commonly used in quantum mechanics to study the behavior of atoms, molecules, and other quantum systems under the influence of external fields such as electromagnetic radiation. It is also used in other fields such as chemistry, condensed matter physics, and atomic and molecular physics.

4. What are the limitations of time-dependent perturbation theory?

Time-dependent perturbation theory is based on certain assumptions and approximations, which may not hold true in all cases. It also becomes increasingly complex for systems with more than two energy levels. Additionally, it may not accurately predict the behavior of systems with strong perturbations or for long periods of time.

5. How is time-dependent perturbation theory applied in practice?

In practice, time-dependent perturbation theory involves solving a series of equations to calculate the probability amplitudes of different energy states of the system at different times. This can be done analytically or numerically using computer simulations. The results can then be compared to experimental data to validate the theory.

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