Time-evolution of quantum state

In summary, the conversation discusses solving a time-evolution problem for a two-level system. The first problem involves finding the probability of the system being in a specific state at a given time, while the second problem introduces an interaction that changes the Hamiltonian and asks for the probability at a different time. Two approaches are discussed, one using the eigenstates and eigenvalues and the other using a time-dependent wave solution. Both methods yield the same result, but the first method may be easier to execute while the second method may be easier to remember.
  • #1
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Homework Statement


A two-level system is setup so the two eigenvalues are given [tex]E_1, E_2[/tex], the Hamiltonian is also given as 2x2 matrix (not shown here). The corresponding eigenstates are easy to solved as

[tex]
|1\rangle =
\left(
\begin{matrix}
1 \\ 0
\end{matrix}
\right), \qquad
|2\rangle =
\left(
\begin{matrix}
0 \\ 1
\end{matrix}
\right)
[/tex]

1) If we know initially the state of the system is in [tex]|2\rangle [/tex], what's the probability of finding the system in [tex]|1\rangle[/tex]?

2) If now we introduce an interaction to change the Hamiltonian

[tex]
H' =
\left(
\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}
\right)
[/tex]
so that the total Hamiltonian is H = H0 + H'

At time is ZERO, the initial state is also in [tex]|2\rangle[/tex], what's probability of finding the system in [tex]|1\rangle[/tex] now?

2. The attempt at a solution
I read some similar examples for solving time-evolution problem but I am quite confuse with the method read in those examples. Here is what I did to solve the problem

For solving problem 1), since we already know the eigenstates and eigenvalues. In addition, the initial state of the system is one eigenstate, so the time-evolution of the system would be

[tex]
|\Psi(t)\rangle = \sum_i c_i \exp(-iE_it/\hbar)|i\rangle = \exp(-iE_2t/\hbar)|2\rangle
[/tex]

But since it doesn't contain [tex]|1\rangle[/tex], hence it is always no chance to find state 1 at time t, that is,

[tex]
\left|\langle 1 |\Psi(t)\rangle\right|^2 = 0
[/tex]

For problem 2), we first solve the eigenvectors and eigenvalues of the total Hamiltonian and get

[tex]
\lambda_1, \quad |\phi_1\rangle ; \qquad \qquad \qquad \lambda_2, \quad |\phi_2\rangle
[/tex]

Now, write [tex]|1\rangle[/tex] and [tex]|2\rangle[/tex] in terms of [tex]|\phi_1\rangle [/tex] and [tex]|\phi_1\rangle [/tex]. Later on, at time t, the state will evolute

[tex]
|\Psi(t)\rangle = \sum_i c_i \exp(-i\lambda_i t/\hbar)|\phi_i\rangle
[/tex]

so the probability to find state [tex]|1\rangle[/tex] will be of the same form [tex]
\left|\langle 1 |\Psi(t)\rangle\right|^2
[/tex] but there [tex]\langle 1 |[/tex] and [tex]|\Psi(t)\rangle[/tex] are all written in terms of the new eigenvectors so the probability is not more zero.

Please tell me if my solution have any problem!

But the way, in some book, they solve the similar problem by assuming the time-dependent wave solution is

[tex]
|\Psi(t)\rangle = \alpha(t) |1\rangle + \beta(t) |2\rangle
[/tex]

Plug these in time-dependent Schrodinger equation and find the following equations

[tex]
i\hbar
\frac{d}{dt}
\left(
\begin{matrix}
{\alpha} \\ {\beta}
\end{matrix}
\right)
=
H
\left(
\begin{matrix}
{\alpha} \\ {\beta}
\end{matrix}
\right)
[/tex]

By solving the simultaneous equations, they find the time-dependent coefficients, so the time-dependent state could be given directly.

I didn't solve the problem with this method, but I wonder if this method is actually equivalent to mine?
 
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  • #2
Yes, the second method is identical to the first method. Or rather, the first method is derived from the second method. The exponential terms that you put in front of the new eigenstates comes from the differential equation that you wrote on the bottom of your response. Either way will work, but I think the first way is easier to do, but can be tricky remembering the details, whereas the second method is tricky to do but easy to remember where to start.
 

1. What is the time-evolution of a quantum state?

The time-evolution of a quantum state refers to the change in a quantum state over time. This change is governed by the Schrödinger equation, which describes how a quantum state evolves in time.

2. How is time-evolution of a quantum state different from classical mechanics?

In classical mechanics, the state of a system can be determined with certainty at any given time. However, in quantum mechanics, the state of a system is described by a wave function that evolves over time and can only predict the probability of finding a particle in a certain state.

3. What factors affect the time-evolution of a quantum state?

The time-evolution of a quantum state is affected by the Hamiltonian of the system, which is the operator that represents the total energy of the system. It is also influenced by any external forces or interactions that the system may experience.

4. Can the time-evolution of a quantum state be reversed?

In general, the time-evolution of a quantum state is irreversible. This is due to the probabilistic nature of quantum mechanics, where the state of a system can only predict the probability of finding a particle in a certain state. However, there are some rare situations where the time-evolution can be reversed, such as in certain quantum systems with time-symmetry.

5. What applications does understanding the time-evolution of quantum states have?

Understanding the time-evolution of quantum states is crucial for many applications in technology, including quantum computing, quantum cryptography, and quantum sensing. It also allows for the study of quantum systems and their behavior, leading to advancements in our understanding of the fundamental laws of nature.

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