Hamiltonian eigenstate problem

  • Thread starter coki2000
  • Start date
91
0
Hi PF members,

I am stuck with a problem about larmor precession. I cannot find the eigenstates of the hamiltonian given as

[itex]H = \frac{\hbar}{2}\begin{pmatrix} \omega_{0} & \omega_{1}\delta(t-t') \\ \omega_{1}\delta(t-t') & \omega_{0} \end{pmatrix}[/itex]

Can anyone help me? Since it has time dependence I cannot figure out how to solve this problem.

Thank you.
 
91
0
Any advice :(
 
134
1
Hi PF members,
[itex]H = \frac{\hbar}{2}\begin{pmatrix} \omega_{0} & \omega_{1}\delta(t-t') \\ \omega_{1}\delta(t-t') & \omega_{0} \end{pmatrix}[/itex]
I'm confused: are those ##\delta##'s Dirac delta functions? If so, then ##\hat{H}(t)## is ill-defined when ##t=t'##. (The integral ##\int \hat{H}(t) dt## can still be defined using step functions.)

I think there might be a typo in there, too. The usual way to write ##\hat{H}## for a spin-1/2 in a ##\mathbf{B}## field is:
##
\hat{H}
= - \tfrac{\hbar\omega_{0}}{2}\hat{\sigma}_z
= \frac{\hbar \omega_{0}}{2}\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}
##
 
91
0
I'm confused: are those ##\delta##'s Dirac delta functions? If so, then ##\hat{H}(t)## is ill-defined when ##t=t'##. (The integral ##\int \hat{H}(t) dt## can still be defined using step functions.)

I think there might be a typo in there, too. The usual way to write ##\hat{H}## for a spin-1/2 in a ##\mathbf{B}## field is:
##
\hat{H}
= - \tfrac{\hbar\omega_{0}}{2}\hat{\sigma}_z
= \frac{\hbar \omega_{0}}{2}\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}
##
Yes, they are dirac-delta functions. Those dirac-deltas stand there for representing a magnetic field pulse in x direction at t = t'
 
Last edited:
91
0
Any ideas?
 
134
1
Yes, they are dirac-delta functions. Those dirac-deltas stand there for representing a magnetic field pulse in x direction at t = t'
When ##t \neq t'##, ##\hat{H}## is ##\tfrac{1}{2}\hbar \omega_0## times the identity matrix. Then every vector is an eigenvector with eigenvalue ##\tfrac{1}{2}\hbar \omega_0##.

When ##t = t'##, ##\hat{H}## is
##
\hat{H} = \frac{\hbar}{2}
\begin{bmatrix}
\omega_0 & \infty \\
\infty & \omega_0
\end{bmatrix}
##
Then ##\hat{H}## is ill-defined. This is why I think the problem has somehow been miscommunicated.
 
91
0
When ##t \neq t'##, ##\hat{H}## is ##\tfrac{1}{2}\hbar \omega_0## times the identity matrix. Then every vector is an eigenvector with eigenvalue ##\tfrac{1}{2}\hbar \omega_0##.

When ##t = t'##, ##\hat{H}## is
##
\hat{H} = \frac{\hbar}{2}
\begin{bmatrix}
\omega_0 & \infty \\
\infty & \omega_0
\end{bmatrix}
##
Then ##\hat{H}## is ill-defined. This is why I think the problem has somehow been miscommunicated.
Yes I understand your point. But problem says that we apply an instantaneous magnetic field in x direction on the precessing particle at time t'. I tried to solve this like the dirac-delta potential problem but it is different I think. Because in this case I cannot find a time dependent wave function for the initial hamiltonian since now potential is time dependent. So we cannot seperate time dependent solution and time-independent solution.
 
1,006
104
When the Hamiltonian is time-dependent, there aren't really "eigenstates." Or rather, for a given t you can find the eigenstates of ##H(t)##, but the eigenstates for different t will in general be different.

However, note that for t < t' and t > t', the Hamiltonian is time-independent. You can find the eigenstates of the time-independent Hamiltonian that governs the system for these times. Then to connect t < t' to t > t' you need to integrate the Schrodinger equation

##i \hbar \frac{d}{dt} \psi(t) = H(t) \psi(t)##

over the infinitesimal interval ##t' - \epsilon < t < t' + \epsilon##. If you take ##\epsilon \to 0## this will give a simple relationship between ##\psi(t - \epsilon)## and ##\psi(t + \epsilon)##.

Mathematically, this procedure is much like the way in which you solve the time-independent Schrodinger equation for the potential ##V(x) = -V_0 \delta(x-x_0)##.
 
91
0
When the Hamiltonian is time-dependent, there aren't really "eigenstates." Or rather, for a given t you can find the eigenstates of ##H(t)##, but the eigenstates for different t will in general be different.

However, note that for t < t' and t > t', the Hamiltonian is time-independent. You can find the eigenstates of the time-independent Hamiltonian that governs the system for these times. Then to connect t < t' to t > t' you need to integrate the Schrodinger equation

##i \hbar \frac{d}{dt} \psi(t) = H(t) \psi(t)##

over the infinitesimal interval ##t' - \epsilon < t < t' + \epsilon##. If you take ##\epsilon \to 0## this will give a simple relationship between ##\psi(t - \epsilon)## and ##\psi(t + \epsilon)##.

Mathematically, this procedure is much like the way in which you solve the time-independent Schrodinger equation for the potential ##V(x) = -V_0 \delta(x-x_0)##.
But in the larmor precession case, eigenvectors of the hamiltonian is all time-independent. How can I know the before and after wave functions. Before and after time t' wave functions are same.
 
1,006
104
Here is the main nontrivial part of the problem:

Suppose the state of the system at ##t = t' - \epsilon## is ##\psi_0##. What is the state at ##t = t' + \epsilon##? Assume ##\epsilon## is infinitesimal.

You can get the answer to this sub-problem by integrating the Schrodinger equation over the time interval in question.

Once you have solved this sub-problem, you should be able to extend it to answer the general question: given the state of the system at any time ##t_1##, what is the state at any other time ##t_2##?
 

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