# Hamiltonian eigenstate problem

#### coki2000

Hi PF members,

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

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

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

Thank you.

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#### NegativeDept

Hi PF members,
$H = \frac{\hbar}{2}\begin{pmatrix} \omega_{0} & \omega_{1}\delta(t-t') \\ \omega_{1}\delta(t-t') & \omega_{0} \end{pmatrix}$
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:
$\hat{H} = - \tfrac{\hbar\omega_{0}}{2}\hat{\sigma}_z = \frac{\hbar \omega_{0}}{2}\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

#### coki2000

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'

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Any ideas?

#### NegativeDept

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.

#### coki2000

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.

#### The_Duck

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 $V(x) = -V_0 \delta(x-x_0)$.

#### coki2000

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.

#### The_Duck

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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