How do you apply perturbation theory to a magnetic field Hamiltonian?

[Danielk010]

TL;DR

Given,
Take with . Determine the energy eigenvalues exactly and
compare with the results of perturbation theory through second order in .

I understand that the first order results would be $\langle \phi_n^0 | -\mu * B | \phi_n^0 \rangle$ = $\langle \phi_n^0 | -\frac{gq}{2mc}\hat{S} * B | \phi_n^0 \rangle$, the second order results would be $\sum_{k \ne n} \frac{|\langle \phi_n^0 | -\mu * B | \phi_n^0 \rangle|^2} {E_n^{(0)} - E_k^{(0)}}$, and the unperturbed hamiltonian would $E_n^{(0)} = (n + \frac{1}{2})\hbar\omega$.

I am confused on how to evaluate braket of phi as the problem is asking for the exact energy eigenvalues. I tried plugging in $B = \frac{-Zev \times r}{cr^3}$, but then I would have the Z, v and r term, which would not give me an exact solution.


Am I on the right track? Is there an equation I am missing? Thank you for any help on this problem. Also do any of you all know how to add equations to the TL;DR?

 

[renormalize]

TL;DR: Given,
Take with . Determine the energy eigenvalues exactly and
compare with the results of perturbation theory through second order in .

Can you edit your post to very clearly state the problem you are trying to solve?

 

[Danielk010]

Sorry, I tried to insert LaTeX in the TL;DR but it did not display correctly. I am also past the time limit to edit a post.

Here is the problem:
The spin Hamiltonian for a spin-1/2 particle in an external magnetic field is
$$\hat{H} = -\hat{\mu} * B = - \frac{gq}{2mc}\hat{S} * B$$
Take $B = B_0k + B_2j\text{, with }B_2 \ll B_0$. Determine the energy eigenvalues exactly and compare with the results of perturbation theory through second ordder in $\frac{B_2}{B_0}$


I made some progress and I asked the TA for my class for help on this problem. For my initial attempt for the first order differential, I did this:

In the end, I got $\frac{\omega_0 \hbar}{2} sin w_0t$ where $\omega_1 = -\frac{gq}{2mc}(B_0 + B_2)$.
I started by setting the hamiltonian to be: $\textbf{Equation 1} = H = -\frac{gq}{2mc} * \hat{S_z}B_0 + \hat{S_y}B_2$.

Assuming, $\omega_0 = -\frac{gq}{2mc}$, we can set the energy to $E_n^1 = \omega_0(<n|\hat{S_z}B_0|n> + <n|\hat{S_yB_1}|n>)$ given the first-order P.T equation.

Since $B_0$ and $B_2$ are constants, you can get $w_1(<S_z> + <S_y>)$. From equations (4.23 $<S_z> = 0$) and (4.30 $<S_y> = \frac{hbar}{2}sin\omega_0t$)

I get $E_n^1 = \frac{\omega_0 \hbar}{2} sin w_0t$. According to the TA, it should be 0, which I am confused on how they got that.

For the exact energy eigenvalues, he mentioned to use the Pauli matrices.
I used Equation 1 and by plugging in the Pauli matrices, I got $\begin{pmatrix} \omega_0 B_0 & -i\omega_0 B_2 \\ i\omega_0 B_2 & -\omega_0 B_0 \end{pmatrix}$. By taking the det. I got $-B_0^2\omega_0^2 + \lambda^2 - B_2^2\omega_0^2$ => $\lambda = \omega_0 \sqrt{B_0^2 + B_2^2}$. According to the TA, it is supposed to be $\pm \hbar \omega [1 + ...]$, which is not what I got.

Sorry for the lengthy post, but what did I do wrong? I am using the A Modem Approach to Quantum Mechanics Second Edition textbook. Please let me know if there is anything else I need to provide.

 

[vvbn6]

Hm, I am doing right now more or less the same problem and more or less the same questions

I can only say that during the week I have pulled myself quite too far into advanced rotation group theory and QFT, which wasn’t necessary.

The docents, who do QFT and solid state physics, are used to this mathematical language very well, for manipulating with large number of spins. For everyone else (with basic algebra course) this is not obvious.

In generally I at all wanted to push into BraKets singlet and triplet states with arrows up and down, and compose a some kind of a spin matrix 3x3, and calculate a determinant.

And after that, even more questionable things have started.

Did topic starter achieve something yesterday?

 

[vvbn6]

I did this:

In the end, I got ω0ℏ2sinw0t where ω1=−gq2mc(B0+B2).

Pardon, is the message broken? The expression after "I did this" is not showing.

 

[div_grad]

@Danielk010, @vvbn6 The problem becomes simple if you recall that spin operators $S_i$ for spin-$\frac{1}{2}$ particle can be written using the Pauli matrices $\sigma_x, \sigma_y, \sigma_z$ as $S_i = \frac{\hbar}{2}\sigma_i$ (for $i=x,y,z$). The Hamiltonian for this problem is
$$
H = -\vec{\mu}\cdot\vec{B} = -\alpha\vec{S}\cdot\vec{B} = -\frac{\alpha\hbar}{2}\left(\sigma_x B_x + \sigma_yB_y + \sigma_z B_z\right)
$$
where $\alpha$ is a constant. @Danielk010 I guess that the $\vec{B}$ field you gave in Post #3 has only the $y$ and $z$ components, $\vec{B} = (0, B_2, B_0)$. If that is the case, then the Hamiltonian above is
$$
H = -\frac{\alpha\hbar}{2}\left(\sigma_yB_y + \sigma_z B_z\right) = -\frac{\alpha\hbar}{2}\begin{pmatrix}B_0 & -iB_2 \\ iB_2 & -B_0\end{pmatrix} \rm{,}
$$
and this is a simple $2\times 2$ matrix that you can easily diagonalize to find the exact eigenvalues of $H$ and their corresponding eigenvectors.

In the perturbative approach you have $B_2 \ll B_0$, so that your perturbation is $V = -\frac{\alpha\hbar}{2}\sigma_yB_2$, and you compute the matrix elements of $V$ using unpertrubed eigenfunctions of $H_0 = -\frac{\alpha\hbar}{2}\sigma_zB_0$, which can be chosen as $\ket{1} = \begin{pmatrix}1 \\ 0\end{pmatrix}$ and $\ket{2} = \begin{pmatrix}0 \\ 1\end{pmatrix}$ since the matrix is already diagonal.