# Tag Info

1

Think of this not as an extremely rigorous way of solving the differential equation, but rather as using your intuition to guess a solution. Often when you are given a differential equation, the solution is not at all obvious, and perhaps the equation isn't even solvable analytically. Instead of giving up, though, sometimes you can identify a parameter ...

2

For eq3 and eq4 it is a simple exercise to show that in general $$\text{det }(\mathbb{1}- \mathbb{M}) = 1 - \text{tr }\mathbb{M} + \frac{1}{2} \text{tr }^2\mathbb{M} - \frac{1}{2}\text{tr }\mathbb{M}^2 + \mathcal{O}\left(\mathbb{M}^3\right)$$ hint: use the Levi-civita identity for the determinant Applying this formula immediately gives you ...

2

They are not comparable. LSZ is actually the justification for the previous derivation, but the first is more easily introduced. You should not take the imaginary time in a physical sense. This is a computational method of regularising infinities and it is perfectly ok.

1

Your second equation is strange. If it is 'per unit time', then you should not have the 't' in the numerator. If it is the total transition probability at time $t$, then you should have the 't'. Anyway, your second equation is the result of a summation over the final states. Written in the way you wrote, $|out\rangle$ should be understood as a ...

1

Since the eigenvalues are not degenerate, the correction to the energy level $E_n$ is just $\langle n | \gamma B | n \rangle$. It's easy to see that the correction $\delta E_n$ is $\gamma \hbar$ for all $n$. The correction is good if $$\frac{\delta E_n}{E_n}\ll 1$$ That's $$\frac{\gamma}{\omega(n+\frac{1}{2})}\ll 1$$ For all $n$. And this is guaranteed ...

-1

The first order correction to the energies are given by the eigenvalues of the secular matrix. If the secular matrix is diagonal, then the diagonal entries are the eigenvalues of that matrix. In your case, the secular matrix is not diagonal, so you need to solve the full eigenvalue problem. The diagonal matrix in your case is calculated to be:  \hbar ...

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