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Your hamiltonian looks pretty wonky, and the notation should give itself away. $F(t)$ is a force, so what is it doing inside the hamiltonian without an $x$? Instead, you might want to consider the ha …

As mentioned in the comments, this is known as the Rotating-Wave Approximation, and it holds because the exponents $(\omega\pm \omega_0)t$ are real, so that the exponentials $|e^{i(\omega\pm \omega_0) …

Ooofff, there's a lot going on here. For one, the choice in the test case in Sakurai's (5.1.5), which is seen more clearly in the equivalent equation (5.1.6), $$ H = \begin{pmatrix} E_1^{(0)} & \lam …

There are two main reasons. The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where …

$ \newcommand{\ket}[1]{\left \lvert #1 \right \rangle} \newcommand{\bra}[1]{\left \langle #1 \right \rvert} \newcommand{\braket}[2]{\left \langle #1 | #2 \right \rangle} \newcommand{\bbraket}[3]{\left …

As mentioned in the comments by Bubble, this is answered in Ground State Energy Calculations for the Quartic Anharmonic Oscillator, Robert Smith. Notes for Math 4901, University of Minnesota, Morr …

No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretica …

Fourier frequencies, and particularly complex ones, are best thought about in terms of oscillating exponentials rather than sines and cosines. That is, you express the function of interest $f(t)$ as s …

The presence of $\omega$ in the second expression should give the game away: what you call the 'generalized' Rabi frequency is the frequency of the oscillations in population when you drive a two-leve …

As I mentioned in the comments, the assertion that $E_n^{(1)}\equiv0$ cannot hold in general since a scalar perturbation does not obey it. For the particular case you mention, a linear perturbation o …

Simply put, force is always the spatial gradient of energy. That's what $$ \mathrm dU=-\mathbf F\cdot\mathrm d\mathbf r $$ really means: if you have an energy that depends on position, then you have a …

To be honest, it depends on the situation, and particularly on how long the pump lasts and what the lifetime of the states is. If you have long-lived states, then sequential processes become possible …

No, your calculation is not correct $-$ you're using a non-canonical transformation in ways which assume that it is canonical. More specifically, it is correct to say that if your hamiltonian maps in …

No, the two explanations are not the same. The Wikipedia article is alarmingly incorrect. To be clear: the splitting of energy levels as atoms are brought together to form a molecule, and the formatio …

Electrons interact because they repulse each other electrostatically, and there is a specific interaction term in the hamiltonian for that. This acts regardless of whether the wavefunctions interact, …