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Considering an atom with two states: $|g\rangle$ and $|e\rangle$, its Hamiltonian, when illuminating with some drive frequency $\omega_d$, which couples two states (according to wiki):

$$H/\hbar=\omega|e\rangle\langle e|+\Omega e^{i\omega_dt}|g\rangle\langle e|+\Omega^* e^{-i\omega_dt}|e\rangle\langle g|,$$ where $\Omega$ is some complex drive strength.

According to wiki, if we enter rotating frame of reference by unitary transformation $U=e^{i\omega t|e\rangle\langle e|}$ (my first question - why it is unitary?), we'll get:

$$H/\hbar\to \Omega\, e^{i(\omega_d-\omega)t} |g\rangle \langle e| + \Omega^*\, e^{i(\omega-\omega_d)t} |e\rangle \langle g|,$$

why is it equivalent to initial Hamiltonian? where is $\omega|e\rangle\langle e|$ term?

Curious
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1 Answers1

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To see why U is unitary expand it as the formal power series it is to find the inverse \begin{align*} U = e^{i\omega t|e\rangle\langle e|} = \sum_{n} \frac{(iwt)^n}{n!} |e\rangle\langle e| \end{align*} because $|e\rangle\langle e|$ is a projector higher powers are just the projector itself. Since it's a projector it's also hermitian so if we now conjugate the power series we get $$ U^\dagger = \sum_{n} \frac{(-i\omega t)^n}{n!} |e\rangle\langle e| = e^{-i\omega t |e\rangle\langle e|} $$ We want to see that $U$ is unitary i.e. $UU^\dagger = \mathbb{1}$, to show that you can apply the Baker-Campbell-Hausdorff formula. To do that we first need to calculate the commutator of the terms inside the exponential $$ \big[i\omega t |e\rangle\langle e|, -i\omega t |e\rangle\langle e|\big] = \omega^2 t^2 \left(|e\rangle\langle e| - |e\rangle\langle e| \right) = 0 $$ Since the commutator is 0 we can actually just treat the exponentials as you'd treat the case where the exponent is just some real number i.e.

\begin{align*} U U^\dagger = e^{a}e^{-a} = e^{a + (-a)} = 1 \end{align*} therefore $U^\dagger = U^{-1}$, so $U$ is indeed unitary. It's also easy to generalise this to the case where $A$ is some arbitrary hermitian operator. Then the exponential $e^{iA}$ will be a unitary operator.
We say that two Hamiltonians that are connected by a unitary transformation are equivalent because unitary transformations preserve the eigenvalues, expectations values and the dot product of any two state vectors (the proofs are easy just plug in the definition of the transformation). This implies that any measurements we make will be the same and therefore we say that they are equivalent.

Regarding the disappearance of the diagonal term:

Under a general unitary transformation the transformed hamiltonian will be $$ H\to H' = UHU^\dagger + i\hbar \frac{\partial U}{\partial t}U^\dagger $$ Where the first term is just a change of basis like in normal linear algebra and the second term has to be added for a timedependet transformation because we want the Schrödinger equation to always be true and the vectors transform as $|\psi'\rangle = U |\psi\rangle$ (this is really the starting point) and the time derivative in the SE then leads to the second term in the transformed hamiltonian.

So just looking at the diagonal term the transformation looks something like \begin{align*} H'/\hbar &= U \omega |e\rangle\langle e| U^\dagger + i i\omega |e\rangle\langle e| U U^\dagger \\ &= \omega |e\rangle\langle e| 1 - \omega |e\rangle\langle e| 1\\ &= 0 \end{align*} where in the first term we used the fact that some operator $A$ will commute with the exponential $e^{iA}$ which can by shown by looking at the powerseries expansion and since $[A,A]=0$ you can simply move one A out to the left.

Wihtedeka
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