Suppose I model an atom as a two-level system with states $|g \rangle$ and $|e\rangle$, with eigenvalue equations $\hat{H_1}|g\rangle = g|g \rangle$ and $\hat{H_1}|e\rangle = e|e \rangle$, and an electromagnetic wave as a harmonic oscillator, with eigenvalue equation $\hat{H_2}|n\rangle = \hbar\omega_p(n+\frac{1}{2})|n\rangle$. The whole system would have a Hamiltonian given by $$ \hat{H_0} = \hat{H_1}\otimes \hat{I_2} + \hat{I_1}\otimes \hat{H_2} = \frac{\hbar\omega_a}{2}(|e \rangle \langle e|-|g \rangle \langle g|)\otimes \hat{I_2} + \hbar\omega_p\hat{I_1}\otimes\left(\hat{a}^†\hat{a}+ \frac{1}{2}\right). $$ A perturbation is turned on:
$$ \delta \hat{H} = \alpha(|g \rangle \langle e|+|e \rangle \langle g|)\otimes (\hat{a}+\hat{a}^ †). $$ Where $\alpha$ is a constant. To find the rotated stated $\bar{\delta H}$, I must compute $\bar{\delta H}= e^{\frac{i\hat{H_0}t}{\hbar}}\delta \hat{H} e^{-\frac{i\hat{H_0}t}{\hbar}}$. I have some trouble understanding how the operators act in these cases. To generalize, suppose $\hat{A}$ is an operator defined in subspace $\zeta_1$ and $\hat{B}$ is an operator defined in subspace $\zeta_2$, then $\hat{A}\otimes\hat{B}$ is an operator defined in the tensor product space $\zeta =\zeta_1 \otimes\zeta_2 $, (As $\delta H$ in this case). If I define $\hat{C} = \hat{D}\otimes\hat{E}$ with $\hat{D}$ in $\zeta_1$ and $\hat{E}$ in $\zeta_2$. I think is reasonable to state: $$ \hat{C}(\hat{A}\otimes\hat{B}) = (\hat{D}\otimes\hat{E})(\hat{A}\otimes\hat{B}) = (\hat{D}\hat{A})\otimes (\hat{E}\hat{B}). $$ (Right?).
In this particular case I can take $\hat{C}$ as $\hat{C} = e^{-\frac{i\hat{H_0}t}{\hbar}}$ which lives in the tensor product space, but since I don't have it in the explicit product of two operators in each subspace, I don't know how to compute $\bar{\delta H}$. So far I have done is to compute the action of $\bar{\delta H}$ on the eigenstates of $\hat{H_0}$, which are $|e \rangle\otimes |n \rangle$ and $|g \rangle \otimes |n \rangle$ for $n=0,1,2,..$.

I would really appreciate any insight on these tensor properties.

  • 1
    $\begingroup$ What is the question, actually? Do you want to compute $e^{iH_0t}$? If so, does this help? This seems like a rather unnecessary long post, tbh. The title of this post is very confusing, too. $\endgroup$ Dec 31, 2022 at 18:24

1 Answer 1


The thing you wrote with the operator $C = D \otimes E$ is correct.

The thing you may have missed is that the operator $H_0$ is the sum of two terms $H_{1,2}$, each of which acts in different subspaces of the full Hilbert space. This means that the two terms $H_{1,2}$ commute with each other. So when we write $\exp (\lambda (A + B))$ for $[A,B]=0$, we can just write this as $ e^{\lambda (A+B)} = e^{\lambda A} \, e^{\lambda B} = e^{\lambda B}\, e^{\lambda A}$. If $[A,B]\neq 0$, then we use the Baker-Campbell-Hausdorff formula.

In other words, $\exp (- i H_0 t) = \exp (- i H_1 t) \, \exp (-i H_2 t)$. This might be all you need to move forward, but just in case, here's the rest:

This whole thing is even easier if you define the Pauli operator $Z = \left| g \middle\rangle \hspace{-0.3mm} \middle\langle g \right| - \left| e \middle\rangle \hspace{-0.3mm} \middle\langle e \right|$ so that $H_1 = \hbar \omega_a Z / 2$ and $\exp (- i H_1 t / \hbar) = \cos (\omega_a t / 2) \, \mathbb{1} - i \, \sin (\omega_a t/2) \, Z$ and $\exp ( - i H_2 t / \hbar ) = \exp( - i \omega_p t \, N / 2 )$ up to a phase that will be cancelled by $\exp ( + i H_2 t / \hbar )$

Now, $\delta H = \alpha \, X \otimes (a + a^{\dagger})$, and we have:

$ e^{i \, H_0 \, t/\hbar} \, \delta H \, e^{-i \, H_0 \, t/\hbar} \, = \, \left[ \, e^{i \, \omega_a \, Z / 2} \, X \, e^{-i \, \omega_a \, Z / 2}\, \right] \otimes \left[ \, e^{i \,\omega_p \,t \, \hat{N}/2} \,\left( \hat{a} + \hat{a}^{\dagger} \right) \, e^{-i \,\omega_p \,t \, \hat{N}/2} \, \right] $

$= \left[\, \left( \cos^2 (\frac{\omega_a \, t}{2} )- \sin^2 (\frac{\omega_a \, t}{2} ) \right) \, X + i \, \sin (\frac{\omega_a \, t}{2} )\, \cos (\frac{\omega_a \, t}{2} ) \, \left[ Z, X \right] \, \right] \otimes \sum\limits_{n=0}^{\infty} \frac{(i \, \omega_p / 2)^n}{n!} \, \left[ \hat{N}, \hat{a} + \hat{a}^{\dagger} \right]_n $

where I used the Campbell identity (one of the BCH guys), where $[A,B]_0 = B$, $[A,B]_1 = [A,B]$, $[A,B]_2 = [A,[A,B]]$, and so on. We then use $[\hat{N},\hat{a}]=-\hat{a}$ and $[\hat{N},\hat{a}^{\dagger}]=\hat{a}^{\dagger}$ to find

$e^{i \, H_0 \, t/\hbar} \, \delta H \, e^{-i \, H_0 \, t/\hbar} \, = \, \left[\, \cos (\omega_a \, t) \, X - \sin (\omega_a \, t )Y \, \right] \otimes \left[ \, e^{- i \, \omega_p \, t/2} \, \hat{a} + e^{i \, \omega_p \, t/2} \, \hat{a}^{\dagger} \right]$

as the answer? Unless I made some algebraic mistakes.

In this case, it's easiest just to deal with the operators. Whenever you deal with tensor-product Hilbert spaces (or many-body quantum mechanics in general), it can be helpful to imagine implicit identities attached to every single- or few-body operator.

  • $\begingroup$ Thank you so much for your answer!. $\endgroup$
    – Spherk
    Jan 5, 2023 at 14:08

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