# How does a general operator in a tensor product space act on operators explicitly written as the tensor product of operators in each subspace?

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.

• 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. Commented Dec 31, 2022 at 18:24

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.

• Thank you so much for your answer!. Commented Jan 5, 2023 at 14:08