# Evolution Operator of $H = Z_\text{A} \otimes X_\text{B}$ on a Coupled Qubit System [closed]

I have a Hamiltonian in the tensor product space of a two-qubit system: \begin{align} H = Z_\text{A} \otimes X_\text{B} \end{align} and am to compute the time evolution operator in order to show that it applies A-controlled X-rotations to B, in the end it should look like: \begin{align} U_\text{AB}(t) = e^{-itH} = |0_\text{A}\rangle \langle 0_\text{A}| \otimes e^{-itX_\text{B}}+ |1_\text{A}\rangle \langle1_\text{A}|\otimes e^{itX_\text{B}}. \end{align}

My approach was to write the exponential in both subspaces and express it in trigonometrical form: \begin{align} U_\text{AB}(t) = e^{-itH} = e^{-itZ_\text{A}} \otimes e^{-itX_\text{B}} = \cos(t) \mathbb{1}_\text{A} - i \sin(t)Z_\text{A} \otimes \cos(t) \mathbb{1}_\text{B} -i \sin(t)X_\text{B}. \end{align}

But I got stuck there and don't know whether my first assumption can actually be made. I feel I'm lacking deeper understanding of how exactly the tensor product works. I would be grateful for any kind of advise.

• What is your question? – Norbert Schuch May 3 at 5:22

The Hamiltonian can be written as $$H = Z_A\otimes X_B = (|0_A\rangle\langle0_A|-|1_A\rangle\langle1_A|)\otimes X_B$$, where we have expanded $$Z_A = |0_A\rangle\langle 0_A|-|1_A\rangle\langle 1_A|$$ in its eigenbasis. the evolution operator can be written as $$U_{AB} = e^{-iHt} = e^{-i(|0_A\rangle\langle 0_A|-|1_A\rangle\langle 1_A|)\otimes X_B t}$$ $$= \sum_n (-it)^n (|0_A\rangle\langle 0_A|-|1_A\rangle\langle 1_A|)^n\otimes X_B^n$$ Now, $$(|0_A\rangle\langle 0_A|)^2 = |0_A\rangle\langle 0_A|0_A\rangle\langle 0_A| = |0_A\rangle\langle 0_A|$$ $$\Rightarrow (|0_A\rangle\langle 0_A|)^n = |0_A\rangle\langle 0_A|$$. Similarly, we have $$(|1_A\rangle\langle 1_A|)^n = |1_A\rangle\langle 1_A|$$.
Also, as $$|0_A\rangle$$ and $$|1_A\rangle$$ are orthogonal states, we have all cross terms vanishing: $$\langle 1_A|0_A\rangle = 0 = \langle 0_A|1_A \rangle$$. This leaves us with the following: $$U_{AB} = \sum_n (-it)^n (|0_A\rangle\langle 0_A|+(-1)^n|1_A\rangle\langle 1_A|)\otimes X_B^n$$ $$= |0_A\rangle\langle 0_A|\otimes\sum_n (-it)^n X_B^n+|1_A\rangle\langle 1_A|\otimes \sum_n (it)^n X_B^n$$ $$= |0_A\rangle\langle 0_A|\otimes e^{-i X_B t}+|1_A\rangle\langle 1_A|\otimes e^{i X_B t}$$ And hence, you get the required time evolution! Here, we have assumed that $$\hbar = 1$$ ofcourse. It can be put back in the appropriate place to get- $$= |0_A\rangle\langle 0_A|\otimes e^{-i X_B t/\hbar}+|1_A\rangle\langle 1_A|\otimes e^{i X_B t/\hbar}$$
• Your mistake is that you assume $$e^{A\otimes B}=e^{A}\otimes e^{B}$$. This is wrong - just as, in analogy, $$e^{ab}=e^a\,e^b$$ is wrong for numbers $$a$$ and $$b$$. (Indeed, if $$A$$ and $$B$$ are just numbers, tensor product boils down to the normal product.)
• A very useful fact is that for any $$M$$ with $$M^2=I$$, $$e^{-iMt} = \cos(t)\,I - i\,\sin(t)\, M$$ which you can easily check from the Taylor series.
• Note that $$M=Z_A\otimes X_B$$ has the property that $$M^2=I$$.