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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.

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  • $\begingroup$ What is your question? $\endgroup$ – Norbert Schuch May 3 at 5:22
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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- $$ = |0_A\rangle\langle 0_A|\otimes e^{-i X_B t/\hbar}+|1_A\rangle\langle 1_A|\otimes e^{i X_B t/\hbar}$$

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  • 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 $M^2=I$.

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