# Tag Info

7

It depends on the Hamiltonian. There are three particular questions whose answers might influence your choice of strategy: Does the Hamiltonian have any particular structure or symmetry? How quickly does the Hamiltonian change in time? What do you know about the initial state in relation to the initial Hamiltonian? Obviously, if the Hamiltonian has any ...

7

Hint: Instead of using the BCH formula in the form usually presented, for example at the top of this Wikipedia page, use this consequence of Hadamard's Lemma: $$\tag{1} e^{iHt}\hat{a}e^{-iHt} = \hat{a} + [iHt,\hat{a}] + \frac{1}{2!}[iHt,[iHt,\hat{a}] + \cdots$$ Now substitute $H$ into the right-hand side and evaluate the commutators between $\hat{a}$ and ...

6

I'm going to define $|n\rangle$ to be "the walker is at site $n$". Now imagine the walk as specified: $$|n\rangle\rightarrow (|n-1\rangle+|n+1\rangle)/\sqrt{2}.$$ You can put some phases in if you want to, it's not going to change my basic argument. Now, imagine this is implemented by a unitary operator. This means that we need $$\langle n-1|U^\... 6 Generally speaking, a realization of a quantum gate involves coherent manipulation of a two-level system (but this is nothing new to you, maybe). For example, you can use two long-lived electronic states in a trapped atom (neutral or ionized in vacuo) and use an applied electric field to implement single-qubit operations (see trapped ions or optical lattices,... 5 Use the differential form of the time evolution,$$dO/dt=i[H, O]\ .$$5 Calculate$$ \begin{align} \hat{U}|00\rangle &= \exp\left(-igt(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)\right)|00\rangle \\ &= \sum_{k=0}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\hat{a}_2)^k|00\rangle \\ &= |00\rangle + \sum_{k=1}^\infty \frac{(-igt)^k}{k!}(\hat{a}^\dagger_2\hat{a}_1+\hat{a}^\dagger_1\...

4

There's more than one way, and I'll suggest two of them here: Expand $\hat{U}$ using the formula for the Taylor series of an exponential ($e^\hat{A}$) centered around $\hat{A}=\hat{0}$, and then you will have a sum of terms where each term no longer involves an exponential operator (i.e. you have just pure creation and annihilation operators and products/...

2

Let $|\psi\rangle$ be an eigenstate of an operator $A$, $A|\psi\rangle=\lambda|\psi\rangle$. Then $$e^A |\psi\rangle = \sum_{k=0}^\infty \frac{A^k}{k!}|\psi\rangle = \sum_{k=0}^\infty \frac{\lambda^k}{k!}|\psi\rangle = e^\lambda |\psi\rangle.$$ In this particular case, $A=-igt(a_2^\dagger a_1+a_1^\dagger a_2)$, of which $|00\rangle$ is an eigenstate with ...

2

Note that $$[(a^\dagger)^n,a] = -n(a^{\dagger})^{n-1}, \qquad [(a^\dagger)^n a^m,a] = -n (a^\dagger)^{n-1}a^m, \qquad [a^n,a]=0.$$ Consider an arbitrary function of the mode operators, that we assume be written in normal formal: $$f(a,a^\dagger) = \sum_{n,m=0}^\infty c_{n,m} (a^\dagger)^n a^m.$$ We know that e^{f(a,a^\dagger)}a e^{-f(a,a^\dagger)} = \sum_{...

1

I'm not sure for this specific problem, and more broadly Hamiltonians are typically in the "eye of the beholder." For example, for quantum chemistry problems, Hamiltonians are really clean mappings from problems that were originally in quantum chemistry. For example, there are "annihilation/creation operators" (discussed more here) that could be converted ...

1

There are two possible answers. Let's say the universe evolves from $t=0$ to $t_f$ then the unitary evolution $U$ from $0$ to $t_f$ induces a CP evolution on the subsystem. To see this, note that the composition of CP maps is CP. Now, the reduced (system) evolution is $Tr_E U\rho_s\otimes\rho_E U^\dagger$ which is a composition of the map \$\rho_s\...

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