- #1
Marioweee
- 18
- 5
- Homework Statement
- We will consider that the ion is in an harmonic trap. The ion has two internal states |g⟩ and |s⟩ and it interacts with a laser that induces a state-dependent force. The quantum dynamics is governed by the Hamiltonian
$$H = H_R + H_f$$
$$H_R = Ω(|s><g| + |g><s|)$$
$$H_f = g(|s><s| − |g><g|)(a^\dagger + a)$$
HR induces Rabi oscillations between internal states. Hf describes an optical force that depends on the state of the
ion. Note that the operator $(a^\dagger + a)$ is proportional to the position of the ion. The force’s sign depends on the internal state of the ion, such that
$$|s⟩ → Hf = g(a^\dagger + a)$$
$$|g⟩ → Hf = −g(a^\dagger + a)$$
1. Imagine the ion is initially in the ground state
|ψ1⟩ = |g⟩|0⟩, where |g⟩ is the electronic ground state and |0⟩ is the ground state of the vibrational degrees of freedom (zero phonons). We consider that there is no applied force and the Rabi term HR acts for a time $t=\frac{\pi}{4\Omega}$ . After that, the state of the ion is
|ψ2⟩ = (a|g⟩ + b|s⟩)|0⟩ Calculate a and b.
2. Now we turn off the Rabi term (Ω = 0) and let the state-dependent force act for a time τ .Show that the state of the ion is
|ψ2⟩ = a|g⟩|α⟩ + b|s⟩| − α⟩
and calculate the value of α.
- Relevant Equations
- -
I have already solved question number 1 by applying the schrödinger equation obtaining that
$$\ket{\psi_2}(t) = \cos(\Omega t)\ket{g} - i \sin (\Omega t)\ket{s}$$
and therefore in ##t=\frac{\pi}{4\Omega}##
$$\ket{\psi_2}(t) = \dfrac{1}{\sqrt{2}}(\ket{g} - i \ket{s})$$
I have some doubts about the second exercise. What I have tried is the following
$$|ψ(t)⟩ = e^{-iH_f t}\dfrac{1}{\sqrt(2)}(|g⟩ -i |s⟩)\ket{0} = \dfrac{1}{\sqrt(2)}(e^{it}|g⟩ -i e^{-it} |s⟩)\otimes e^{-igt(a^\dagger + a)}\ket{0}$$
I think that
$$e^{-ig(a^\dagger + a)}\ket{0}$$
should be something proportional to ##\ket{\alpha} = \ket{-igt}## (because of the displacement operator). But the displacement operator proportional to ##a^\dagger - a## so im a bit confused. Moreover, the ##e^{it}## and ##e^{-it}## terms make me think something is wrong.
$$\ket{\psi_2}(t) = \cos(\Omega t)\ket{g} - i \sin (\Omega t)\ket{s}$$
and therefore in ##t=\frac{\pi}{4\Omega}##
$$\ket{\psi_2}(t) = \dfrac{1}{\sqrt{2}}(\ket{g} - i \ket{s})$$
I have some doubts about the second exercise. What I have tried is the following
$$|ψ(t)⟩ = e^{-iH_f t}\dfrac{1}{\sqrt(2)}(|g⟩ -i |s⟩)\ket{0} = \dfrac{1}{\sqrt(2)}(e^{it}|g⟩ -i e^{-it} |s⟩)\otimes e^{-igt(a^\dagger + a)}\ket{0}$$
I think that
$$e^{-ig(a^\dagger + a)}\ket{0}$$
should be something proportional to ##\ket{\alpha} = \ket{-igt}## (because of the displacement operator). But the displacement operator proportional to ##a^\dagger - a## so im a bit confused. Moreover, the ##e^{it}## and ##e^{-it}## terms make me think something is wrong.
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