- #1

physicsclaus

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- TL;DR Summary
- Hello everyone, I want to create a cat state by transforming a system of ground state to the system of quantum supposition of coherent states with the following steps. However, I do not know how to really achieve this experimentally and what I should consider.

I am so new to circuit quantum electrodynamics. As far as I know, there are few things I could manipulate, like resonator, qubit resonance frequencies, Hamiltonian, coupling strength, Hilbert-space cutoff, dissipation rate, but they do not make sense to me and I do not how they can relate to my desired way of transformation. I think I know the basic mathematics to transform the state in bracket notation, the physics concept is the most difficult part. I have to admit that I have quite a lot of misunderstanding on doing this transformation, I hope you could bear with me and answer my questions.

Besides, I do not understand Wigner function of cavity and why it is useful to Schrodinger Cat States.

I tried to read some wiki pages and some lecture notes, but I cannot comprehend them at this learning stage. I hope someone could me some digests.

Here is my reference: S. M. Girvin, Schrodinger Cat States in Circuit QED, arXiv:1710.03179 [quant-ph]

https://arxiv.org/abs/1710.03179v1

I mostly read section 1.4, other parts are quite hard for me to grasp the concept. I appreciate if someone can provide me more comprehensible learning materials.

\section{My First Step}

Initially, the system is in the ground state: $$\ket{\Psi_i} = \ket{0} \otimes \ket{g}$$

\section{My Second Step}

Then, I want to apply a drive pulse to the qubit to place it in an equal superposition of ground and excited states:

$$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{0} \otimes [\ket{g}+\ket{e}]$$

However, I am not sure how to transform the initial state to superposition.

I propose that we can shine cohere electromagnetic radiation at frequency $\omega$ s.t. $E = \hbar\omega$ is the energy difference between the states, which causes Rabi oscillation between ground state ans excited state over time. But how the rabi frequency can be fixed? Is it by induced dipole moment and the intensity of the driving field? How can we induce the dipole moment then?

In quantum circuit, we simply apply a Hadamard gate to it. Thinking of Bloch Sphere, I think I need to apply a $\frac{\pi}{2}$ pulse to the qubit to make the state transform from $\ket{0}$ to be along with the x-axis.

\section{My Third Step}

After I have the superposition state, I want to apply a drive pulse to the cavity to displace it by an amount $2\alpha$, that I want to drive it into a coherent state $\ket{2\alpha}$, which is coupled to the ground state.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} + \ket{0} \otimes \ket{e}]$$

To achieve, I think I can apply a displacement operator to generate the coherent state.

I consider the context of quantum harmonic oscillator, I have the Hamiltonian characterized as follow:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega\hat{x}^2 = \hbar\omega(\hat{a}^\dag \hat{a}+\frac{1}{2}) = \hbar\omega(\hat{N}+\frac{1}{2})$$

, which can also be represented in terms of ladder operator or number operator.

The eigenvalue equation of the Hamiltonian of the quantum harmonic oscillator

$$\hat{H} \ket{n} = E_n \ket{n}$$

The eigenvalue is quantized, so I have

$E_n = \hbar\omega(n+\frac{1}{2})$, for $n=0,1,2,3,\dots$

The associated eigenstate n can be constructed by acting n times of the raising operator on the ground state

$\ket{n} = \frac{(\hat{a}^\dag)^n}{\sqrt{n!}}\ket{0}$

I can have a lowering operator, which can move between eigenstates. So the lowering operator can lower the energy state by one quantum. Similarly, I can have the raising operator to have the following relation

$$\hat{a}\ket{n} = \sqrt{n}\ket{n-1}, \hat{a}^\dag\ket{n} = \sqrt{n+1}\ket{n+1}$$

Now I define my displacement operator as $\hat{D}$, displaced by an amount $\alpha$

$$\hat{D}(\alpha) = e^{\alpha \hat{a}^\dag - \alpha^*\hat{a}}$$

However, I only speculate and I am not sure if it is defined like this.

Assume it is unitary (I know I can prove it, but I don't want to make my post to lengthy)

I can express the unitary transformation of the displacement operator as:

$$\hat{D}^\dag(\alpha)\hat{a}\hat{D} (\alpha)= \hat{D}^\dag(\alpha)(\hat{D}(\alpha)\hat{a}+ \alpha\hat{D}(\alpha))$$

with the commutation relation

$[\hat{a}, \hat{D}(\alpha)] = \alpha\hat{D}(\alpha) \implies \hat{a} \hat{D}(\alpha) = \hat{D}(\alpha) \hat{a} + \alpha \hat{D}(\alpha)$

Then I can have

$$\hat{D}^\dag(\alpha)\hat{a}\hat{D} (\alpha) = \hat{D}^\dag(\alpha)\hat{D}(\alpha)\hat{a} + \hat{D}^\dag(\alpha)\hat{D}(\alpha)\alpha = \hat{a} + \alpha$$

Similarly, I have

$$\hat{D}^\dag(\alpha)\hat{a}^\dag\hat{D} (\alpha) = \hat{a}^\dag + \alpha^*$$

Considering the coherent state:

$$\hat{a}\ket{\alpha} = \alpha\ket{\alpha}$$

I have

$$\hat{a}(\hat{D}(\alpha)\ket{0}) = \alpha(\hat{D}(\alpha)\ket{0}) = \alpha \ket{0+\alpha} $$

, which confirms that $\hat{D}(\alpha)\ket{0}$ is the eigenstate of lowering operator with eigenvalue $\alpha$, so we can generate coherent state $\ket{\alpha}$ by the application of the displacement operator on the ground state. That is how I can have $\ket{0} \rightarrow \ket{2\alpha}$.

However, I do not know how the operation works in experimental setup.By the way, is this one an entangled state and why?

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} + \ket{0} \otimes \ket{e}]$$

In my opinion, the cat's life is correlated to the existence of poison. But how can I sure if it is not a product state?

Should there be a phase and make it like this?:

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} \pm \ket{0} \otimes \ket{e}]$$

I know there is a term called quantum back action, but I am not quite certain how this phenomenon related to this situation.

I think I have quite a lot of misunderstanding related to microwave engineering. For instance, let's say the qubit is in ground state why the cavity resonance frequency is $\omega_c$. When the qubit is in the excited state, what would be the cavity frequency? I do not know the terms like line-width, cavity response, susceptibility, strong-dispersive region, etc. And why does the cavity state can displace from $\ket{0}$ to $\ket{\alpha}$ if and only if the qubit is in ground state. Why the cavity remains in the vacuum state if the qubit is in the excited state? How can this relation correspond to the line-width? \section{My Fourth Step}

Next, I wan to have a state that the qubit can flip the excited state back to the ground state when the cavity is in the ground state by applying a drive pulse.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} + \ket{0}] \otimes \ket{g}$$

It seems like a CNOT gate in quantum circuit, but I do not know how it operate in physical experimentation.

If my previous step is an entangled state, how can I disentangle the qubit from the cavity? Why we need to flip the qubit if and only if the cavity is in the vacuum state, I mean why it works like that? Why we need to have a large amplitude of $\alpha$?

What is strong-dispersive coupling? Why do we apply a $\pi$ pulse to the qubit can cause a quantized light shift to the frequency and how this transition can make a product state?

\section{My Fifth Step}

I want to transform the state by displacing an amount of $-\alpha$ by driving a pulse to the cavity.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{\alpha}+ \ket{-\alpha}] \otimes \ket{g}$$

This is actually a cat state by definition. I think it is not an entangled state but a product state, because the ground state can be separated.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{\alpha} \otimes \ket{g} \pm \ket{-\alpha} \otimes \ket{g}]$$

So the cavity is in a quantum superposition of two different coherent states. For the $\pm$ sign, is there anything to deal with the photon number parity of the state or the spatial parity symmetry? Why the photon number parity reverses the position and momentum of the oscillator? Why we need to consider even and odd parity cat states?

I do not understand how this displacement relate to the phase space to produce the cat state. I know there few concepts to look at, like quantum state tomography, quantum jump spectroscopy, photon number distribution, even or odd cats, Wigner function fringes. But I lacks these concepts for discussion.

\section{other way}

I know there is another way to produce the cat states with non-deterministic parity. How the measurement back action create a cat state?

$$\ket{\alpha} - \frac{1}{\sqrt{2}}[\ket{\Psi_+}+ \ket{\Psi_-}]$$

Can someone provide me anything I need to consider to get into my final state?

Is it that I need to adjust the frequency of the driving pulse so to transform every step of the state? Is there any other thing I need to consider?

Besides, I do not understand Wigner function of cavity and why it is useful to Schrodinger Cat States.

I tried to read some wiki pages and some lecture notes, but I cannot comprehend them at this learning stage. I hope someone could me some digests.

Here is my reference: S. M. Girvin, Schrodinger Cat States in Circuit QED, arXiv:1710.03179 [quant-ph]

https://arxiv.org/abs/1710.03179v1

I mostly read section 1.4, other parts are quite hard for me to grasp the concept. I appreciate if someone can provide me more comprehensible learning materials.

\section{My First Step}

Initially, the system is in the ground state: $$\ket{\Psi_i} = \ket{0} \otimes \ket{g}$$

\section{My Second Step}

Then, I want to apply a drive pulse to the qubit to place it in an equal superposition of ground and excited states:

$$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{0} \otimes [\ket{g}+\ket{e}]$$

However, I am not sure how to transform the initial state to superposition.

I propose that we can shine cohere electromagnetic radiation at frequency $\omega$ s.t. $E = \hbar\omega$ is the energy difference between the states, which causes Rabi oscillation between ground state ans excited state over time. But how the rabi frequency can be fixed? Is it by induced dipole moment and the intensity of the driving field? How can we induce the dipole moment then?

In quantum circuit, we simply apply a Hadamard gate to it. Thinking of Bloch Sphere, I think I need to apply a $\frac{\pi}{2}$ pulse to the qubit to make the state transform from $\ket{0}$ to be along with the x-axis.

\section{My Third Step}

After I have the superposition state, I want to apply a drive pulse to the cavity to displace it by an amount $2\alpha$, that I want to drive it into a coherent state $\ket{2\alpha}$, which is coupled to the ground state.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} + \ket{0} \otimes \ket{e}]$$

To achieve, I think I can apply a displacement operator to generate the coherent state.

I consider the context of quantum harmonic oscillator, I have the Hamiltonian characterized as follow:

$$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega\hat{x}^2 = \hbar\omega(\hat{a}^\dag \hat{a}+\frac{1}{2}) = \hbar\omega(\hat{N}+\frac{1}{2})$$

, which can also be represented in terms of ladder operator or number operator.

The eigenvalue equation of the Hamiltonian of the quantum harmonic oscillator

$$\hat{H} \ket{n} = E_n \ket{n}$$

The eigenvalue is quantized, so I have

$E_n = \hbar\omega(n+\frac{1}{2})$, for $n=0,1,2,3,\dots$

The associated eigenstate n can be constructed by acting n times of the raising operator on the ground state

$\ket{n} = \frac{(\hat{a}^\dag)^n}{\sqrt{n!}}\ket{0}$

I can have a lowering operator, which can move between eigenstates. So the lowering operator can lower the energy state by one quantum. Similarly, I can have the raising operator to have the following relation

$$\hat{a}\ket{n} = \sqrt{n}\ket{n-1}, \hat{a}^\dag\ket{n} = \sqrt{n+1}\ket{n+1}$$

Now I define my displacement operator as $\hat{D}$, displaced by an amount $\alpha$

$$\hat{D}(\alpha) = e^{\alpha \hat{a}^\dag - \alpha^*\hat{a}}$$

However, I only speculate and I am not sure if it is defined like this.

Assume it is unitary (I know I can prove it, but I don't want to make my post to lengthy)

I can express the unitary transformation of the displacement operator as:

$$\hat{D}^\dag(\alpha)\hat{a}\hat{D} (\alpha)= \hat{D}^\dag(\alpha)(\hat{D}(\alpha)\hat{a}+ \alpha\hat{D}(\alpha))$$

with the commutation relation

$[\hat{a}, \hat{D}(\alpha)] = \alpha\hat{D}(\alpha) \implies \hat{a} \hat{D}(\alpha) = \hat{D}(\alpha) \hat{a} + \alpha \hat{D}(\alpha)$

Then I can have

$$\hat{D}^\dag(\alpha)\hat{a}\hat{D} (\alpha) = \hat{D}^\dag(\alpha)\hat{D}(\alpha)\hat{a} + \hat{D}^\dag(\alpha)\hat{D}(\alpha)\alpha = \hat{a} + \alpha$$

Similarly, I have

$$\hat{D}^\dag(\alpha)\hat{a}^\dag\hat{D} (\alpha) = \hat{a}^\dag + \alpha^*$$

Considering the coherent state:

$$\hat{a}\ket{\alpha} = \alpha\ket{\alpha}$$

I have

$$\hat{a}(\hat{D}(\alpha)\ket{0}) = \alpha(\hat{D}(\alpha)\ket{0}) = \alpha \ket{0+\alpha} $$

, which confirms that $\hat{D}(\alpha)\ket{0}$ is the eigenstate of lowering operator with eigenvalue $\alpha$, so we can generate coherent state $\ket{\alpha}$ by the application of the displacement operator on the ground state. That is how I can have $\ket{0} \rightarrow \ket{2\alpha}$.

However, I do not know how the operation works in experimental setup.By the way, is this one an entangled state and why?

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} + \ket{0} \otimes \ket{e}]$$

In my opinion, the cat's life is correlated to the existence of poison. But how can I sure if it is not a product state?

Should there be a phase and make it like this?:

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} \otimes \ket{g} \pm \ket{0} \otimes \ket{e}]$$

I know there is a term called quantum back action, but I am not quite certain how this phenomenon related to this situation.

I think I have quite a lot of misunderstanding related to microwave engineering. For instance, let's say the qubit is in ground state why the cavity resonance frequency is $\omega_c$. When the qubit is in the excited state, what would be the cavity frequency? I do not know the terms like line-width, cavity response, susceptibility, strong-dispersive region, etc. And why does the cavity state can displace from $\ket{0}$ to $\ket{\alpha}$ if and only if the qubit is in ground state. Why the cavity remains in the vacuum state if the qubit is in the excited state? How can this relation correspond to the line-width? \section{My Fourth Step}

Next, I wan to have a state that the qubit can flip the excited state back to the ground state when the cavity is in the ground state by applying a drive pulse.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{2\alpha} + \ket{0}] \otimes \ket{g}$$

It seems like a CNOT gate in quantum circuit, but I do not know how it operate in physical experimentation.

If my previous step is an entangled state, how can I disentangle the qubit from the cavity? Why we need to flip the qubit if and only if the cavity is in the vacuum state, I mean why it works like that? Why we need to have a large amplitude of $\alpha$?

What is strong-dispersive coupling? Why do we apply a $\pi$ pulse to the qubit can cause a quantized light shift to the frequency and how this transition can make a product state?

\section{My Fifth Step}

I want to transform the state by displacing an amount of $-\alpha$ by driving a pulse to the cavity.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{\alpha}+ \ket{-\alpha}] \otimes \ket{g}$$

This is actually a cat state by definition. I think it is not an entangled state but a product state, because the ground state can be separated.

$$\ket{\psi} = \frac{1}{\sqrt{2}}[\ket{\alpha} \otimes \ket{g} \pm \ket{-\alpha} \otimes \ket{g}]$$

So the cavity is in a quantum superposition of two different coherent states. For the $\pm$ sign, is there anything to deal with the photon number parity of the state or the spatial parity symmetry? Why the photon number parity reverses the position and momentum of the oscillator? Why we need to consider even and odd parity cat states?

I do not understand how this displacement relate to the phase space to produce the cat state. I know there few concepts to look at, like quantum state tomography, quantum jump spectroscopy, photon number distribution, even or odd cats, Wigner function fringes. But I lacks these concepts for discussion.

\section{other way}

I know there is another way to produce the cat states with non-deterministic parity. How the measurement back action create a cat state?

$$\ket{\alpha} - \frac{1}{\sqrt{2}}[\ket{\Psi_+}+ \ket{\Psi_-}]$$

Can someone provide me anything I need to consider to get into my final state?

Is it that I need to adjust the frequency of the driving pulse so to transform every step of the state? Is there any other thing I need to consider?