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David
David
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I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right>$ and damping with $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right>$ and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right>$ and damping with $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right>$ and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

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David
David
  • 53
  • 5

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right>$\left|1\right>$ and damping with $d=\left| 0\right> \left< 1\right|$" $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right>$\left|0\right>$ and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$"$d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right> and damping with $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right> and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right>$ and damping with $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right>$ and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.

Source Link
David
David
  • 53
  • 5

Lindblad operators seemingly not working as jump operators

I am currently trying to understand the Lindblad superoperator $$\mathcal{L}[d]\rho = d \rho d^\dagger - (d^\dagger d \rho + \rho d^\dagger d)/2$$ for a simple 2 level system.

In the literature i found, the operator d is seen as a jump operator that can be written in the form $$d = \left|a\right>\left<b\right|.$$

Modeling a system with Hamiltonian $H=0$ (Markov Approximation) and one damping operator we get the Lindblad Master equation $$\dot{\rho} = \gamma \mathcal{L}[d]\rho$$ where $\gamma$ is the damping rate.

For example, with $d=\left| 0 \right>\left<1\right|$ the system will get pulled from $\left| 1 \right>$ to $\left|0 \right>$, just like i would expect when viewing $d$ as a jump operator.

Path through the Bloch sphere when starting at <span class=$\left|1\right> and damping with $d=\left| 0\right> \left< 1\right|$" />

This is no longer the case when using $d = \left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$. I would expect that the system is pulled from $\left| 0 \right>$ to $\frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}}$. The system will instead get damped towards $\left|1\right>$.

Path through the Bloch sphere when starting at <span class=$\left|0\right> and damping with $d=\left( \frac{\left| 0 \right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0 \right|$" />

What is going here? Is this view of d as a jump operator only true for basis states? And if so, why? I have read here, that there are not formal constraints on the form of $d$, so how can an operator like $d = \left( \frac{\left| 0\right> +\left| 1\right>}{\sqrt{2}} \right) \left< 0\right|$ be interpretated?

Thank you for your help.