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I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)=\exp[-\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t')]\Psi(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• This is just a gauge transformation $A\rightarrow A,\phi=0\rightarrow \frac{e}{2m_ec^2}A^2$. The physical quantities should be unchanged after the transformation. Is the floquet quasienergy a physical quantity? I am asking this because after the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)\rightarrow\Psi(r,t)=\exp[\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t')]\tilde{\Psi}(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• This is just a gauge transformation $A\rightarrow A,\phi=0\rightarrow -\frac{e}{2m_ec^2}A^2$. The physical quantities should be unchanged after the transformation. Is the floquet quasienergy a physical quantity? I am asking this because after the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)=\exp[-\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t')]\Psi(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• Since this transformation is just a gauge transformation(see my comments). I heard that the physical quantities is unchanged by the gauge transformation. If this is true, is the floquet quasienergy a physical quantity in my second question? What about the quantity in my third question?

• After the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

Please help if you have answer(s), also any comments are welcomed.

I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)=\exp[-\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t')]\Psi(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• This is just a gauge transformation $A\rightarrow A,\phi=0\rightarrow \frac{e}{2m_ec^2}A^2$. The physical quantities should be unchanged after the transformation. Is the floquet quasienergy a physical quantity? I am asking this because after the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)=\exp(-\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t'))\Psi(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• Since this transformation is just a gauge transformation(see my comments). I heard that the physical quantities is unchanged by the gauge transformation. If this is true, is the floquet quasienergy a physical quantity in my second question? What about the quantity in my third question?

• After the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

Please help if you have answer(s), also any comments are welcomed.

I am reading an article about Bloch-Floquet state. My questions is in Part II.B and Appendix A of this paper, I will describe them below.

The original Schordinger equation we consider is:

$$i\hbar\frac{\partial}{\partial t} \tilde{\Psi}(r,t)=\tilde{H}(t)\tilde{\Psi}(r,t)$$

where:

$$\tilde{H}(t)=\frac{1}{2m_e}(\frac{\hbar}{i}\nabla+\frac{e\vec{A}(t)}{c})^2+V_c(r) $$

with $A(t)$ periodic in time and $V_c(r)$ periodic in space.

According to Floquet theorem, this time-periodic Hamiltonian has the wave function in the form:

$$\tilde{\Psi}(r,t)=e^{-i\tilde{\epsilon}(k)t/\hbar}e^{ik\cdot r}\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$$ with $\tilde{\phi}_{\tilde{\epsilon},k}(r,t)$ periodic in space and time, we call $\tilde{\epsilon}(k)$ the Bloch-Floquet quasienergy.

The author did a following transform to avoid dealing with the square term of $A(t)$:

$$\tilde{\Psi}(r,t)=\exp[-\frac{ie^2}{2m_e\hbar c^2}\int^t dt'A^2(t')]\Psi(r,t)$$

Substitute this to the original Schordinger equation we arrive at:

$$i\hbar\frac{\partial}{\partial t} \Psi(r,t)=H(t)\Psi(r,t)$$

where:

$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ is the field free Hamiltonian. Finally, the author claimed that the physics quantities are invariant under such a transformation and give an example of the invariance of the current density(I can see the identity).

My question is:

• Since this transformation is just a gauge transformation(see my comments). I heard that the physical quantities is unchanged by the gauge transformation. If this is true, is the floquet quasienergy a physical quantity in my second question? What about the quantity in my third question?

• After the transformation, the author is using this new Schordinger equation to calculate the Bloch-Floquet quasienergy, is these qusienergies same as the ones obtained using the original Schordinger equation?

• If I calculate this quantity $\int d\vec{r}\Psi^*H\Psi$ and $\int d\vec{r}\tilde{\Psi}^*\tilde{H}\tilde{\Psi}$ , what's the physical meaning of them? One can easily see that they are not equal. Also what's the difference and relation between this quantity and the floquet energy?

Please help if you have answer(s), also any comments are welcomed.