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:

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


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

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:

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


$$H(t)=H_0+\frac{e}{m_ec}\vec{A}(t)\cdot \frac{\hbar}{i}\nabla$$ $H_0$ 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?


1 Answer 1


This is a unitary transformation, which may be called "to go to an interaction picture". Its general formalism is, suppose the Schrodinger equation is \begin{split} H|\psi\rangle=i\hbar\partial_t|\psi\rangle \end{split} One can apply a unitary transformation $T$ to the state: \begin{split} |\psi\rangle=T|\psi'\rangle \end{split} then the new state $|\psi'\rangle$ satisfies the Schrodinger equation \begin{equation} H'|\psi'\rangle=i\hbar|\psi'\rangle \end{equation} with \begin{split} H'=T^{-1}HT-i\hbar T^{-1}\partial_t T \end{split}

To put it in a more familiar form, if the the Hamiltonian is $H=H_0+V$ in Schrodinger picture, then in the interaction picture you may choose $|\psi_I\rangle=\exp(iH_0t/\hbar)|\psi_S\rangle$ and the wave function in the interaction picture satisfies \begin{equation} H'|\psi_I\rangle=i\hbar\partial_t|\psi_I\rangle \end{equation} with $H'=\exp(iH_0t/\hbar)V\exp(-iH_0t/\hbar)$.

This is how the paper gets the second form of the Hamiltonian. Because we are just going to a specific interaction picture, physical quantities should not change as long as you do things consistently (by undoing the transformation). Please notice, however, interaction picture is mainly designed to solve the dynamics, the eigenenergies may change if you do not undo the transfomation after solving the time evolution. For example, in the example given above, if you do not undo the transformation and use $H'$ to calculate eigenenergies, you can only get eigenvalues of $V$, and $\int d\vec r\psi^*H'\psi$ is not the expectation value of the full Hamiltonian, it is just the expectation value of the interacting Hamiltonian. In this sense, using the new Hamiltonian, I do not think the author will always get the same eigenenergies of the original Hamiltonian.

About quasienergies derived from Floquent theorem, in general they are not the eigenenergies of the Hamiltonian, which can be seen in at least two ways. On the one hand, time dependent Hamiltonians just do not have time independent eigenvalues. On the other hand, eigenenergies (divided by $\hbar$) are the eigenfrequencies of time evolution, but quasienergies do not have this property since $\tilde\phi_{\epsilon,k}(r,t)$ is in general time dependent. These quasienergies provide a analog of eigenenergies in the periodically perturbed system, in the sense that one you know all of these quasienergies and Floquent states, you can get the full time evolution by expanding the wave function in this basis.

In this particular problem, it is also a gauge transformation, as you argued. However, if $\vec A$ was space dependent, the transformation you wrote could not keep electromagnetic field invariant, so it is not a gauge transformation in that case.

  • $\begingroup$ Yes, $\vec{A}$ is spatial independent. In your answer, talking about $\int d\vec{r}\Psi^* H' \Psi$ is meaningless, since they are not even in the same representation. However, one can easily show that $\int d\vec{r}\Psi_I^* H' \Psi_I=\int d\vec{r}\Psi^* H \Psi$, which is not true in my case. So what exactly is the meaning of this quantity in my case? Should I say that $\int d\vec{r}\tilde{\Psi}^* \tilde{H} \tilde{\Psi}$ is the total energy of the system? What is meaning of the same expression after the transform? $\endgroup$ Commented Apr 23, 2014 at 1:36
  • $\begingroup$ So do you think the Floquet quasi-energies before and after are same? $\endgroup$ Commented Apr 23, 2014 at 1:51
  • $\begingroup$ Maybe I misunderstood you but now I am confused with your notation, so please allow me to confirm: when you were asking about $\int d\vec r\psi^*H\psi$, does $\psi$ mean the wave function in interaction picture and does $H$ mean the Hamiltonian in the interaction picture? I just meant to say $\int d\vec r\psi_I^*H'\psi_I$ is the interaction energy and $\int d\vec r\tilde\psi^*\tilde H\tilde \psi$ is the total energy. The safest way to see what an expression in the interaction picture means is to go back to the Schrodingr picture. $\endgroup$ Commented Apr 23, 2014 at 1:56
  • $\begingroup$ I do not think the quasienergies are the same. Let us just take a simple limit that the period of $\vec A$ is extremely long so that $H$ is almost time independent, then before transformation, you are solving the Hamiltonian in Schrodinger picture, while after transformation, you are solving the "interaction Hamiltonian", which in general should have different eigenenergies. $\endgroup$ Commented Apr 23, 2014 at 1:57
  • $\begingroup$ The notation with tilde is your Schordinger picture, the notation without tilde is your interaction picture. I have improved my question in my main text, see if I have made something clearer. $\endgroup$ Commented Apr 23, 2014 at 2:01

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