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Hello, I have the following problem:
A system in the lab frame is described by a time dependent rotating potential ##V(\vec{r},t)##.
So ##H_{lab}=\frac{\boldsymbol{p}^{2}}{2m} + V(\vec{r},t)##.
My book says that the Hamiltonian in the rotating frame is given by
##H_{rot}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r}) - w\boldsymbol{L'_z}##
Where ##\boldsymbol{p'}## and ##L'_z## are operators in the rotating frame. So to clarify, in this frame the potential is stationary.
(Let's take this as an ansatz! Assume this is the correct rotating Hamiltonian for all reasoning to follow.)
So in the rotating frame a ##\Psi'## will solve the Schrodinger equation with the adjusted Hamiltonian.
How to conver from ##\Psi'## to the wavefunction in the lab frame?Edit:
I see some papers proposing the following: (in hbar=1 units)
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt}\Psi_{lab}(\vec{r'},t)##
(1)
Let me show why I don't see how this is correct:
Writing out the time evolution for ##\Psi'## in the rotating frame Schrodinger equation gives:
##\Psi'(\vec{r'},t)=e^{-i\boldsymbol{H''}t + i\boldsymbol{L'_z}wt} \Psi'(\vec{r},0) ##
Where ##\boldsymbol{H''}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r'})##
Now if ##\boldsymbol{L'z}## and ##\boldsymbol{H''}## commute which I'm pretty sure they do then:
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi'(\vec{r},0) ##
Since ##\Psi'(\vec{r},0) = \Psi_{lab}(\vec{r},0)## we have this final equation :
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0) ##
Eq(2)
Now now! Compare eq(1) and eq(2), it does look like:
##\Psi_{lab}(\vec{r},t)=e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0) ##
Which is a result I can not believe. If you take a look at ##\boldsymbol{H''}## it has a time INDEPENDENT potential in there. This would mean that to find the evolution in the Lab frame all we have to do is ignore the time dependence of the potential and just let it evolve as if the potential was stationary in the lab frame all along.
What is wrong?
A system in the lab frame is described by a time dependent rotating potential ##V(\vec{r},t)##.
So ##H_{lab}=\frac{\boldsymbol{p}^{2}}{2m} + V(\vec{r},t)##.
My book says that the Hamiltonian in the rotating frame is given by
##H_{rot}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r}) - w\boldsymbol{L'_z}##
Where ##\boldsymbol{p'}## and ##L'_z## are operators in the rotating frame. So to clarify, in this frame the potential is stationary.
(Let's take this as an ansatz! Assume this is the correct rotating Hamiltonian for all reasoning to follow.)
So in the rotating frame a ##\Psi'## will solve the Schrodinger equation with the adjusted Hamiltonian.
How to conver from ##\Psi'## to the wavefunction in the lab frame?Edit:
I see some papers proposing the following: (in hbar=1 units)
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt}\Psi_{lab}(\vec{r'},t)##
(1)
Let me show why I don't see how this is correct:
Writing out the time evolution for ##\Psi'## in the rotating frame Schrodinger equation gives:
##\Psi'(\vec{r'},t)=e^{-i\boldsymbol{H''}t + i\boldsymbol{L'_z}wt} \Psi'(\vec{r},0) ##
Where ##\boldsymbol{H''}=\frac{\boldsymbol{p'}^{2}}{2m} + V(\vec{r'})##
Now if ##\boldsymbol{L'z}## and ##\boldsymbol{H''}## commute which I'm pretty sure they do then:
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi'(\vec{r},0) ##
Since ##\Psi'(\vec{r},0) = \Psi_{lab}(\vec{r},0)## we have this final equation :
##\Psi'(\vec{r'},t)=e^{i\boldsymbol{L'_z}wt} e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0) ##
Eq(2)
Now now! Compare eq(1) and eq(2), it does look like:
##\Psi_{lab}(\vec{r},t)=e^{-i\boldsymbol{H''}t} \Psi(\vec{r},0) ##
Which is a result I can not believe. If you take a look at ##\boldsymbol{H''}## it has a time INDEPENDENT potential in there. This would mean that to find the evolution in the Lab frame all we have to do is ignore the time dependence of the potential and just let it evolve as if the potential was stationary in the lab frame all along.
What is wrong?
Last edited: