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I've been working on this for quite a while now and can't seem to figure it out. I am concerned with the dynamics of a general two-state system, specifically the case of a particle in a constant magnetic field coupled with a time dependent magnetic field.

Let's start with the simple case, just a particle in a static magnetic field, say $\vec{B} = B_0 \hat{k}$. Then, the Hamiltonian for the system will come from the interaction of the dipole moment with the field, i.e. $$\hat{H} = -\hat{\mu} \cdot \vec{B}$$ We know that the magnetic dipole moment is proportional to the intrinsic spin of the system as $$\hat{\mu} = \gamma \hat{S},$$ where $\gamma$ is the gyromagnetic ratio. Ok, our Hamiltonian is $$\hat{H} = -\gamma B_0 \hat{S}_z = \omega_0 \hat{S}_z,$$ where $\omega_0$ is known as the Larmor frequency. The dynamics of this is very easy, the spin just undergoes a precession about the $z$-axis. If we start in the state, say $$|\psi(0)> = |+x> = \frac{1}{\sqrt{2}}(|+z> + |-z>),$$ then the evolution will be $$|\psi(t)> = e^{-i \omega_0 t/2}\frac{1}{\sqrt{2}}(|+z> + e^{i \omega_0 t}|-z>).$$ We see that the probability of a "flip" in state to $|-x>$ is given by $$|<-x|\psi(t)>|^2 = |\frac{1}{2}(<+z|-<-z|)(|+z> + e^{i \omega_0 t}|-z>)|^2 = \boxed{\sin^2\Big(\frac{\omega_0}{2}t\Big)}.$$ Ok, we see that it flips at times $T_n = \frac{\pi}{\omega_0}(2n-1)$, where $n$ is just a positive integer. This is all good and well. Now we add a time dependent magnetic field of the form $$\vec{B} = B_1(\cos(\omega t) \hat{i} + \sin(\omega t) \hat{j}).$$ This is an oscillating magnetic field on top of the previous magnetic field in the $z$ direction. Our new Hamiltonian becomes (I am dropping the hats on the operators...) $$H = H_0 + H_1 = \omega_0 S_z + \omega_R(\cos(\omega t) S_x + \sin(\omega t) S_y),$$ where $\omega_R$ is known as the Rabi frequency. This is where I am stuck. We now have a Hamiltonian with spins in every direction (time dependent too), so we cannot simply construct the time evolution operator and act on the initial state as we did before. We have a couple of more options to choose from. I tried constructing the matrix representation of this operator in terms of the basis states (which we choose to be $|\pm z>$). The resulting Hamiltonian will be $$H \doteq \frac{\hbar}{2} \begin{pmatrix} \omega_0 & \omega_R e^{-i\omega t} \\ \omega_R e^{i \omega t} & -\omega_0 \end{pmatrix}$$ In fact, this is the most general Hamiltonian for a two-state system. But I am still stuck in that I have no clue how to find the eigenvalues and eigenstates of this Hamiltonian. I get as far as $$ \frac{\hbar}{2} \begin{pmatrix} \omega_0 & \omega_R e^{-i\omega t} \\ \omega_R e^{i \omega t} & -\omega_0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = i\hbar \begin{pmatrix} \dot{a} \\ \dot{b} \end{pmatrix},$$ which is just the Schrodinger equation for the state $|\psi(t)> = a(t) |+z> + b(t) |-z>$. But I still can't solve these sets of differential equations because of the time dependence.

Ok, so then I tried to get rid of this time dependence by doing a unitary transformation on the Hamiltonian, i.e. $$H \rightarrow UHU^{\dagger}.$$ This is equivalent to transforming into the rotating frame of reference, rotating with the magnetic field. From a video I found online from MIT's Atomic and Optical Physics course (Lecture 4, minute 50), the professor said to pick the unitary transformation as $$ U = \begin{pmatrix} e^{i\omega t/2} & 0 \\ 0 & e^{-i \omega t/2} \end{pmatrix}$$ But this confuses me as well. This is just the representation of the rotation operator in the $z$ direction in the $z$ basis, rotating an angle $\theta = \omega t$. Upon transforming the Hamiltonian, I get $$ \begin{align*} H' &= \frac{\hbar}{2} \begin{pmatrix} e^{i\omega t/2} & 0 \\ 0 & e^{-i \omega t/2} \end{pmatrix} \begin{pmatrix} \omega_0 & \omega_R e^{-i\omega t} \\ \omega_R e^{i \omega t} & -\omega_0 \end{pmatrix} \begin{pmatrix} e^{-i\omega t/2} & 0 \\ 0 & e^{i \omega t/2} \end{pmatrix} \\ &= \frac{\hbar}{2} \begin{pmatrix} \omega_0 & \omega_R \\ \omega_R & -\omega_0 \end{pmatrix}. \end{align*} $$ Yet again this is not what we want, the whole $\omega$ term drops out! In the lecture, the professor said that upon doing this transformation, you are supposed to get $$ H' = \frac{\hbar}{2} \begin{pmatrix} \delta & \omega_R \\ \omega_R & -\delta \end{pmatrix}, $$ where $\delta = \omega - \omega_0$, but I am just not getting this.

But whatever, let's say that there is actually a unitary transformation that will give me this time independent Hamiltonian. Ok, we can plug this in to the Schrodinger equation and get $$ \frac{\hbar}{2} \begin{pmatrix} \delta & \omega_R \\ \omega_R & -\delta \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = i\hbar \begin{pmatrix} \dot{a} \\ \dot{b} \end{pmatrix} $$ But when I try to solve this, I end up getting some nonsense.

The very last thing I tried was to find the energy eigenstates of this new Hamiltonian, since the evolution would be simple in that case. If we knew the energy states, then we can write our initial state in that basis, i.e. $$ |\psi(t=0)> = c_0 |E+> + c_1 |E->. $$ Then, the evolution is simple; you just tack on the exponentials as $$|\psi(t)> = c_0 e^{-iE_{+}t/\hbar} |E+> + c_1 e^{-iE_{-}t/\hbar}|E->.$$ However, that process is extremely tedious and after finding the energy eigenvalues $E_{\pm} = \pm \sqrt{\delta^2 + \omega_R^2}$, the rest would take a really long time to do, if it was even correct. I feel like there must be a simpler way of solving this problem. Any help at all would be much appreciated, thank you.

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  • $\begingroup$ "In fact, this is the most general Hamiltonian for a two-state system." -> Just so you know, this is not true. For example, you could have terms like $\omega_R\cos{\omega t}$ as the off-diagonal terms (if you did not use magnetic fields in both the x and y direction, you would get something like this). In this case the unitary transformation you have used does not fully remove the time-dependence, and one must use the rotating wave approximation to get a stationary Hamiltonian: en.wikipedia.org/wiki/Rotating_wave_approximation $\endgroup$ – Rococo Jan 23 '17 at 5:51
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(I hope I did this right...)

Start from your original $H$ and $\vert\psi(t)\rangle$ and make the change $\vert \phi(t)\rangle = U(t)\vert\psi(t)\rangle$ as you suggested. The Schrodinger equation then becomes \begin{align} i\hbar \frac{\partial }{\partial t}\vert \psi(t)\rangle &= H(t) \vert \psi(t)\rangle\\ i\hbar \frac{\partial }{\partial t} U^{\dagger}(t)\vert\phi(t)\rangle &= H(t) U^{\dagger}(t)\vert \phi(t)\rangle \end{align} which in turn yields $$ i\hbar U^{\dagger}(t) \frac{-i\omega}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right) \vert \phi(t)\rangle + i\hbar U^\dagger(t)\frac{\partial }{\partial t}\vert\phi(t) \rangle = H U^\dagger(t)\vert \phi(t)\rangle\, . $$ Multiplying the $U(t)$ from the left gives \begin{align} i\hbar\frac{\partial}{\partial t}\vert\phi(t)\rangle&= -\frac{\hbar\omega}{2}\left(\begin{array}{cc} 1 & 0 \\ 0 &-1 \end{array}\right)\vert\phi(t)\rangle + H'\vert\phi(t)\rangle\, , \\ &= \frac{\hbar\omega}{2}\left(\begin{array}{cc} -1 & 0 \\ 0 &1 \end{array}\right)\vert\phi(t)\rangle + \frac{\hbar}{2} \left(\begin{array}{cc}\omega_0 &\omega_r \\ \omega_r &-\omega_0 \end{array}\right)\vert\phi(t)\rangle\, ,\\ &=\frac{\hbar}{2} \left(\begin{array}{cc}\omega_0-\omega &\omega_r \\ \omega_r &-(\omega_0 -\omega) \end{array}\right)\vert\phi(t)\rangle \end{align} As you can see, there is an additional term, which I eventually moved to the right, that comes from the time derivative of $U(t)$.


Nota: this is similar to the Coriolis-like terms that occur when going to a rotating frame, the origins of which are also in the time derivative of the rotation matrix linking the space-fixed and the body-fixed frames.

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  • $\begingroup$ this is it. Thank you! I didn't know that you had to change the states as well. $\endgroup$ – Josh Pilipovsky Jan 23 '17 at 1:16
  • $\begingroup$ It's change of basis so it will induce both a change in the basis states and in the operators. $\endgroup$ – ZeroTheHero Jan 23 '17 at 1:21
  • $\begingroup$ So in the general case, if I want to move to a rotating frame, the transformation will be a rotation matrix? $\endgroup$ – Josh Pilipovsky Jan 23 '17 at 1:22
  • $\begingroup$ In this case it's an $SU(2)$ matrix rather than $SO(3)$ but aside from covering group technicalities yes. The crux of it here is that, if the transformation $U$ is time dependent, there will be additional pieces that come from the time derivative of this $U$, just like there are extra terms that appear in a rotating frame. $\endgroup$ – ZeroTheHero Jan 23 '17 at 1:24
  • $\begingroup$ Hmmm... not so sure about my last comment... certainly in general this will be a unitary matrix... not sure any more if it's necessarily a rotation... methinks of a $3\times 3$ system where the matrices would be 3-dimensional, and the transformation might very well be $U(3)$ in that case... possibly depends on the Hamiltonian... I'd have to think about it a little more. $\endgroup$ – ZeroTheHero Jan 23 '17 at 1:26
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You have the correct end result, although typically I have seen the 'Rabi frequency' is written as $\Omega$.

In general, time dependent Hamiltonians are tricker to solve than time-independent Hamiltonians - exactly as you pointed out, this is the motivation for moving to the 'rotating frame'. The unitary transformation that you wrote is the formal correct transformation although it may not be intuitive to you. Here is a slightly different formulation (entirely equivalent):

Your time dependent Hamiltonian is $$\mathcal{H}/\hbar = \begin{pmatrix}\omega_0 & \Omega e^{-i \omega t} \\ \Omega e^{i \omega t} & 0 \end{pmatrix}$$ where I have changed what I call 0 energy to make one of the spin eigenstates at 0 energy and the other one at some fixed offset which I have relabeled as $\omega_0$ (this is entirely equivalent to your system but with my definition of $\omega_0$ off by a factor of 2, and I did this because it is identical physics but simplifies the problem).

As you wrote, \begin{eqnarray*} \dot{a} &=& -i(\omega_0 a+ \Omega e^{-i \omega t} b) \\ \dot{b} &=& -i \Omega e^{i \omega t} a \end{eqnarray*} Now, as you noted we have some pesky time dependent terms. Let's get rid of the annoying time dependence in the equation for $\dot{b}$ by defining a new variable that already includes this time dependence: $\tilde{a}= e^{i \omega t} a$. This of course greatly simplifies $\dot{b} = -i \Omega \tilde{a}$.

Now we rewrite the other equation in terms of $\tilde{a}$: $$ \dot{\tilde{a}} = i \omega \tilde{a} + e^{i \omega t} \dot{a} = i \omega \tilde{a} - ie^{i \omega t}(\omega_0 a + \Omega e^{-i \omega t}b) $$ $$ \to \dot{\tilde{a}} = i (\omega - \omega_0) \tilde{a} - i \Omega b $$ We have now come to our final time-independent equations: \begin{eqnarray*} \dot{\tilde{a}} &=& -i \Omega b + i(\omega - \omega_0) \tilde{a} \\ \dot{\tilde{b}} &=& -i \Omega \tilde{a} \end{eqnarray*}

You can see the connection between this system of equations in these new variables and the transformed Hamiltonian that you wrote - that is because the change of variables is formally equivalent to a unitary transformation of the Hamiltonian.

The physics of this Hamiltonian (given most simply as $\begin{pmatrix} \delta & \Omega \\ \Omega & 0 \end{pmatrix}$) is very deep and important in modern atomic physics. The simplest case is when $\delta = 0$, known as 'resonance' - in this case the spin oscillates between the two original eigenstates with frequency given by $\Omega$ - this is known as Rabi oscillation.

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  • $\begingroup$ So what you are saying is that we transform the state to get rid of the time dependence? Also, how is the unitary transformation that I defined correct if it gives the wrong answer? Thank you. $\endgroup$ – Josh Pilipovsky Jan 20 '17 at 18:45

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