# How do we perform transverse measurements in a two level system?

In quantum mechanics any two level system can be mapped onto effective spin variables. If the system is defined by two energy levels, $|E_1\rangle$ and $|E_2\rangle$, the Hamiltonian is

$$H = \left(\begin{array}{cc} E_1 & 0 \\ 0 & E_2 \end{array}\right) \, .$$

This can be recast as

$$H = \frac{E_1+E_2}{2} 1\!\!1 + \frac{E_1-E_2}{2} \sigma_3\, .$$

$1\!\!1$ is the identity operator and $\sigma_3$ is the third pauli matrix. This is very nice because we can now apply our knowledge and intuition of spin dynamics to any two level system.

My question is the following: for a real spin I can measure expectation values of $\sigma_{1,2}$ by physically rotating my system and align its $x$ or $y$ axis with my measuring apparatus. How do I do this (in the lab!) with a pseudo-spin system? Imagine I am looking at a cold atom and I want to know if the system is in an symmetric or anti-symmetric superposition of $|E_1\rangle$ and $|E_2\rangle$, how do I do it?

## 1 Answer

You do exactly the same thing: you "rotate" the state and then measure along whatever axis your measurement apparatus happens to measure. The only difference here is that the "rotation" does not necessarily correspond to a rotation in space like it does for a true spin.

What follows is a detailed description of how we do rotations of a generic 2 level system. These rotations, plus measurement along a fixed axis, yield effective measurements along any axis.

# Generic example

Consider a harmonic oscillator system with $H=\hbar \omega_0(n + 1/2)$. Suppose we subject this thing to an external force $$F(t) = F_d \cos(\omega t + \phi).$$ It's probably known to you that if $\omega_d=\omega_0$ then the driving will cause the system to undergo transitions between the various states. So, let's work in the case $\omega_d = \omega_0$. What is the Hamiltonian caused by this driving force? The work done by a force is $\text{force}\times \text{distance}$ so the Hamiltonian term is $$H_d = -F(t)x = -F_d \cos(\omega_0 t + \phi) x$$ where the minus sign comes because an external force pointed to the right means the system has less potential energy if it goes to the right. We can rewrite the position operator as (see any intro textbook) $$x = x_0(a + a^\dagger)$$ where $x_0$ is a characteristic length scale in the problem. Using this, the driving Hamiltonian becomes $$H_d = -F_d x_0 \cos(\omega_0 t + \phi) (a + a^\dagger).$$ giving a full Hamiltonian $$H = \hbar \omega_0(n + 1/2) - F_d x_0 \cos(\omega_0 t + \phi) (a + a^\dagger) .$$ This is tricky because we have both the original Hamiltonian $H_0$ and a time dependent $V(t)$. To fix this we will go into a rotating frame.

# The rotating frame

Suppose we have a system with Hamiltonian $$H = H_0 + V(t)$$ There are several alternate "pictures" one can use to simplify the problem. You've probably heard of the "Heisenberg picture" and perhaps the "interaction picture". Here we develop a third picture called the "rotating frame". Consider the propagator of $H_0$ $$U_0(t) = \exp[-itH_0/\hbar] .$$ If the time dependent $V(t)$ were absent, then the time evolution of the system would because $$|\Psi_0(t)\rangle = U_0(t)|\Psi(0)\rangle.$$ The idea of the rotating frame is to undo the part of the evolution due to $U_0$. Define a new state $$|\Psi'(t)\rangle \equiv R(t) |\Psi_0(t)\rangle .$$ where $R(t) \equiv U_0(t)^\dagger = \exp[itH_0/\hbar]$. In other words, $R$ undoes $U_0$. Now let's track the evolution of this new thing \begin{align} i\hbar d_t |\Psi'(t)\rangle &= i\hbar d_t (R(t) |\Psi_0(t)\rangle) \\ &= i\hbar \dot{R}(t) |\Psi_0(t)\rangle + i\hbar R(t) d_t |\Psi_0(t)\rangle \\ &= i\hbar \dot{R}(t) R^\dagger(t)|\Psi'(t)\rangle + R(t)(H_0 + V(t))|\Psi_0(t)\rangle \\ &= [i\hbar (iH_0/\hbar)R(t)R^\dagger(t) + H_0 + R(t)V(t)R^\dagger(t) ] |\Psi'(t)\rangle \\ &= [R(t) V(t) R^\dagger(t)]|\Psi'(t)\rangle . \end{align} We now have a simple Schrodinger equation where the effective "rotating frame Hamiltonian" is $$H_r = R(t)V(t)R^\dagger(t).$$ The point is that the original Hamiltonian is completely gone. This leaves only a time dependent part which makes life somewhat easier.

# Back to the example

We had $$H = \hbar \omega_0(n + 1/2) - F_d x_0 \cos(\omega_0 t + \phi) (a + a^\dagger) .$$ Let's use the time independent part to make a rotating frame. The rotation operator $R$ is $$R(t) = \exp[i\omega_0t (n + 1/2)]$$ and the time depedent part of the Hamiltonian is $$V(t) = -F_d x_0 \cos(\omega_0 t + \phi) (a + a^\dagger).$$ If we form the rotating frame Hamiltonian we find (not doing algebra here) $R(t)aR^\dagger(t) = a e^{-i\omega_0 t}$, which leads to $$H_r = R(t)V(t)R^\dagger(t) = -F_d x_0 \cos(\omega_0 t + \phi) (a e^{-i\omega_0 t} + a^\dagger e^{i\omega_0 t}).$$ If we now use $$\cos(\omega t + \phi) = \frac{1}{2} \left[ e^{i(\omega_0 t + \phi)} + e^{-i(\omega_0 t + \phi} \right]$$ we get $$H_r = -\frac{F_d x_0}{2} \left[ ae^{i\phi} + ae^{-i(2\omega_0 t + \phi)} + a^\dagger e^{-i\phi} + a^\dagger e^{i(2\omega_0 t + \phi)} \right]$$ The two time dependent terms oscillate at high frequency and are neglected in the so-called "rotating wave approximation". This leaves $$H_r = -\frac{F_d x_0}{2} \left[ ae^{i\phi} + a^\dagger e^{-i\phi} \right] . \qquad (*)$$ Now suppose that our system weren't quite a harmonic oscillator so that only the first energy gap has energy spacing $\hbar \omega_0$. In that case our drive is not on resonance with any other levels and it's an ok approximation to consider only the lowest two levels which are on resonance with the drive. If we truncate the $a$ and $a^\dagger$ operators to the lowest two levels, they become $$a = \left( \begin{array}{cc} 0&1\\0&0 \end{array}\right) \qquad a^\dagger = \left( \begin{array}{cc} 0&0\\1&0 \end{array} \right).$$ Substituting this into $(*)$ gives $$H_r = -\frac{F_d x_0}{2} \left( \begin{array}{cc} 0 & e^{i\phi} \\ e^{-i\phi} & 0 \end{array} \right) = -\frac{F_d x_0}{2}(\cos(\phi)\sigma_x - \sin(\phi)\sigma_y).$$ This is the key result. Here we see $H_r$ to be a rotation of the state around an axis in the $xy$ plane. The axis angle is determined by $\phi$ which was just the phase of our initial drive signal. This means that by controlling the phase of our driving force, we can rotate the state about any axis in the $xy$ plane! Of course, except in the case of a real spin, this "rotation" is not a rotation in real 3D space. It is a rotation in the Hilbert space of the two level system, which, as you know, looks like the surface of a sphere (called the "Bloch sphere"). We have shown here how to make rotations around the $x$ and $y$ axes in that imaginary spherical space.

In order to do rotations about the $z$ axis, you just change the energy spacing between the levels. At this point I bet you can calculate exactly how that works using the formalism presented already.

In the case of electron transitions between various atomic orbitals, the physics is exactly as presented here. Usually the force comes from the electric field of a laser or RF generator acting on the charged electron. If the frequency of the laser or RF field is matched to one of the electron's transitions, i.e. $\omega_{\text{drive}} \approx \Delta E/\hbar$, then the rotating frame argument presented here goes through to show that the electron will Rabi oscillate between the two levels involved in that transition.

Homework: Explicitly compute the propagator induced by the $H_r$ we found.

Note: The rotating wave approximate is good as long as the frequency of rotation about the Bloch sphere induced by the drive is less than $|E_1 - E_2|/\hbar$.