# 2.3: Two-Level Systems


Let’s use the time-propagator in a model calculation that we will refer to often. It is common to reduce or map quantum problems onto a two level system (2LS). We will pick the most important states for our problem and find strategies for discarding or simplifying the influence of the remaining degrees of freedom. Consider a 2LS with two unperturbed or “zeroth order” states $$| \varphi _ {a} \rangle$$ and $$| \varphi _ {b} \rangle$$ with energies $${\varepsilon} _ {a}$$ and $${\varepsilon} _ {b}$$, which are described by a zero-order Hamiltonian $$H_0$$:

\begin{align} \hat {H} _ {0} &= | \varphi _ {a} \rangle \varepsilon _ {a} \left\langle \varphi _ {a} | + | \varphi _ {b} \right\rangle \varepsilon _ {b} \langle \varphi _ {b} | \\[4pt] &= \left( \begin{array} {l l} {\varepsilon _ {a}} & {0} \\[4pt] {0} & {\varepsilon _ {b}} \end{array} \right) \label{1.33} \end{align}

These states interact through a coupling $$V$$ of the form

\begin{align} \hat {V} &= | \varphi _ {a} \rangle V _ {a b} \left\langle \varphi _ {b} | + | \varphi _ {b} \right\rangle V _ {b a} \langle \varphi _ {a} | \\[4pt] &= \left( \begin{array} {l l} {0} & {V _ {a b}} \\[4pt] {V _ {b a}} & {0} \end{array} \right) \label{1.34} \end{align}

The full Hamiltonian for the two coupled states is $$\hat{H}$$:

\begin{align} \hat {H} & = \hat {H} _ {0} + \hat {V} \\[4pt] & = \left( \begin{array} {c c} {\varepsilon _ {a}} & {V _ {a b}} \\[4pt] {V _ {b a}} & {\varepsilon _ {b}} \end{array} \right) \label{1.35} \end{align}

The zero-order states are $$| \varphi _ {a} \rangle$$ and $$| \varphi _ {b} \rangle$$. The coupling mixes these states, leading to two eigenstates of $$\hat{H}$$, $$| \varphi _ {+} \rangle$$ and $$| \varphi _ {-} \rangle$$, with corresponding energy eigenvalues $${\varepsilon} _ {+}$$ and $${\varepsilon} _ {-}$$, respectively.

We will ask: If we prepare the system in state $$| \varphi _ {a} \rangle$$, what is the time-dependent probability of observing it in $$| \varphi _ {b} \rangle$$? Since $$| \varphi _ {a} \rangle$$ and $$| \varphi _ {b} \rangle$$ are not eigenstates of $$\hat{H}$$, and since our time-propagation will be performed in the eigenbasis using Equation \ref{1.29}, we will need to find the transformation between these bases.

We start by searching for the eigenvalues of the Hamiltonian (Equation \ref{1.35}). Since the Hamiltonian is Hermitian, ($$H _ {i j} = H _ {j i}^{*}$$), we write

$V _ {a b} = V _ {b a}^{*} = V e^{- i \varphi} \label{1.36}$

$\hat {H} = \left( \begin{array} {c c} {\varepsilon _ {a}} & {V e^{- i \varphi}} \\[4pt] {V e^{+ i \varphi}} & {\varepsilon _ {b}} \end{array} \right) \label{1.37}$

Often the couplings we describe are real, and we can neglect the phase factor $$\phi$$. Now we define variables for the mean energy and energy splitting between the uncoupled states

$E = \frac {\varepsilon _ {a} + \varepsilon _ {b}} {2}$

$\Delta = \frac {\varepsilon _ {a} - \varepsilon _ {b}} {2} \label{1.39}$

We can then obtain the eigenvalues of the coupled system by solving the secular equation

$\operatorname {det} ( H - \lambda I ) = 0$

giving

$\varepsilon _ {\pm} = E \pm \Omega \label{1.41}$

Here I defined another variable

$\Omega = \sqrt {\Delta^{2} + V^{2}} \label{1.42}$

To determine the eigenvectors of the coupled system $$| \varphi _ {\pm} \rangle$$, it proves to be a great simplification to define a mixing angle $$\theta$$ that describes the relative magnitude of the coupling relative to the zero-order energy splitting through

$\tan 2 \theta = \frac {V} {\Delta} \label{1.43}$

We see that the mixing angle adopts values such that $$0 \leq \theta \leq \pi / 4$$. Also, we note that

$\sin 2 \theta = V / \Omega \label{1.44}$

$\cos 2 \theta = \Delta / \Omega \label{1.45}$

In this representation the Hamiltonian (Equation \ref{1.37}) becomes

$\hat {H} = E \overline {I} + \Delta \left( \begin{array} {l l} {1} & {\tan 2 \theta e^{- i \varphi}} \\[4pt] {\tan 2 \theta e^{+ i \varphi}} & {- 1} \end{array} \right) \label{1.46}$

and we can express the eigenvalues as

$\varepsilon _ {\pm} = E \pm \Delta \sec 2 \theta \label{1.47}$

Next we want to find $$S$$, the transformation that diagonalizes the Hamiltonian and which transforms the coefficients of the wavefunction from the zero-order basis to the eigenbasis. The eigenstates can be expanded in the zero-order basis in the form

$| \varphi _ {\pm} \rangle = c _ {a} | \varphi _ {a} \rangle + c _ {b} | \varphi _ {b} \rangle \label{1.48}$

So that the transformation can be expressed in matrix form as

$\left( \begin{array} {l} {\varphi _ {+}} \\[4pt] {\varphi _ {-}} \end{array} \right) = S \left( \begin{array} {l} {\varphi _ {a}} \\[4pt] {\varphi _ {b}} \end{array} \right)\label{1.49}$

To find $$S$$, we use the Schrödinger equation $$\hat {H} | \varphi _ {\pm} \rangle = \varepsilon _ {\pm} | \varphi _ {\pm} \rangle$$

substituting Equation \ref{1.48}. This gives

$S = \left( \begin{array} {l l} {\cos \theta} & {e^{- i \varphi / 2}} & {\sin \theta} & {e^{i \varphi / 2}} \\[4pt] {- \sin \theta} & {e^{- i \varphi / 2}} & {\cos \theta} & {e^{i \varphi / 2}} \end{array} \right) \label{1.50}$

Note that $$S$$ is unitary since $$S^{\dagger} = S^{- 1}$$ and $$\left( S^{T} \right)^{*} = S^{- 1}$$. Also, the eigenbasis is orthonormal:

$\left\langle \varphi _ {+} | \varphi _ {+} \right\rangle + \left\langle \varphi _ {-} | \varphi _ {-} \right\rangle = 1.$

Now, let’s examine the eigenstates in two limits:

1. Weak coupling ($$| {V} / \Delta | \ll 1$$). Here $$\theta \approx 0$$, and $$| \varphi _ {+} \rangle$$ corresponds to $$| \varphi _ {a} \rangle$$ weakly perturbed by the $$V_{ab}$$ interaction. $$| \varphi _ {-} \rangle$$ corresponds to $$| \varphi _ {b} \rangle$$. In another way, as $$\theta \rightarrow 0$$, we find $| \varphi _ {+} \rangle \rightarrow | \varphi _ {a} \rangle$ and $| \varphi _ {-} \rangle \rightarrow | \varphi _ {b} \rangle.$
2. Strong coupling ($$| V / \Delta | \gg 1$$). In this limit $$\theta = \pi / 4$$, and the $$a/b$$ basis states are indistinguishable. The eigenstates are symmetric and antisymmetric combinations: $| \varphi _ {\pm} \rangle = \frac {1} {\sqrt {2}} ( | \varphi _ {b} \rangle \pm | \varphi _ {a} \rangle ) \label{1.51}$ Note from Equation \ref{1.50} that the sign of $$V$$ dictates whether $$| \varphi _ {+} \rangle$$ or $$| \varphi _ {-} \rangle$$ corresponds to the symmetric or antisymmetric eigenstate. For negative $$V \gg \Delta$$, $$\theta = - \pi / 4$$, and the correspondence in Equation \ref{1.51} changes to $$\mp$$.

We can schematically represent the energies of these states with the following diagram. Here we explore the range of $${E} _ {\pm}$$ available given a fixed coupling $$V$$ and varying the splitting $$\Delta$$.

This diagram illustrates an avoided crossing effect. The strong coupling limit is equivalent to a degeneracy point ($$\Delta = 0$$) between the states $$| \varphi _ {a} \rangle$$ and $$| \varphi _ {b} \rangle$$. The eigenstates completely mix the unperturbed states, yet remain split by the strength of interaction $$2V$$. We will return to the discussion of avoided crossings when we describe potential energy surfaces and the adiabatic approximation, where the dependence of $$V$$ and $$\Delta$$ on position $$R$$ must be considered.

Now we can turn to describing dynamics. The time evolution of this system is given by the time-propagator

$U (t) = | \varphi _ {+} \rangle e^{- i \omega , t} \left\langle \varphi _ {+} | + | \varphi _ {-} \right\rangle e^{- i \omega t} \langle \varphi _ {-} | \label{1.52}$

where $$\omega _ {\pm} = \varepsilon _ {\pm} / \hbar$$. Since $$\varphi _ {a}$$ and $$\varphi _ {b}$$ are not the eigenstates, preparing the system in state $$\varphi _ {a}$$ will lead to time evolution! Let’s prepare the system so that it is initially in $$\varphi _ {a}$$.

$| \psi ( 0 ) \rangle = | \varphi _ {a} \rangle \label{1.53} \nonumber$

Evaluating the time-dependent amplitudes of initial and final states with the help of $$S$$, we find

\begin{align*} c _ {a} (t) & = \left\langle \varphi _ {a} | U (t) | \varphi _ {a} \right\rangle \\[4pt] & = e^{- i E t} \left[ \cos^{2} \theta e^{i \Omega _ {R} t} + \sin^{2} \theta e^{- i \Omega _ {R} t} \right] \label{1.54} \\[4pt] c _ {b} (t) & = \left\langle \varphi _ {b} | U (t) | \varphi _ {a} \right\rangle \\[4pt] & = 2 \sin \theta \cos \theta e^{- i E t} \sin \Omega _ {R} t \label{1.55} \end{align*}

So, what is the probability that it is found in state $$| \varphi _ {b} \rangle$$ at time $$t$$?

\begin{aligned} P _ {b a} (t) & = \left| c _ {b} (t) \right|^{2} \\[4pt] & = \frac {V^{2}} {V^{2} + \Delta^{2}} \sin^{2} \Omega _ {R} t \end{aligned} \label{1.56}

where

$\Omega _ {R} = \frac {1} {\hbar} \sqrt {\Delta^{2} + V^{2}} \label{1.57}$

$$\Omega _ {R}$$, the Rabi Frequency, represents the frequency at which probability amplitude oscillates between $$\varphi _ {a}$$ and $$\varphi _ {b}$$ states.

Notice for the weak coupling limit ($$V \rightarrow 0$$), $$\varphi _ {\pm} \rightarrow \varphi _ {a , b}$$ (the eigenstates resemble the stationary states), and the time-dependence disappears. In the strong coupling limit ($$V \gg \Delta$$), amplitude is exchanged completely between the zero-order states at a rate given by the coupling: $$\Omega _ {R} \rightarrow V / \hbar$$. Even in this limit it takes a finite amount of time for amplitude to move between states. To get $$P=1$$ requires a time $$\tau$$:

$\tau = \pi / 2 \Omega _ {R} = \hbar \pi / 2 V. \nonumber$