next up previous contents
Next: Important Quantum Mechanical Model Up: The Two-Level System: Time-Evolution Previous: The Two-Level System: Time-Evolution   Contents

Time Evolution of States

To start with, we simply recall our first axiom of quantum mechanics here:

Axiom 1: A quantum mechanical system is described by a vector $ \vert\Psi(t)\rangle\equiv \Psi(t)$ of a Hilbert space $ {\cal H}$. The time evolution of $ \Psi(t)$ is determined by the Schrödinger equation

$\displaystyle i\hbar\frac{\partial}{\partial t} \Psi(t) = \hat{H} \Psi(t)$     (211)

The Hamilton operator $ \hat{H}$ is an operator corresponding to the total energy of the system. The solutions of the stationary Schrödinger equation at fixed energy,
$\displaystyle \hat{H}\phi=E\phi$     (212)

are called stationary states, the possible energies $ E$ eigenenergies.

We now specialize everything to our two-level system:

Hilbert space $ {\cal H} = C^2$ 2d complex vector space
Basis vectors $ \vert L\rangle = \left(\begin{array}{c}
1 \\
0
\end{array}\right), \vert R\rangle= \left(\begin{array}{c}
0 \\
1
\end{array}\right)$ particle left or right
Arbitrary vector $ \vert\psi\rangle = c_0 \vert L\rangle + c_1 \vert R\rangle$, $ \vert c_0\vert^2+\vert c_1\vert^2=1$ particle in arbitrary state
Hamiltonian \begin{displaymath}\hat{H}=\left(
\begin{array}{cc}
\varepsilon_L & T\\
T^* & \varepsilon_R
\end{array}\right)\end{displaymath} two-dimensional matrix
Stationary states $ \hat{H}\vert 1\rangle = \varepsilon_1 \vert 1\rangle$, $ \hat{H}\vert 2\rangle = \varepsilon_2 \vert 2\rangle$ The two energy eigenstates

Time evolution means the following: Suppose the state of the quantum system at time $ t=0$ is $ \vert\Psi(t=0)\rangle =\vert\Psi_0\rangle$. Then, the solution of the time-dependent Schrödinger equation $ i\hbar\partial_t \Psi(t) = \hat{H} \Psi(t)$ gives us the state $ \vert\Psi(t)\rangle$ at a later time $ t>0$.

CASE 1: Time evolution of a stationary state

If the initial state $ \vert\Psi_0\rangle$ is a stationary state, the time-evolution is simple:

$\displaystyle \vert\Psi(t=0)\rangle$ $\displaystyle =$ $\displaystyle \vert 1\rangle \leadsto \vert\Psi(t>0)\rangle = \vert 1\rangle e^{-i\varepsilon_1 t/\hbar}$  
$\displaystyle \vert\Psi(t=0)\rangle$ $\displaystyle =$ $\displaystyle \vert 2\rangle \leadsto \vert\Psi(t>0)\rangle = \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}$ (213)

because
$\displaystyle i\hbar \frac{\partial}{\partial t}\vert\Psi(t)\rangle =
i\hbar \frac{\partial}{\partial t} \vert 1\rangle e^{-i\varepsilon_1 t/\hbar}$ $\displaystyle =$ $\displaystyle \underline{\varepsilon_1 \vert 1\rangle} e^{-i\varepsilon_1 t/\hb...
...hat{H}
\vert 1\rangle} e^{-i\varepsilon_1 t/\hbar}=\hat{H} \vert\Psi(t)\rangle.$ (214)

(the same for $ \vert\Psi(t=0)\rangle = \vert 2\rangle $). The time-evolution is `trivial' and just given by the phase factor $ e^{-iEt/\hbar}$, where $ E$ is the eigenenergy of the stationary state.

CASE 2: Time evolution of a superposition of the two energy eigenstates

$\displaystyle \vert\Psi(t=0)\rangle$ $\displaystyle =$ $\displaystyle \alpha_1 \vert 1\rangle + \alpha_2 \vert 2\rangle
\leadsto \vert\...
...{-i\varepsilon_1 t/\hbar}
+ \alpha_2 \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}$ (215)

because
$\displaystyle i\hbar \frac{\partial}{\partial t}\vert\Psi(t)\rangle$ $\displaystyle =$ $\displaystyle i\hbar \frac{\partial}{\partial t}\left[\alpha_1
\vert 1\rangle e...
...repsilon_1 t/\hbar}
+\alpha_2 \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}\right]$  
  $\displaystyle =$ $\displaystyle \alpha_1{\varepsilon_1 \vert 1\rangle} e^{-i\varepsilon_1 t/\hbar}
+\alpha_2{\varepsilon_2 \vert 2\rangle} e^{-i\varepsilon_2 t/\hbar}$  
  $\displaystyle =$ $\displaystyle \hat{H} \alpha_1 \vert 1\rangle e^{-i\varepsilon_1 t/\hbar}
+\hat...
...lpha_2 \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}
=\hat{H} \vert\Psi(t)\rangle.$ (216)

To obtain the time evolution of an arbitrary initial state $ \vert\Psi(t=0)\rangle$, we therefore have to do the following: decompose $ \vert\Psi(t=0)\rangle$ into a linear combination of energy eigenstates ($ \vert 1\rangle$ and $ \vert 2\rangle$). Then, simply dress the stationary states in this linear combination with the phase factors $ e^{-i\varepsilon_1t/\hbar}$ and $ e^{-i\varepsilon_2t/\hbar}$. The linearity of the Schrödinger equation makes it that simple: the time evolution of a sum of stationary states is the sum of the time-evolved linear components of this sum.

The calculational effort is in the determination of the coefficients $ \alpha_1$ and $ \alpha_2$: we discuss this in the special

CASE 3: The initial state is $ \vert\Psi(t=0)\rangle = \vert L\rangle$, describing the particle in the left well. Note that we don't have simply a time evolution given by a phase factor:

$\displaystyle \vert\Psi(t=0)\rangle = \vert L\rangle \leadsto \vert\Psi(t>0)\rangle {\bf\ne} e^{-iEt/\hbar} \vert L\rangle$   $\displaystyle \mbox{\rm for some $E$}$$\displaystyle .$      

The state $ \vert L\rangle$ is not an eigenstate of the Hamiltonian $ \hat{H}$, therefore its time evolution is not given by a simple phase factor. What we have to do is clear: we have to find the decomposition of the vector $ \vert L\rangle$ into a linear combination of energy eigenstates $ \vert 1\rangle$ and $ \vert 2\rangle$,
$\displaystyle \vert\Psi(t=0)\rangle = \vert L\rangle = \alpha_1 \vert 1\rangle ...
...-i\varepsilon_1 t/\hbar}
+ \alpha_2 \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}.$     (217)

For simplicity, we do this for the special case of negative $ T$ and identical energy parameters $ \varepsilon_L=\varepsilon_R$ in the Hamiltonian $ \hat{H}$. We had calculated already before in (3.41),
$\displaystyle \varepsilon_L=\varepsilon_R=:\varepsilon_0,\quad
T=-\vert T\vert\leadsto \vert 1\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\vert L\rangle + \vert R\rangle \right],\quad
\varepsilon_1=\varepsilon_0 - \vert T\vert$  
$\displaystyle \vert 2\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[ -\vert L\rangle + \vert R\rangle \right],\quad
\varepsilon_2= \varepsilon_0 + \vert T\vert.$ (218)

We solve this equation for $ \vert L\rangle$,
$\displaystyle \vert L\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[ \vert 1\rangle - \vert 2\rangle \right]\leadsto
\alpha_1 = \frac{1}{\sqrt{2}},\quad \alpha_2 = -\frac{1}{\sqrt{2}}.$ (219)

By this we have found our desired result, that is the time evolution of the initial state $ \vert L\rangle$,
$\displaystyle \vert\Psi(t=0)\rangle = \vert L\rangle$ $\displaystyle \leadsto$ $\displaystyle \vert\Psi(t)\rangle = \frac{1}{\sqrt{2}}\left[ \vert 1\rangle e^{-i\varepsilon_1 t/\hbar}
- \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}\right].$ (220)

We have expressed $ \vert\Psi(t)\rangle$ in the basis of the energy eigenstates $ \vert 1\rangle$ and $ \vert 2\rangle$. We now would like to express $ \vert\Psi(t)\rangle$ in the basis of the left and right states, $ \vert L\rangle$ and $ \vert R\rangle$: this is simple because we know
$\displaystyle \vert 1\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[\vert L\rangle + \vert R\rangle \right],\...
...rt 2\rangle = \frac{1}{\sqrt{2}}\left[ -\vert L\rangle + \vert R\rangle \right]$  
$\displaystyle \leadsto \underline{\vert\Psi(t)\rangle}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left[ \vert 1\rangle e^{-i\varepsilon_1 t/\hbar}
- \vert 2\rangle e^{-i\varepsilon_2 t/\hbar}\right] =$  
  $\displaystyle =$ $\displaystyle \frac{1}{{2}}\left\{ \left[\vert L\rangle + \vert R\rangle \right...
...[-\vert L\rangle + \vert R\rangle \right] e^{-i\varepsilon_2 t/\hbar}\right\} =$  
  $\displaystyle =$ $\displaystyle \frac{1}{{2}}\left\{\vert L\rangle \left[e^{-i\varepsilon_1 t/\hb...
... \left[e^{-i\varepsilon_1 t/\hbar} - e^{-i\varepsilon_2 t/\hbar}\right]\right\}$  
  $\displaystyle =:$ $\displaystyle \underline{c_0(t)\vert L\rangle + c_1(t)\vert R\rangle}$  
$\displaystyle c_0(t)$ $\displaystyle :=$ $\displaystyle \frac{1}{{2}}\left[e^{-i\varepsilon_1 t/\hbar} + e^{-i\varepsilon...
...{1}{{2}}\left[e^{-i\varepsilon_1 t/\hbar} - e^{-i\varepsilon_2 t/\hbar}\right].$ (221)

The coefficients $ c_0(t)$ and $ c_1(t)$ have a clear physical meaning: let us recall
$\displaystyle \vert c_0\vert^2(t)=p_0(t)$ $\displaystyle =$ the probability to find the particle in the left well  
$\displaystyle \vert c_1\vert^2(t)=p_1(t)$ $\displaystyle =$ the probability to find the particle in the right well$\displaystyle .$ (222)

Obviously, these probabilities are a function of time now: $ \vert\Psi(t)\rangle$ is not stationary. We calculate
$\displaystyle p_0(t)$ $\displaystyle =$ $\displaystyle \frac{1}{{4}}\left\vert e^{-i\varepsilon_1 t/\hbar} + e^{-i\varep...
...2
=\frac{1}{2}\left\{ 1 + \cos [(\varepsilon_1-\varepsilon_2)t/\hbar)] \right\}$  
  $\displaystyle =$ $\displaystyle \cos^2 [\frac{(\varepsilon_1-\varepsilon_2)t}{2\hbar}]$  
$\displaystyle p_1(t)$ $\displaystyle =$ $\displaystyle \frac{1}{{4}}\left\vert e^{-i\varepsilon_1 t/\hbar} - e^{-i\varep...
...2
=\frac{1}{2}\left\{ 1 - \cos [(\varepsilon_1-\varepsilon_2)t/\hbar)] \right\}$  
  $\displaystyle =$ $\displaystyle \sin^2 [\frac{(\varepsilon_1-\varepsilon_2)t}{2\hbar}]$ (223)

and therefore
$\displaystyle \cos^2 [\frac{(\varepsilon_1-\varepsilon_2)t}{2\hbar}]$ $\displaystyle =$ the probability to find the particle in the left well  
$\displaystyle \sin^2 [\frac{(\varepsilon_1-\varepsilon_2)t}{2\hbar}]$ $\displaystyle =$ the probability to find the particle in the right well$\displaystyle .$ (224)

As a function of time, the probabilities oscillate with an angular frequency that is given by the energy splitting $ \Delta=\varepsilon_2-\varepsilon_1=2\vert T\vert$, divided by $ \hbar$. At the initial time $ t=0$, the particle is in the left well (the probability $ p_0(0)=1$), but for $ t>0$ this probability starts to oscillate: the particle tunnels from the left well into the right well, back into the left well and so forth.


next up previous contents
Next: Important Quantum Mechanical Model Up: The Two-Level System: Time-Evolution Previous: The Two-Level System: Time-Evolution   Contents
Tobias Brandes 2004-02-04