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Theoretical Example of Superposition

Unique to wave theories is the concept of superposition of states. Indeed, the interference of states is often used as a defining characteristic of quantum systems such that the destruction of a pre-existing interference pattern is identified with a transition into the classical realm (``the appearance of the classical world'' [15]).

In demonstrating the mathematics of decoherence, the simplest system we could possibly start with must entail at least two states capable of superposition. We consider a simple harmonic oscillator in the state,

$\displaystyle \vert\psi\rangle = \vert\alpha_1\rangle + \vert\alpha_2\rangle$ (2.1)

Generically, we define $ \alpha$ as an eigenvalue of the annhilation operator state of the oscillator,

$\displaystyle \vert\alpha\rangle = \exp(-\frac{\vert\alpha\vert^2}{2})\sum^{\infty}_{n=0}\frac{\alpha^n}{\sqrt{n!}}\vert n\rangle$ (2.2)

These so called coherent states represent a quantum system which is very close to being in a classical state (see appendix). Let us ignore, for now, the need for normalization (or if preferred, assume the system is normalized as is). It is well known that the Hamilton for such a system is simply given by

$\displaystyle H = \hbar \omega a^{\dagger}a$

The unitary evolution of this wave function is,

$\displaystyle \vert\psi(t)\rangle = \exp(-i\omega a^{\dagger}at)(\vert\alpha_1\rangle + \vert\alpha_2 \rangle)$ (2.3)

Note that

$\displaystyle \exp(-i\omega a^{\dagger}at)\exp(-\frac{\vert\alpha\vert^2}{2})\s...
...2})\sum^{\infty}_{n=0}\frac{\alpha^n}{\sqrt{n!}}\exp(-i\omega tn)\vert n\rangle$

It can be written,

$\displaystyle \vert\psi(t)\rangle = (\vert\alpha_1 \exp(-i\omega t)\rangle + \vert\alpha_2 \exp(-i\omega t)\rangle)$ (2.4)

The density matrix is

$\displaystyle \rho(t)=\sum_{i,j=1}^{2}\vert\alpha_i \exp(-i\omega t)\rangle \langle \alpha_j \exp(-i\omega t)\rangle\vert$ (2.5)

To get a clear picture of the interference this superposition entails, we can take a look at this system in the coordinate representation. We have that,

$\displaystyle a\vert\alpha\rangle = \alpha\vert\alpha\rangle = \frac{1}{\sqrt{2\hbar \omega}}(\omega q + ip)\vert\alpha\rangle \nonumber$

We can thus obtain,

$\displaystyle \langle q'\vert\alpha \rangle = \left(\frac{\omega}{\pi \hbar}\ri...
...{\frac{2\omega}{\hbar}}\alpha q' - \frac{\vert\alpha\vert^2+\alpha^2}{2}\right)$ (2.6)

Let $ \vert\alpha_i \exp(-i\omega t)\rangle \equiv \vert\alpha_i(t)\rangle$ . Then we can write,

$\displaystyle \langle q' \vert \rho(t) \vert q'\rangle = \langle q' \vert\left(\sum_{i,j}\vert \alpha_i\rangle \langle \alpha_j \vert\right) \vert q'\rangle$

$\displaystyle \langle q'\vert\rho(t)\vert q'\rangle = \vert\langle q'\vert\alph...
...^2 + 2 Re\langle q' \vert \alpha_1(t)\rangle \langle \alpha_2(t)\vert q'\rangle$ (2.7)

The latter follows from the fact that the two cross terms will be conjugates, and for any complex number z, $ z + z^*=2Re(z)$ . We can partition this function into real and imaginary parts as follows. We write

$\displaystyle \alpha(t)=\alpha \exp(-i\omega t) = \alpha (\cos \omega t - i\sin \omega t)$

We note that
$\displaystyle \frac{\alpha^2 + \vert\alpha\vert^2}{2}$ $\displaystyle =$ $\displaystyle \frac{\alpha(\cos^2\omega t - \sin^2 \omega t - 2 i \cos \omega t \sin \omega t + \cos^2 \omega t + \sin^2 \omega t)}{2}$  
  $\displaystyle =$ $\displaystyle \alpha^2\left(\cos^2 \omega t - \frac{i}{2}\sin 2 \omega t\right)$ (2.8)


$\displaystyle \langle q' \vert \alpha \rangle = \left(\frac{\omega}{\pi \hbar}\...}{\hbar}}\alpha q' \sin \omega t + \frac{\alpha^2}{2}\sin 2 \omega t \right)

It is now conventional to chose a simple case in which $ \alpha_1 \equiv \alpha$ and $ \alpha_2 \equiv - \alpha$ ,

$\displaystyle \vert\langle q' \vert \alpha_{1,2}\rangle \vert^2 = \sqrt{\frac{\...
...ega}{\hbar}}q' \pm \sqrt{2}\alpha \cos \omega t\right)^2\right)\equiv I_{1,2}^2$ (2.9)

The convenient $ \pm\alpha$ form makes calculation of the cross term easy, and it is,

$\displaystyle 2 Re\langle q' \vert \alpha_1(t)\rangle \langle \alpha_2(t)\vert ...
...rac{2\omega}{\hbar}}\alpha q' \sin \omega t\right) \equiv 2I_1I_2\cos \theta(t)$ (2.10)

$\displaystyle \langle q'\vert\rho(t)\vert q'\rangle=\overbrace{\overbrace{I_1^2...
...erbrace{\overbrace{2I_1I_2\cos \theta(t)}^{\mbox{quantum interference}}}^{\chi}$ (2.11)

If Decoherence is to bring us into the classical realm, $ \chi$ must be destroyed. Let us be more explicit here. Equation 2.11 gives us only partial information, i.e. the probabilities for any given position q'. To represent the density matrix in proper form, we need integrate over all positions, i.e. the density matrix is

$\displaystyle \int^{\infty}_{-\infty}\left(\begin{array}{cc}I_1^2&I_1I_2\cos \theta(t)\\ I_2I_1 \cos \theta(t)&I_2^2\end{array}\right)dq'$ (2.12)

A classical density matrix would not have off diagonal terms. The total probability of finding the particle in one state or the other is given by

$\displaystyle Tr\left(\int^{\infty}_{-\infty}\left(\begin{array}{cc}I_1^2&I_1I_...
...d{array}\right)dq'\right)=\int^{\infty}_{-\infty}\left(I_1^2 +I_2^2\right)dq'=1$ (2.13)

Experimental demonstration of the effect of superposition has made much progress in the last decade, and we discuss one such example in the next section.

next up previous contents
Next: Experimental Example of Superposition Up: Is Quantum Decoherence the Previous: What is Decoherence?   Contents
tim jones 2007-04-11