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Probability Interpretation

Now that we know how to represent our state vector as a superposition of states, but we know that we can only measure the state vector to be in one of the base states, we must determine what happens when we measure the state vector. We know from quantum physics that given an initial condition the state vector will evolve in time in accordance with Schrödenger's equation:

\begin{displaymath}ih\frac{\partial \left\vert X(t) \right\rangle }{\partial t} = H(t)\left\vert X(t) \right\rangle\end{displaymath}

Where $i$ is the square root of negative one, $h$ is $1.0545*10^{-34}$ Js, and $H$ is the Hamiltonian operator, which is determined by the physical characteristics of the system being evolved.

In Dirac notation this expression is:

\begin{displaymath}ih\frac{\partial w_{i}(t)}{\partial t} = \sum_{j}H(t)_{ij} * w_{j}(t)\end{displaymath}

The evolution of the state vector through time would appear to be continuous, but these equations only apply to a quantum mechanical system evolving in isolation from the environment. The only way to observe the state of the state vector is to in some way cause the quantum mechanical system to interact with the environment. When the state vector is observed it makes a sudden discontinuous jump to one of the eigenstates. When the quantum mechanical system interacts with the outside environment and measured the state vector is said to have collapsed. (Williams, Clearwater)

To perform any useful calculations we must be able so say something about which base state a quantum mechanical system will collapse into. The probability that the state vector will collapse into the $j$'th eigenstate is given by $\left\vert w_{j}\right\vert^{2}$ which is defined to be $a_{j}^{2} + b_{j}^{2}$ when $w_{j} = a_{j} + i*b_{j}$ and $w_{j}$ is the complex projection of the state vector onto the $j$'th eigenstate. In general the chance of choosing any given state is

\begin{displaymath}Prob(j) = \frac{\vert w_{j}\vert^{2}}{\sum_{k = 0}^{ n-1} \left\vert w_{k}\right\vert^{2}}\end{displaymath}

but as mentioned earlier we will insist on having the state vector of length one, and in this case the probability expression simplifies to $Prob(j) = \vert w_{j}\vert^{2}$.

We now know how to describe an $n$ state quantum system, which can be placed in an arbitrary superposition of states. We also know how measuring this superposition yield a given base state with a given probability. This is all that we need to understand about our quantum memory register to be able to simulate its behavior.


next up previous contents
Next: Quantum Parallelism Up: The Quantum Computer Previous: The Quantum Memory Register   Contents
Matthew Hayward 2008-04-26