September 29th, 2008

The No Clone Theorem

Included here is two proofs of the No Clone theorem in Quantum Mechanics. To properly understand them, you should at least be familiar with bracket notation.

Proof 1
Given a state \left|\psi_1\right> and a blank sheet of paper \left|X\right>, we place them both into our Quantum Xerox machine, and we get out \left|\psi_1\right>\left|\psi_1\right>. Repeating with a second state, \left|\psi_2\right>\left|X\right> we get \left|\psi_2\right>\left|\psi_2\right>. So far, so good.

The problem happens when we take an admixture of states \alpha\left|\psi_1\right> + \beta\left|\psi_2\right> and throw that into our quantum Xerox machine.

(\alpha\left|\psi_1\right> + \beta\left|\psi_2\right>)\left|X\right> should yield \alpha\left|\psi_1\right> + \beta\left|\psi_2\right> \alpha\left|\psi_1\right> + \beta\left|\psi_2\right>.

But it does not. Instead, it yields \alpha\left|\psi_1\right> + \beta\left|\psi_2\right>\left|X\right> \rightarrow \alpha\left|\psi_1\right>\left|\psi_1\right> + \beta\left|\psi_2\right>\left|\psi_2\right>

Our Quantum Xerox machine can copy lines (states) parallel to |\psi_1> or |\psi_2> but completely screws up an sort of linear combination. It’s as if our machine could copy horizontal or vertical lines just fine, but cannot do diagonal lines at all. Shame Xerox does not service quantum machines.

Proof 2

Let U be a unitary time operator U(t) acting on |\psi> and |X> such that U(t)|\psi_1>|X> = |\psi_1>|\psi_1>; and ditto for |\psi_2>.

<\psi_1|<\psi_2|U^{\dagger}U|\psi_1>|\psi_2> = <\psi_1|<\psi_2|\psi_1>|\psi_2> which is true iff |\psi_1> = |\psi_2> or |\psi_1> is orthogonal to |\psi_2>.

The first proof is highly inspired by David Griffiths.

Share and Enjoy:
  • Digg
  • del.icio.us
  • Facebook
  • Mixx
  • Google Bookmarks
  • Fark
  • Reddit
  • Yahoo! Bookmarks
(RSS 2.0, Both comments and pings are currently closed.

Comments are closed.

Your Ad Here