Oct
17
2008

No Cloning Theorem & Heisenberg’s Uncertainty Principle

A demonstration of the link between the no cloning theorem and Heisenberg’s uncertainty principle.

In quantum information theory the no cloning theorem states that it is not possible to copy an unknown quantum state or quantum bit known as a qubit. The simplest proof I have come across to date goes a little something like this.

Assume there exists an unitary operator U (or a quantum gate) that can clone the unknown state

\left|\phi\right>=\alpha\left|0\right>+\beta\left|1\right>

We will show that acting on the tensor product of the unknown state \left|\phi\right> and a prepared state \left|0\right> with U, in two equivalent ways, gives an inconsistent result.

1. Acting on the unexpanded form of our unknown state

U\left|\phi\right>\left|0\right>=\left|\phi\right>\left|\phi\right>

which can be written as

U \left|\phi\right>\left|0\right>=(\alpha\left|0\right>+\beta\left|1\right>)(\alpha\left|0\right>+\beta\left|1\right>)
U \left|\phi\right>\left|0\right>=\alpha^2\left|00\right>+\alpha\beta\left|01\right>+\beta\alpha\left|10\right>+\beta^2\left|11\right>

2. Acting on the expanded form of our unknown state

U(\alpha\left|0\right>+\beta\left|1\right>)\left|0\right>=\alpha\left|00\right>+\beta\left|11\right>

From this we see that

U(\alpha \left| 0 \right>+\beta\left| 1 \right>)\left| 0 \right>\neq U\left|\phi\right>\left|0\right>

An inconsistency. This proves that there cannot exist a unitary operator that clones an unknown quantum state.

This is good.

Corollary

When a measurement is made on a quantum state, the state itself changes and information contained within the state is lost forever. It is therefore not possible to know simultaneously certain properties of a quantum state; such pairs of properties are known as conjugate variables. This premise of quantum mechanics is known as Heisenberg’s uncertainty principle (HUP). If we could clone an unknown quantum state we could simultaneously measure these conjugate variables and hence would be in complete violation of HUP.

Another way of looking at this is if we could measure all properties of an unknown quantum state simultaneously we could indeed recreate that state i.e. clone it.

Therefore we see that the no cloning theorem and Heisenberg’s uncertainty principle are two just two sides of the same coin.

Nice.

__________________________________

Unitary operators

A unitary operator has the property

U^\dagger U = I = identity \: matrix

In quantum computation we talk of a quantum gate acting on a quantum bit or qubit. A quantum gate can be thought of as an operator for the purposes of this discussion.

Since

\left|\varphi\right>=\alpha\left|0\right>+\beta\left|1\right>

has normalization condition

|\alpha|^2+|\beta|^2=1

It only makes sense that the post-operated state

\left|\varphi '\right>=\alpha '\left|0\right>+\beta '\left|1\right>

has the normalization condition

|\alpha '|^2+|\beta '|^2=1

This is accomplished by demanding the operator U is unitary. Amazingly it turns out that the unitary constraint is the only constraint placed on quantum gates, that is, any unitary matrix qualifies as a quantum gate.

Quantum Gates

We can represent a qubit in matrix form

\left|\varphi\right> =
\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}
=\alpha\left|0\right>+\beta\left|1\right>

The quantum NOT gate, X, for example, which acts to change 0 –> 1 and 1 –> 0, takes the matrix form

X =
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}

It operates on a single qubit as

X
\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}
 =
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}
=
\begin{bmatrix}
\beta\\
\alpha
\end{bmatrix}

And it is unitary

X^\dagger X =
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
=
\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}
=I

4 Comments »

  • Cool - i need to know what unitary matrices are and why we use them for the Orch OR model of consciousness!
    A few things that are unclear to me:
    1)Are you defining U|0> = |phi> in these proofs?

    2)How does beta|1>|0> = beta|11> in approach 2?

    Cheers

    James

    Comment | October 21, 2008
  • Hey James,

    Perhaps I have been a little unclear here.

    1. No U|0> =! |phi>, the point here is that we assume there exists an operator U that is both unitary and acts upon the tensor product |phi>|0> to give |phi>|phi> i.e.

    U|phi>|0> = |phi>|phi>

    The details of U are not of interest to the proof. What does matter is that U acts on an unknown state |phi> and a prepared state |0> (we know this to be |0> because we have prepared it, i.e. we measured it before hand and thus its wave function collapsed such that it is now |0>)

    2. You have misread the proof here.

    Remember what the operator U does…

    U|phi>|0> = |phi>|phi>
    U|red tractor>|0> = |red tractor>|red tractor> = |(red tractor) (red tractor)>

    Hence where we have

    U(a|0> +B|1>)|0> = Ua|0>|0> + UB|1>|0>
    –>
    U(a|0> +B|1>)|0> = a|00> + B|11>

    Does this make sense?

    There is an explanation of unitary operators beneath, but if it is unclear I can do more extensive post on it.

    Matt

    Comment | October 21, 2008
  • ok, but |1>|0> are two known states, unlike |phi>. So is the assumption that U turns the latter component of the product into the former component?

    Comment | October 23, 2008
  • The details of U are not of interest to us. We are only assuming that there exists an operator U that acts on the tensor product of |phi>|0> –> |phi>|phi>. This is just a notation convention, it is the alteration of the whole tensor product that is important.

    However, none of this really matters, because as we showed, U cannot exist as a unitary operator and therefore cannot be a real quantum operator (or quantum gate).

    Comment | October 23, 2008

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