Ex5 2
$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 5.2}$
For which sets of states is there a cloning operator? If the set has a cloning operator, give the operator. If not, explain your reasoning.
a) $\{ \ket 0, \ket 1 \}$,
b) $\{ \ket +, \ket - \}$,
c) $\{ \ket 0, \ket 1, \ket +, \ket - \}$,
d) $\{ \ket 0\ket +, \ket 0\ket -, \ket 1\ket +, \ket 1\ket - \}$,
e) $\{ a\ket 0 + b\ket 1 \}$, where $|a|^2 + |b|^2 = 1$.
page revision: 0, last edited: 18 Nov 2012 18:35
I'm not sure what is being asked in this question. Is the point to consider if an arbitrary vector from a vector spaced spanned by the given vectors can be cloned? If each of the vectors can be cloned with the same operator? I'm pretty sure everyway I read this question the no cloning principle says that the operator doesn't exist.
a)
(1)The unitary transformation
will map $\ket{00} \mapsto \ket{00}$ and $\ket{10} \mapsto \ket{11}$, and so in this sense it is a cloning operator for the set $\{\ket 0, \ket 1\}$.
b)
The same matrix in the basis $\{\ket{++}, \ket{+-}, \ket{-+}, \ket{--}\}$ can be used as a cloning operator for the set $\{\ket +, \ket -\}$.
c)
(2)A necessary condition for a set of states to have a cloning operator is that the states are mutually orthogonal. To show this, let's assume $\ket{\psi}$ and $\ket{\varphi}$ have a cloning operator $U$. Then
This means that $\braket{\psi}{\varphi}$ is either 1 or 0. If we have $\braket{\psi}{\varphi} = 1$, then because of the normalised states: $||\psi||² = 1$ and $||\varphi||² = 1$, we have
(3)and the positive definiteness of the inner product shows $\ket{\psi} - \ket{\varphi} = 0$. That is, the states are identical.
In conclusion, since the given set of states is not orthogonal, the set cannot have a cloning operator.
d)
This set is orthogonal and a cloning operator can be constructed.
e)
This set describes arbitrary states and the no-cloning theorem says that a cloning operator for the set does not exist.