Exercise 3.3 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf {Exercise\ 3.3}$
Show that the state

(1)

\begin{align} \ket{W_n} = \frac{1}{\sqrt{n}} (\ket{0\dots 001} + \ket{0\dots 010} + \ket{0\dots 100} + \cdots + \ket{1\dots 000}) \end{align}

is entangled, with respect to the decompostion into the $n$ qubits, for every $n>1$.

Proof by Contradiction

twhittle 21 Dec 2013 03:39

Assume $|W_n\rangle$ is not entangled and n > 1. Then $|W_n\rangle$ can be written as:

(1)

\begin{align} a_1|0\rangle+b_1|1\rangle\otimes a_2|0\rangle+b_2|1\rangle\otimes\dotsc\otimes a_n|0\rangle+b_n|1\rangle \end{align}

From the definition of $|W_n\rangle$ and (1), it follows that:

(2)

\begin{align} a_1\dotsc a_{n-1}b_n=a_1\dotsc b_{n-1}a_n=\dotsc=b_1\dotsc a_{n-1}a_n = \frac{1}{\sqrt n} \end{align}

From (2) we can say that $b_i\neq0\text{ }\forall i\in [1,n]$ (3) and $a_j\neq0\text{ }\forall j\in [1,n]$ (4). We also know from the tensor product that the coefficients for all states that do not appear in $|W_n\rangle$ must be zero. For instance, the coefficient of $|1\dotsc 11\rangle$ given by $b_1\dotsc b_{n-1}b_n$ must equal 0. In order for this to be true, at least one $b_i=0\text{, }i\in[1,n]$. This is a contradiction to (3) above, and thus $|W_n\rangle$ is entangled.

Last edited on 16 Jan 2014 04:29 by twhittle Show more