Exercise 3.11 Discussion

$\def\abs#1{\vert #1 \vert}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.11}$

Let $\ket\psi$ be an $n$-qubit state. Show that the sum of the distances from $\ket\psi$ to the standard basis vectors $\ket{j}$ is bounded below by a positive constant that depends only on $n$,

(1)\begin{align} \sum_j \abs{\ket\psi - \ket{j}} \geq C, \end{align}

where $\abs{\vec{v}}$ indicates the length of the enclosed vector. Specify such a constant $C$ in terms of $n$.

page revision: 0, last edited: 11 Dec 2011 23:23

For $n=1$, the stated theorem is clearly not true – the distance can be zero. For the rest of this answer we assume $n > 1$.

Let’s label the standard basis vectors $\ket{b_0} \ldots \ket{b_{n-1}}$, and let $\ket{b_n}$ be another name for $\ket{b_0}$ – this will simplify the proof notation.

For any state $\ket{\psi}$, the sum of the distances from $\ket{\psi}$ to the basis vectors is

(1)We can now use the triangle inequality:

(2)It is easy to see that the distance between any two standard basis vectors is $\sqrt{2}$. Hence

(3)This is a positive lower bound dependent on $n$. (It is not a

tightlower bound.)ReplyOptions