Exercise 3.9 Discussion

$\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\mathbf{Exercise\ 3.9}$

a) Show that any $n$-qubit quantum state can be represented by a vector of the form

(1)\begin{align} a_0\ket{0\dots 00} + a_1\ket{0\dots 01} + \dots+a_{2^n-1}\ket{1\dots 11} \end{align}

where the first non-zero $a_i$ is real and non-negative.

b) Show that this representation is unique in the sense that any two different vectors of this form represent different quantum states.

page revision: 0, last edited: 11 Dec 2011 23:18

Consider a state represented by a vector

(1)with arbitrary $a_0 \dots a_{2^n-1}$. Consider $a_i$ starting with $i=0$.

If $a_i = 0$, repeat the process with $a_{i+1}$. We are guaranteed to eventually encounter $a_k \neq 0$ because the vector is by definition unit length.

If $a_i$ is real and positive, the vector is already in the desired form.

Otherwise, $a_i$ can be represented as $a_i = re^{i\phi}$, with $r$ a non-negative (in fact, positive) real. Multiply $a_0, \dots, a_{2^n-1}$ by $e^{-i\phi}$. The result is by definition a vector representing an equivalent state, with amplitudes $b_0 \dots b_{i-1} = 0$ and $b_i = r$, where r is a positive real. Therefore, the result is in the desired canonical form.

ReplyOptionsSuppose there are two state vectors $\ket{a}$ and $\ket{b}$, both of the canonical form described above, and $\ket{a} \neq \ket{b}$ while $\ket{a} \sim \ket{b}$.

Because $\ket{a} \sim \ket{b}$, there exist $e^{i\phi}$ such that $\ket{a} = e^{i\phi}\ket{b}$.

Let's consider the first non-zero $b_k$. Because $\ket{a} = e^{i\phi}\ket{b}$, $a_k = e^{i\phi}b_k$.

Because $\ket{b}$ is canonical, $b_k$ is a positive real number.

If $e^{i\phi} \neq 1$, $a_k = e^{i\phi}b_k$ is not a positive real number. For all $i < k: a_i = 0$, because the corresponding $b_i = 0$ and $a_i = e^{i\phi}b_i$. Therefore, the first non-zero amplitude in $\ket{a}$ is not a positive real number. This contradicts the assumption that $\ket{a}$ is canonical.

If $e^{i\phi} = 1$, then $\ket{a} = 1\ket{b} = \ket{b}$. This contradicts the assumption that $\ket{a} \neq \ket{b}$.

ReplyOptions