 #1
Haorong Wu
 394
 85
 Homework Statement:
 Prove ##H^{\otimes n} \cdot H^{\otimes n} = I##
 Relevant Equations:

##H^{\otimes n} = \frac 1 {\sqrt {2^n}} \sum_{x,y} {\left ( 1 \right )} ^{x \cdot y} \left  x \right > \left < y \right ##
where ##x \text { and } y \text { are from } 00 \dots 00 \text { to } 11 \dots 11##, and ##x \cdot y = x_1 y_1 + x_2 y_2 + \cdots + x_n y_n##
From the properties of tensor product, ##H^{\otimes n} \cdot H^{\otimes n} =\left ( H_1 \cdot H_1 \right ) \otimes \left ( H_2 \cdot H_2 \right ) \otimes \cdots \otimes \left ( H_n \cdot H_n \right ) =I \otimes I \otimes \cdots \otimes I =I## where ##H_i## acts on the ##i^{th}## qubit.
But I want to try another way from the definition of ##H^{\otimes n} ##:
##H^{\otimes n} \cdot H^{\otimes n} \\ =\left [ \frac 1 {\sqrt {2^n}} \sum_{x,y} {\left ( 1 \right )} ^{x \cdot y} \left  x \right > \left < y \right  \right ] \left [ \frac 1 {\sqrt {2^n}} \sum_{i,j} {\left ( 1 \right )} ^{i \cdot j} \left  i \right > \left < j \right \right ] \\ = \frac 1 {2^n} \sum_x \sum_y \sum _j {\left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} \left  x \right > \left < j \right  \\ = \frac 1 {2^n} \sum_x \sum_j \left ( \sum _y {\left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} \right ) \left  x \right > \left < j \right ##
I'm stuck because I have no idea how to properly calculate ## \sum _y { \left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} ##. The answer should be ##0## if ##x \neq j##, and ##1## otherwise.
Any advice? Thanks!
But I want to try another way from the definition of ##H^{\otimes n} ##:
##H^{\otimes n} \cdot H^{\otimes n} \\ =\left [ \frac 1 {\sqrt {2^n}} \sum_{x,y} {\left ( 1 \right )} ^{x \cdot y} \left  x \right > \left < y \right  \right ] \left [ \frac 1 {\sqrt {2^n}} \sum_{i,j} {\left ( 1 \right )} ^{i \cdot j} \left  i \right > \left < j \right \right ] \\ = \frac 1 {2^n} \sum_x \sum_y \sum _j {\left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} \left  x \right > \left < j \right  \\ = \frac 1 {2^n} \sum_x \sum_j \left ( \sum _y {\left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} \right ) \left  x \right > \left < j \right ##
I'm stuck because I have no idea how to properly calculate ## \sum _y { \left ( 1 \right )} ^{x \cdot y} {\left ( 1 \right )} ^{y \cdot j} ##. The answer should be ##0## if ##x \neq j##, and ##1## otherwise.
Any advice? Thanks!