- #1

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- 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!