Prove that the product of two n qubits Hadamard gates is identity

In summary, the conversation discusses an alternative method for calculating ##H^{\otimes n} \cdot H^{\otimes n}## using the definition of ##H^{\otimes n}## and determining the sum of a series. The solution involves bringing the ##y## power inside to give ##(-1^y)## and exploring the behavior of ##x+j## to find a pattern. The end result is that ##H^{\otimes n} \cdot H^{\otimes n} = I##.
  • #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!
 
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  • #2
I don't know the best solution here but perhaps this will help.

Try bringing the y power inside to give ##(-1^y)## and play with combinations of x,j, and y being even or odd to see if you can find a pattern in the series.
 
  • #3
jedishrfu said:
I don't know the best solution here but perhaps this will help.

Try bringing the y power inside to give ##(-1^y)## and play with combinations of x,j, and y being even or odd to see if you can find a pattern in the series.
Thanks, jedishrfu. I'll try it.
 
  • #4
It might not be the right approach but until someone posts otherwise here its something that I'd try.

I noticed that when y is even ##(-1^y)## evaluates to 1 and when odd a -1 so now you can look at how x+j behaves.
 
  • #5
I think you could prove this by induction. ##HH=I## is trivial. Then you just need to write
##H^{\oplus n+1}## in terms of ##H^{\oplus n}## and do the matrix mulitiplication.
 
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Likes Haorong Wu and jedishrfu
  • #6
tnich said:
I think you could prove this by induction. ##HH=I## is trivial. Then you just need to write
##H^{\oplus n+1}## in terms of ##H^{\oplus n}## and do the matrix mulitiplication.

Brilliant! Thanks, tnich!
 

1. What is a Hadamard gate?

A Hadamard gate is a quantum logic gate that operates on a single qubit, transforming it into a superposition state. It is represented by the matrix [1/sqrt(2) 1/sqrt(2); 1/sqrt(2) -1/sqrt(2)], where sqrt(2) is the square root of 2.

2. What does it mean for two Hadamard gates to be "product"?

The product of two Hadamard gates means that the output of one Hadamard gate is used as the input for the second Hadamard gate. This is equivalent to applying both gates in succession.

3. What is the significance of two n qubits Hadamard gates?

The significance of two n qubits Hadamard gates is that they can be used to create a quantum circuit that can perform operations on multiple qubits simultaneously. This is important in quantum computing because it allows for more complex calculations to be performed.

4. How can we prove that the product of two n qubits Hadamard gates is identity?

To prove that the product of two n qubits Hadamard gates is identity, we can use the properties of matrix multiplication. The product of two Hadamard gates is a matrix with all entries equal to 1/2^n, where n is the number of qubits. This is equivalent to the identity matrix, which has all diagonal entries equal to 1 and all other entries equal to 0.

5. Why is it important to prove that the product of two n qubits Hadamard gates is identity?

Proving that the product of two n qubits Hadamard gates is identity is important because it shows that the gates are reversible, meaning that they can be undone by applying the same gates in reverse order. This is a fundamental property of quantum computing and is necessary for performing accurate and reliable calculations.

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