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Understanding a quantum algorithm to estimate inner products
Compute average value of two-qubit systemHow exactly is the stated composite state of the two registers being produced using the $R_zz$ controlled rotations?How can I calculate the inner product of two quantum registers of different sizes?How is measurement modelled when using the density operator?Evaluate the given quantum circuitWhat happens with first phase factor in QFT?Swap Test for vector difference - how are different sized inputs combined?Estimation of Z in the quantum Euclidean algorithmUnderstanding the oracle in Deutsch's algorithmDoes the dilation in Naimark's theorem produce a state?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
While reading the paper "Compiling basic linear algebra subroutines for quantum computers", here, in the Appendix, the author/s have included a section on quantum inner product estimation.
Consider two vectors $x,y in mathbbC^n, x= (x_1, dots , x_n), y= (y_1, ldots, y_n)$, we want to estimate the inner product $langle x | y rangle$. Assume we are given a state $|psi rangle = frac 1 sqrt 2 big(|0 rangle |x rangle + |1 rangle |y rangle big)$, after applying a Hadamard transform to the first qubit, the result is:
$$|psi rangle = frac 1 2 big(|0 rangle (|x rangle + |y rangle) + |1 rangle(|x rangle - |y rangle) big).$$
The author then states that after measuring the first qubit in the computational basis, the probability to measure $|0 rangle$ is given by $p = frac 1 2 big(1 + mathrmRe(langle x | y rangle) big)$. I do not understand this statement. From what I understand, after applying a partial measurement to the first qubit, the probability of measuring $|0 rangle$ is given by
$frac 1 4 sqrt sum_i=0^n(overline(x_i+y_i)(x_i+y_i)) ^2$ (in other words the norm of the vector squared), so I am not sure why these formulas are equivalent, or if I am mistaken.
algorithm quantum-state mathematics measurement
edited Jun 5 at 10:47
glS
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
$endgroup$
add a comment |
$begingroup$
While reading the paper "Compiling basic linear algebra subroutines for quantum computers", here, in the Appendix, the author/s have included a section on quantum inner product estimation.
Consider two vectors $x,y in mathbbC^n, x= (x_1, dots , x_n), y= (y_1, ldots, y_n)$, we want to estimate the inner product $langle x | y rangle$. Assume we are given a state $|psi rangle = frac 1 sqrt 2 big(|0 rangle |x rangle + |1 rangle |y rangle big)$, after applying a Hadamard transform to the first qubit, the result is:
$$|psi rangle = frac 1 2 big(|0 rangle (|x rangle + |y rangle) + |1 rangle(|x rangle - |y rangle) big).$$
The author then states that after measuring the first qubit in the computational basis, the probability to measure $|0 rangle$ is given by $p = frac 1 2 big(1 + mathrmRe(langle x | y rangle) big)$. I do not understand this statement. From what I understand, after applying a partial measurement to the first qubit, the probability of measuring $|0 rangle$ is given by
$frac 1 4 sqrt sum_i=0^n(overline(x_i+y_i)(x_i+y_i)) ^2$ (in other words the norm of the vector squared), so I am not sure why these formulas are equivalent, or if I am mistaken.
algorithm quantum-state mathematics measurement
edited Jun 5 at 10:47
glS
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
$endgroup$
$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11
add a comment |
$begingroup$
While reading the paper "Compiling basic linear algebra subroutines for quantum computers", here, in the Appendix, the author/s have included a section on quantum inner product estimation.
Consider two vectors $x,y in mathbbC^n, x= (x_1, dots , x_n), y= (y_1, ldots, y_n)$, we want to estimate the inner product $langle x | y rangle$. Assume we are given a state $|psi rangle = frac 1 sqrt 2 big(|0 rangle |x rangle + |1 rangle |y rangle big)$, after applying a Hadamard transform to the first qubit, the result is:
$$|psi rangle = frac 1 2 big(|0 rangle (|x rangle + |y rangle) + |1 rangle(|x rangle - |y rangle) big).$$
The author then states that after measuring the first qubit in the computational basis, the probability to measure $|0 rangle$ is given by $p = frac 1 2 big(1 + mathrmRe(langle x | y rangle) big)$. I do not understand this statement. From what I understand, after applying a partial measurement to the first qubit, the probability of measuring $|0 rangle$ is given by
$frac 1 4 sqrt sum_i=0^n(overline(x_i+y_i)(x_i+y_i)) ^2$ (in other words the norm of the vector squared), so I am not sure why these formulas are equivalent, or if I am mistaken.
algorithm quantum-state mathematics measurement
edited Jun 5 at 10:47
glS
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
$endgroup$
While reading the paper "Compiling basic linear algebra subroutines for quantum computers", here, in the Appendix, the author/s have included a section on quantum inner product estimation.
Consider two vectors $x,y in mathbbC^n, x= (x_1, dots , x_n), y= (y_1, ldots, y_n)$, we want to estimate the inner product $langle x | y rangle$. Assume we are given a state $|psi rangle = frac 1 sqrt 2 big(|0 rangle |x rangle + |1 rangle |y rangle big)$, after applying a Hadamard transform to the first qubit, the result is:
$$|psi rangle = frac 1 2 big(|0 rangle (|x rangle + |y rangle) + |1 rangle(|x rangle - |y rangle) big).$$
The author then states that after measuring the first qubit in the computational basis, the probability to measure $|0 rangle$ is given by $p = frac 1 2 big(1 + mathrmRe(langle x | y rangle) big)$. I do not understand this statement. From what I understand, after applying a partial measurement to the first qubit, the probability of measuring $|0 rangle$ is given by
$frac 1 4 sqrt sum_i=0^n(overline(x_i+y_i)(x_i+y_i)) ^2$ (in other words the norm of the vector squared), so I am not sure why these formulas are equivalent, or if I am mistaken.
algorithm quantum-state mathematics measurement
algorithm quantum-state mathematics measurement
edited Jun 5 at 10:47
glS
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
edited Jun 5 at 10:47
glS
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
edited Jun 5 at 10:47
glS
5,1281944
edited Jun 5 at 10:47
glS
5,1281944
edited Jun 5 at 10:47
glS
5,1281944
5,1281944
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
asked Jun 5 at 0:30
IntegrateThisIntegrateThis
1726
1726
$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11
add a comment |
$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11
$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You just need to do a bit more algebra: Note that
$$ sum_i=0^n (overlinex_i+y_i)(x_i+y_i)=langle x+y|x+yrangle$$
and then you can distribute the right-hand side to get
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle.$$
Since $| xrangle$ and $| yrangle$ are normalized, we know that $langle x|xrangle=langle y|yrangle=1$. We also know a property of inner products:
$$langle x|yrangle=overlinexrangle$$
Further, if you add a complex number to its conjugate, you get twice its real part, so we have
$$langle x|yrangle+langle y|xrangle=2 Re(langle x|yrangle)$$
Thus, altogether, we have
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle=2+2Re(langle x|yrangle)$$
I think the probability you computed is off by a factor of 2: There is a $frac12$ as a coefficient of $|0rangle$, so since the norm is squared, this will give a factor of $frac14$. This gives the answer you are supposed to get.
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
$endgroup$
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
add a comment |
Your Answer
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1 Answer
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oldest
votes
1 Answer
1
active
oldest
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active
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votes
$begingroup$
You just need to do a bit more algebra: Note that
$$ sum_i=0^n (overlinex_i+y_i)(x_i+y_i)=langle x+y|x+yrangle$$
and then you can distribute the right-hand side to get
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle.$$
Since $| xrangle$ and $| yrangle$ are normalized, we know that $langle x|xrangle=langle y|yrangle=1$. We also know a property of inner products:
$$langle x|yrangle=overlinexrangle$$
Further, if you add a complex number to its conjugate, you get twice its real part, so we have
$$langle x|yrangle+langle y|xrangle=2 Re(langle x|yrangle)$$
Thus, altogether, we have
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle=2+2Re(langle x|yrangle)$$
I think the probability you computed is off by a factor of 2: There is a $frac12$ as a coefficient of $|0rangle$, so since the norm is squared, this will give a factor of $frac14$. This gives the answer you are supposed to get.
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
$endgroup$
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
add a comment |
$begingroup$
You just need to do a bit more algebra: Note that
$$ sum_i=0^n (overlinex_i+y_i)(x_i+y_i)=langle x+y|x+yrangle$$
and then you can distribute the right-hand side to get
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle.$$
Since $| xrangle$ and $| yrangle$ are normalized, we know that $langle x|xrangle=langle y|yrangle=1$. We also know a property of inner products:
$$langle x|yrangle=overlinexrangle$$
Further, if you add a complex number to its conjugate, you get twice its real part, so we have
$$langle x|yrangle+langle y|xrangle=2 Re(langle x|yrangle)$$
Thus, altogether, we have
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle=2+2Re(langle x|yrangle)$$
I think the probability you computed is off by a factor of 2: There is a $frac12$ as a coefficient of $|0rangle$, so since the norm is squared, this will give a factor of $frac14$. This gives the answer you are supposed to get.
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
$endgroup$
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
add a comment |
$begingroup$
You just need to do a bit more algebra: Note that
$$ sum_i=0^n (overlinex_i+y_i)(x_i+y_i)=langle x+y|x+yrangle$$
and then you can distribute the right-hand side to get
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle.$$
Since $| xrangle$ and $| yrangle$ are normalized, we know that $langle x|xrangle=langle y|yrangle=1$. We also know a property of inner products:
$$langle x|yrangle=overlinexrangle$$
Further, if you add a complex number to its conjugate, you get twice its real part, so we have
$$langle x|yrangle+langle y|xrangle=2 Re(langle x|yrangle)$$
Thus, altogether, we have
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle=2+2Re(langle x|yrangle)$$
I think the probability you computed is off by a factor of 2: There is a $frac12$ as a coefficient of $|0rangle$, so since the norm is squared, this will give a factor of $frac14$. This gives the answer you are supposed to get.
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
$endgroup$
You just need to do a bit more algebra: Note that
$$ sum_i=0^n (overlinex_i+y_i)(x_i+y_i)=langle x+y|x+yrangle$$
and then you can distribute the right-hand side to get
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle.$$
Since $| xrangle$ and $| yrangle$ are normalized, we know that $langle x|xrangle=langle y|yrangle=1$. We also know a property of inner products:
$$langle x|yrangle=overlinexrangle$$
Further, if you add a complex number to its conjugate, you get twice its real part, so we have
$$langle x|yrangle+langle y|xrangle=2 Re(langle x|yrangle)$$
Thus, altogether, we have
$$langle x|xrangle+langle x|yrangle+langle y|xrangle+langle y|yrangle=2+2Re(langle x|yrangle)$$
I think the probability you computed is off by a factor of 2: There is a $frac12$ as a coefficient of $|0rangle$, so since the norm is squared, this will give a factor of $frac14$. This gives the answer you are supposed to get.
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
answered Jun 5 at 2:31
Sam JaquesSam Jaques
4987
4987
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
add a comment |
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
$begingroup$
Ok, thanks, very helpful.
$endgroup$
– IntegrateThis
Jun 5 at 2:36
add a comment |
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$begingroup$
While reading about the complex inner product, also referred to as a "Hermitian Form" I came across the fact that $cos theta = frac Re langle x, y rangle $, so I'm sure this has something to do with it, where $theta$ is the angle between $x,y$.
$endgroup$
– IntegrateThis
Jun 5 at 2:03
$begingroup$
Well actually the complex inner product is a "type" of hermitian form.
$endgroup$
– IntegrateThis
Jun 5 at 2:11